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Gaussian Elimination & Row Echelon Form

The Organic Chemistry Tutor · YouTube · 18:40

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The explanation, unpacked.

Reviewed learning material · Video analysis · English
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This 180-second whiteboard segment introduces Gaussian elimination for a three-variable linear system. The presenter writes the system x+y−z=−2x+y-z=-2, 2x−y+z=52x-y+z=5, and -x+2y+2z=12y+2z=1, converts it into an augmented matrix, marks target entries for row echelon form, and performs the row operation R1+R3→R3R_1+R_3\to R_3. The resulting intermediate matrix is [1 1 -1 | -2; 2 -1 1 | 5; 0 3 1 | -1]. The clip ends as the next step toward eliminating the 2 in row 2 is announced but not completed. This 180-second whiteboard segment continues a Gaussian elimination example on the augmented matrix [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]]. The presenter first applies the elementary row operation -2R_1+R2R_2, computing the new second row entry by entry to obtain [[1,1,-1|-2],[0,-3,3|9],[0,3,1|-1]]. He then identifies the remaining 3 in position (3,2) as the next quantity to eliminate and applies R2+R3R_2+R_3, producing [[1,1,-1|-2],[0,-3,3|9],[0,0,4|8]]. The clip emphasizes how row replacement works, why particular multipliers or row sums are chosen, and how repeated elimination moves the matrix toward row echelon form. This 180-second whiteboard segment works one concrete linear-algebra example. Starting from the upper-triangular augmented matrix [[1, 1, -1 | -2], [0, -3, 3 | 9], [0, 0, 4 | 8]], the instructor normalizes the second and third pivots with -1/3 R2R_2 and 1/4R31/4 R_3 to obtain a unit-pivot echelon form. He then labels the columns x, y, z, translates the rows into equations, corrects two mistakes (reading the last row as x instead of z, and dropping the minus sign in the first-row constant), and finishes by back substitution to get x=1x = 1, y=−1y = -1, z=2z = 2. This 180-second whiteboard clip finishes one 3×33\times 3 linear system by back substitution, giving (x,y,z)=(1,-1,2), then uses the resulting upper-triangular matrix to explain row echelon form as having a diagonal of 1s with zeros below and nonzero entries above allowed. It starts a second system, 2x+y−z=12x+y-z=1, 3x+2y+z=103x+2y+z=10, 2x−y+2z=62x-y+2z=6, converts it to an augmented matrix, applies R3R_3←R3−R1R_3-R_1 to obtain [[2,1,-1|1],[3,2,1|10],[0,2,-3|-5]], and warns that the next elimination must preserve the new zero. The clip ends before the second example is solved. This 180-second whiteboard segment demonstrates Gaussian elimination on the augmented matrix [[2,1,-1|1],[3,2,1|10],[0,2,-3|-5]]. The presenter first explains the row replacement -3R_1+2R22R_2, copies rows 1 and 3 unchanged, and computes the new row 2 entry by entry to obtain [0,1,5,17]. The board then moves to the next pivot column, announces -2R_2+R3R_3 to eliminate the 2 in position (3,2), copies the first two rows, and computes only the first entry of the new row 3 as 0 before the clip ends. The emphasis is on systematic row operations, preserving unchanged rows, and transforming the augmented column along with the coefficient columns. This 180-second whiteboard segment finishes one Gaussian-elimination step on a 3x4 augmented matrix by applying the row operation -2R_2+R3R_3. The presenter computes the new third-row entries column by column, obtaining 0 in column 2, -13 in column 3, and -39 in the augmented column, so the matrix becomes upper triangular. The clip then explains that once these zeros are in place, the system can be solved directly without further reduction. The columns are labeled x, y, z, and the three rows are rewritten as 2x+y−z=12x+y-z=1, y+5z=17y+5z=17, and -13z=-39. Back substitution proceeds from the bottom upward: z=3z=3 is found first, with an explicit correction of a sign error in division; y=2y=2 follows from substituting z into the second equation; finally the first equation is reduced to 2x=22x=2, indicating x=1x=1, though that last simplification is not fully shown before the segment ends. This segment shows the final steps of solving a system of linear equations using back-substitution. Given the value of z, the instructor substitutes it into the second equation to find y, and then substitutes both y and z into the first equation to find x. The complete solution is presented as an ordered triplet.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Introduction to Gaussian elimination0:10Writing the three-equation system0:48Converting the system to an augmented matrix1:24Identifying row echelon form targets1:44Applying R1+R3R_1 + R_3 to row 32:48Announcing the next elimination step3:00Set up the next row operation3:18Apply -2R_1+R2R_2 entry by entry4:56Choose the next elimination target5:17Apply R2+R3R_2+R_3 and write the final matrix6:00Starting augmented matrix6:15Identify pivots to normalize6:30Apply -1/3 R2R_2 and 1/4R31/4 R_37:23Rewrite as equations in x, y, z7:45Correct variable and sign errors8:12Back substitution to x=1x = 1, y=−1y = -1, z=2z = 29:00Final answer for the first system9:16What row echelon form looks like here9:34Introducing a new 3-equation system10:27Converting to an augmented matrix10:47Applying R3−R1R_3 - R_111:37Preserving the zero while preparing the next step12:00Initial augmented matrix and elimination target12:35Applying -3R_1+2R22R_2 column by column14:05Starting the next step with -2R_2+R3R_315:00Applying -2R_2+R3R_3 to the third row15:59Upper-triangular matrix completed16:09Converting the matrix into equations with x, y, z16:46Solving the bottom equation for z17:14Back substitution for y17:38Back substitution for x18:00Solving for y and x18:11Final Solution

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The segment opens by naming the method: Gaussian elimination will be used on a system of three linear equations in three variables.

The system is written as x+y−z=−2x+y-z=-2, 2x−y+z=52x-y+z=5, and -x+2y+2z=12y+2z=1. These three equations provide the coefficients and constants that will be organized into a matrix.

To begin the matrix method, each equation is converted into one row of an augmented matrix. The coefficients of x, y, and z are placed before a vertical bar, and the right-hand constants are placed after it, giving [1 1 -1 | -2; 2 -1 1 | 5; -1 2 2 | 1].

The next goal is row echelon form. In this example, the presenter marks diagonal positions that should become 1 and marks lower-left positions that should become 0, identifying the entries that row operations will target.

The first displayed elimination step is R1+R3→R3R_1+R_3\to R_3. This means row 3 is replaced by the entrywise sum of row 1 and row 3, while rows 1 and 2 remain unchanged.

Computing the new row 3 gives 1+(−1)=01+(-1)=0, 1+2=31+2=3, -1+2=12=1, and -2+1=−11=-1. Therefore the matrix becomes [1 1 -1 | -2; 2 -1 1 | 5; 0 3 1 | -1].

At the end of the clip, the presenter identifies the next target: turning the 2 in the first column of row 2 into 0. The specific next row operation is announced only partially before the segment ends.

The board shows the augmented matrix [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]] and the next planned operation -2R_1+R2R_2. The purpose is to cancel the 2 in row 2, column 1 using the pivot 1 in row 1, column 1.

The presenter names the transformation a matrix row operation and begins computing the new row 2 column by column. For column 1, he uses -2(1)+2=02=0, so the first entry of the new row 2 is 0.

For column 2, he combines the row 1 entry 1 with the row 2 entry -1 to get -2(1)+(-1)=-3. This gives the second entry of the new row 2.

For column 3, he uses the row 1 entry -1 and the row 2 entry 1, obtaining -2(-1)+1=31=3. For column 4, he uses -2 and 5, obtaining -2(-2)+5=95=9.

He states that everything else stays the same, so rows 1 and 3 are copied unchanged. The intermediate matrix becomes [[1,1,-1|-2],[0,-3,3|9],[0,3,1|-1]].

Looking at the new matrix, the presenter points out that the entry 3 in row 3, column 2 still needs to become 0. He chooses the operation R2+R3R_2+R_3 and applies the change to row 3.

Rows 1 and 2 are kept the same. For the new row 3, he adds corresponding entries: 0+0=00+0=0 in column 1, -3+3=03=0 in column 2, 3+1=43+1=4 in column 3, and 9+(−1)=89+(-1)=8 in column 4.

The final displayed augmented matrix is [[1,1,-1|-2],[0,-3,3|9],[0,0,4|8]], which is now upper triangular and ready for the next stage of solving the system by back substitution.

The board shows a3×4a 3\times 4 augmented matrix with a vertical separator before the constants column: row 1 is 1, 1, -1 | -2; row 2 is 0, -3, 3 | 9; row 3 is 0, 0, 4 | 8. The narrator says that if he wanted, he could already convert this back to equations and use back substitution, but his immediate goal is to put it into row echelon form.

He circles the three diagonal entries 1, -3, and 4 to indicate the pivots. The first pivot is already 1, so only the second and third pivots need to be changed into 1.

A long arrow is drawn to a new matrix, with the operation labels -1/3 R2R_2 above the arrow and 1/4R31/4 R_3 below it. These are elementary row scalings applied simultaneously to different rows.

Row 1 is copied unchanged in structure. For row 2, each entry is multiplied by -1/3: 0 stays 0, -3 becomes 1, 3 becomes -1, and 9 becomes -3. For row 3, each entry is multiplied by 1/41/4: 0 stays 0, the next 0 stays 0, 4 becomes 1, and 8 becomes 2.

At this stage the displayed right-hand matrix is [[1, 1, -1 | 2], [0, 1, -1 | -3], [0, 0, 1 | 2]], but the first-row constant will later be corrected to -2 because the minus sign was omitted during transcription.

Red labels x, y, z are written above the first three columns of the transformed matrix. The narrator explains that column 1 corresponds to x, column 2 to y, and column 3 to z.

Using those column labels, the rows are translated back into equations. The intended system after correction is x+y−z=−2x + y - z = -2, y−z=−3y - z = -3, and z=2z = 2.

The narrator first misreads the last row as giving x=2x = 2, then immediately corrects himself: because the nonzero pivot is in the third column, the equation is z=2z = 2, not x=2x = 2. He also notices that the first-row constant should be -2, since he forgot to transfer the minus sign earlier.

Back substitution begins from the bottom equation. With z=2z = 2 known, substitute into y−z=−3y - z = -3 to get y−2=−3y - 2 = -3. Adding 2 to both sides yields y=−1y = -1.

Now substitute y=−1y = -1 and z=2z = 2 into the first equation x+y−z=−2x + y - z = -2. This gives x+(−1)−2=−2x + (-1) - 2 = -2, which simplifies to x−3=−2x - 3 = -2. Adding 3 to both sides gives x=1x = 1.

The final solution is boxed in sequence on the board: z=2z = 2, y=−1y = -1, and x=1x = 1. Thus the worked example ends with the ordered solution (x, y, z) = (1, -1, 2).

The board already shows the reduced equations z=2z = 2, y−z=−3y - z = -3, and x+y−z=−2x + y - z = -2. Substituting upward gives y=−1y = -1 and then x=1x = 1, so the solution is written as the ordered triple (1, -1, 2).

The instructor then points to the matrix [[1,1,-1|-2],[0,1,-1|-3],[0,0,1|2]] and describes this shape as row echelon form: there is a diagonal of 1s and zeros beneath it. He explicitly notes that the entries above the diagonal do not matter for this description.

A new example begins with the system 2x+y−z=12x + y - z = 1, 3x+2y+z=103x + 2y + z = 10, and 2x−y+2z=62x - y + 2z = 6. The speaker says the method is Gaussian elimination with back substitution, but he does not intend to reduce all the way to full row echelon form; he only needs enough zeros below the first pivot to solve the system.

The system is translated into the augmented matrix [[2,1,-1|1],[3,2,1|10],[2,-1,2|6]], with each row containing the coefficients of x, y, z followed by the constant term.

To eliminate the first entry of row 3, the operation R3R_3 ← R3−R1R_3 - R_1 is applied. Entrywise, this gives 2−2=02-2=0, 1−(−1)=21-(-1)=2, -1-2=-3, and 1−6=−51-6=-5, so the matrix becomes [[2,1,-1|1],[3,2,1|10],[0,2,-3|-5]].

The instructor then highlights the remaining 3 in row 2 column 1 and the newly created 0 in row 3 column 1, warning that the next subtraction must be chosen carefully so it does not destroy that zero. He starts talking about changing the second row, but the clip ends before the next operation is completed.

The board shows a3×4a 3\times 4 augmented matrix with a vertical bar before the last column: row 1 is 2, 1, -1 | 1; row 2 is 3, 2, 1 | 10; row 3 is 0, 2, -3 | -5. The presenter circles the entries involved in the first elimination, indicating that the goal is to turn the 3 in row 2, column 1 into 0 by combining row 1 and row 2.

Before doing arithmetic, the unchanged rows are rewritten. The speaker states that rows 1 and 3 will not change, so the new matrix will keep [2, 1, -1 | 1] on top and [0, 2, -3 | -5] on the bottom while only row 2 is recomputed.

The row operation is written explicitly as -3R_1+2R22R_2. This means the new second row is formed by taking -3 times the old first row plus 2 times the old second row, entry by entry across all four columns.

For column 1, the calculation is -3(2)+2(3)=02(3)=0. For column 2, it is -3(1)+2(2)=+12(2)=+1. For column 3, it is -3(-1)+2(1)=52(1)=5. For the augmented column, it is -3(1)+2(10)=172(10)=17. These four results give the new row 2 as [0, 1, 5 | 17].

With the intermediate matrix now displayed, the presenter shifts to the next pivot column. The entry 1 in row 2, column 2 is treated as the new pivot, and the 2 in row 3, column 2 is identified as the next quantity to eliminate.

The next operation is announced as -2R_2+R3R_3, meaning row 3 will be replaced by -2 times row 2 plus row 3. Rows 1 and 2 are copied unchanged into the new matrix, and the computation of the third row begins.

Only the first entry of the new row 3 is completed in this clip: since both relevant entries in column 1 are 0, the calculation is -2(0)+0=00=0. The remaining entries of row 3 are not reached before the segment ends.

The board starts from an intermediate augmented matrix whose third row still has a 2 under the pivot 1 in column 2. The operation written on screen is -2R_2+R3R_3, meaning row 3 is replaced by -2 times row 2 plus row 3.

For column 2, the presenter uses the entries 1 in row 2 and 2 in row 3. The arithmetic shown is -2(1)+2=02=0, so the unwanted entry below the pivot becomes zero.

For column 3, the same operation is applied to 5 and -3. The board writes -2(5)+(-3)=-13, giving the new third-row coefficient for z.

For the augmented column, the operation is applied to 17 and -5. The calculation -2(17)+(-5)=-39 is written out, completing the transformed third row.

At this point the matrix is [21−1∣1015∣1700−13∣−39]\begin{bmatrix}2&1&-1&|&1\\0&1&5&|&17\\0&0&-13&|&-39\end{bmatrix}. The lower-left zero block is visually emphasized, showing that the coefficient part is now upper triangular.

The presenter states that once these zeros are obtained, the system can be solved directly without further reducing to row echelon form. The columns are then labeled x, y, z.

Each row is translated into an equation: row 1 gives 2x+y−z=12x+y-z=1, row 2 gives y+5z=17y+5z=17, and row 3 gives -13z=-39.

The solution begins with the bottom equation because it contains only z. Dividing both sides of -13z=-39 by -13 isolates z.

The speaker briefly misspeaks by saying the quotient is negative three, then corrects the sign: since a negative divided by a negative is positive, z=3z=3. This value is boxed in blue.

Next, z=3z=3 is substituted into the second equation y+5z=17y+5z=17. This becomes y+5(3)=17y+5(3)=17, then y+15=17y+15=17, and subtracting 15 from both sides yields y=2y=2.

Finally, y=2y=2 and z=3z=3 are substituted into the first equation 2x+y−z=12x+y-z=1. The board shows 2x+2−(3)=12x+2-(3)=1, which simplifies to 2x−1=12x-1=1.

Adding 1 to both sides gives 2x=22x=2. The clip ends here, so the last simplification to x=1x=1 is implied by the visible work but not fully written before the segment stops.

We have the system of equations 2x+y−z=12x + y - z = 1 and y+5z=17y + 5z = 17, and we already know that z=3z = 3. To find y, we substitute z=3z = 3 into the second equation: y+5(3)=17y + 5(3) = 17. This simplifies to y+15=17y + 15 = 17. Subtracting 15 from both sides gives us y=2y = 2.

Next, to find x, we substitute the known values y=2y = 2 and z=3z = 3 into the first equation: 2x+2−3=12x + 2 - 3 = 1. Combining the constants on the left side yields 2x−1=12x - 1 = 1. Adding 1 to both sides results in 2x=22x = 2. Finally, dividing by 2 gives x=1x = 1.

Now we have the values for all three variables. The complete solution to the system is the ordered triplet (1, 2, 3), meaning x=1x = 1, y=2y = 2, and z=3z = 3.

Knowledge cards

01

Gaussian elimination setup for a 3-variable system

The video starts with a linear system in three unknowns. Writing the equations first makes the coefficients and constants available for matrix conversion.

{x+y−z=−22x−y+z=5−x+2y+2z=1\begin{cases} x + y - z = -2 \\ 2x - y + z = 5 \\ -x + 2y + 2z = 1 \end{cases}
02

Augmented matrix construction

An augmented matrix places the coefficients of the variables on the left of a vertical bar and the constants on the right. Each row corresponds to one equation.

[11−1−22−115−1221]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ -1 & 2 & 2 & 1 \end{array}\right]
03

Row echelon form target in this example

The presenter marks entries intended to become 1 on the diagonal and entries below the first pivot intended to become 0. The clip shows the target pattern rather than a full formal definition.

04

Elementary row operation R1+R3→R3R_1 + R_3 \to R_3

This operation replaces row 3 by the sum of row 1 and row 3. Rows 1 and 2 are copied unchanged, and the vertical bar remains fixed between coefficients and constants.

R3←R1+R3R_3 \leftarrow R_1 + R_3
05

Result after the first elimination step

Entrywise addition gives the new third row: 1+(−1)=01+(-1)=0, 1+2=31+2=3, -1+2=12=1, and -2+1=−11=-1. The transformed matrix has a zero below the first pivot position.

[11−1−22−115031−1]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ 0 & 3 & 1 & -1 \end{array}\right]
06

Next announced target

The clip ends after identifying the next goal: convert the 2 in the first column of row 2 into 0 by applying a change to row 2. The actual operation is not completed in the provided segment.

07

Elementary row replacement -2R_1+R2R_2

This card records the first demonstrated operation. The video replaces row 2 by -2 times row 1 plus row 2 so that the leading 2 in column 1 is canceled. The arithmetic is shown entrywise: -2(1)+2=02=0, -2(1)+(-1)=-3, -2(-1)+1=31=3, and -2(-2)+5=95=9. Rows 1 and 3 are left unchanged.

R2←−2R1+R2R_2 \leftarrow -2R_1+R_2
08

Intermediate matrix after the first elimination

After applying -2R_1+R2R_2 to [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]], the matrix becomes [[1,1,-1|-2],[0,-3,3|9],[0,3,1|-1]]. This step shows the standard Gaussian-elimination pattern of creating a zero below the first pivot while preserving the other rows.

[11−1−20−339031−1]\begin{bmatrix}1&1&-1&-2\\0&-3&3&9\\0&3&1&-1\end{bmatrix}
09

Choosing the next zero target

The presenter then focuses on the remaining nonzero entry below the second pivot, namely the 3 in position (3,2). He explicitly says this number needs to be zero, motivating the next row operation.

10

Second row operation R2+R3R_2+R_3

To eliminate the (3,2) entry, the video adds row 2 to row 3 and writes the result in row 3. The computations are 0+0=00+0=0, -3+3=03=0, 3+1=43+1=4, and 9+(−1)=89+(-1)=8. Rows 1 and 2 remain unchanged.

R3←R2+R3R_3 \leftarrow R_2+R_3
11

Final upper-triangular augmented matrix

The segment ends with the matrix [[1,1,-1|-2],[0,-3,3|9],[0,0,4|8]]. This is the result of two successive elimination steps and has the staircase zero pattern associated with progress toward row echelon form.

[11−1−20−3390048]\begin{bmatrix}1&1&-1&-2\\0&-3&3&9\\0&0&4&8\end{bmatrix}
12

From upper-triangular matrix to row echelon form

The example starts with an augmented matrix that is already upper triangular but whose pivots are not all 1. The instructor marks the diagonal entries 1, -3, and 4 and explains that row echelon form here means turning each pivot into 1. Since the first pivot is already 1, only rows 2 and 3 require scaling.

13

Elementary row scaling operations used

Two row operations are written on the arrow between matrices: -1/3 R2R_2 and 1/4R31/4 R_3. These scale the entire second and third rows respectively, producing the unit pivots needed for the echelon form used in back substitution.

R2←−13R2,R3←14R3R_2 \leftarrow -\frac{1}{3}R_2,\quad R_3 \leftarrow \frac{1}{4}R_3
14

Resulting echelon-form augmented matrix

After the row scalings, the displayed matrix becomes [[1, 1, -1 | -2], [0, 1, -1 | -3], [0, 0, 1 | 2]] once the omitted minus sign in the first-row constant is corrected. This is the form from which the equations are read.

[11−1∣−201−1∣−3001∣2]\begin{bmatrix}1&1&-1&|&-2\\0&1&-1&|&-3\\0&0&1&|&2\end{bmatrix}
15

Column labels determine which variable each row solves for

The red labels x, y, z above the first three columns show that column position, not row position alone, determines the variable. The last row has its pivot in the z-column, so it gives z=2z = 2. The instructor explicitly corrects an initial mistake of calling that value x.

16

Back substitution procedure

Once the system is written as x+y−z=−2x + y - z = -2, y−z=−3y - z = -3, z=2z = 2, the solution proceeds upward from the last equation. First z=2z = 2 is read directly. Then z is substituted into y−z=−3y - z = -3 to get y=−1y = -1. Finally y and z are substituted into the first equation to get x=1x = 1.

17

Common transcription pitfalls in this example

Two errors are demonstrated and corrected on screen: assigning the last-row value to the wrong variable, and dropping the minus sign when copying the first-row constant from -2 to the rewritten equation. Both corrections are necessary to reach the final solution x=1x = 1, y=−1y = -1, z=2z = 2.

18

Back substitution gives (1,-1,2)

From the reduced equations z=2z=2, y−z=−3y-z=-3, and x+y−z=−2x+y-z=-2, substitute upward: z=2z=2, then y−2=−3y-2=-3 so y=−1y=-1, then x+(−1)−2=−2x+(-1)-2=-2 so x=1x=1. The solution is the ordered triple (1,-1,2).

z=2, y−z=−3, x+y−z=−2⇒(x,y,z)=(1,−1,2)z=2,\ y-z=-3,\ x+y-z=-2 \Rightarrow (x,y,z)=(1,-1,2)
19

Row echelon form in this clip

The video characterizes row echelon form by a diagonal of 1s and zeros below that diagonal. Nonzero entries above the diagonal are allowed, as shown in the matrix with leading 1s at positions (1,1), (2,2), and (3,3).

[11−1∣−201−1∣−3001∣2]\begin{bmatrix}1&1&-1&|&-2\\0&1&-1&|&-3\\0&0&1&|&2\end{bmatrix}
20

Gaussian elimination without full reduction

For the second example, the instructor says he will not convert all the way to row echelon form. The immediate goal is only to create zeros below the first pivot so that elimination and back substitution can finish the solution.

21

System to augmented matrix

The system 2x+y−z=12x+y-z=1, 3x+2y+z=103x+2y+z=10, 2x−y+2z=62x-y+2z=6 is represented by placing the coefficients of x, y, z in three columns and the constants in the augmented column.

[21−1∣1321∣102−12∣6]\begin{bmatrix}2&1&-1&|&1\\3&2&1&|&10\\2&-1&2&|&6\end{bmatrix}
22

First elimination step R3−R1R_3 - R_1

Subtract row 1 from row 3 to zero out the first entry of row 3. The computations are 2−2=02-2=0, 1−(−1)=21-(-1)=2, -1-2=-3, and 1−6=−51-6=-5, leaving rows 1 and 2 unchanged.

R3←R3−R1: [21−1∣1321∣1002−3∣−5]R_3\leftarrow R_3-R_1:\ \begin{bmatrix}2&1&-1&|&1\\3&2&1&|&10\\0&2&-3&|&-5\end{bmatrix}
23

Do not destroy the zero you just made

After creating the 0 in row 3 column 1, the next elimination target is the 3 in row 2 column 1. The clip warns that choosing the wrong row combination could make the existing zero disappear, so the order and choice of row operations matter.

24

Augmented matrix used in the example

The clip works from a3×4a 3\times 4 augmented matrix whose first three columns are coefficients and fourth column is constants: [21−113211002−3−5]\left[\begin{array}{ccc|c}2&1&-1&1\\3&2&1&10\\0&2&-3&-5\end{array}\right]. Every row operation is applied across the whole row, including the constants column.

[21−113211002−3−5]\left[\begin{array}{ccc|c}2&1&-1&1\\3&2&1&10\\0&2&-3&-5\end{array}\right]

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 32

x

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The variable x is written in the first equation and appears as the first column of the augmented matrix.

Symbol

x

Meaning

First unknown variable in the system of three linear equations.

Domain

Real-valued unknown; no explicit domain stated in the video.

y

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The variable y is written in the equations and corresponds to the second column of the augmented matrix.

Symbol

y

Meaning

Second unknown variable in the system of three linear equations.

Domain

Real-valued unknown; no explicit domain stated in the video.

z

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The variable z is written in the equations and corresponds to the third column of the augmented matrix.

Symbol

z

Meaning

Third unknown variable in the system of three linear equations.

Domain

Real-valued unknown; no explicit domain stated in the video.

[A | b]

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    A bracketed array with a vertical bar is drawn to the right of the system.

  2. Audio
    Observation

    The narrator says to convert the system into an augmented matrix and use a vertical bar to separate the left side from the right side.

Symbol

[A | b]

Meaning

Augmented matrix formed by placing the coefficients of x, y, z on the left of a vertical bar and the constants on the right.

Domain

Matrix representation of a linear system.

R1+R3→R3R_1 + R_3 \to R_3

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The notation R1+R3R_1 + R_3 is written above the arrow leading to the next matrix.

  2. Audio
    Observation

    The narrator says he will add row one and row three together and apply that change to row three.

Symbol

R1+R3→R3R_1 + R_3 \to R_3

Meaning

Elementary row operation replacing row 3 by the sum of row 1 and row 3.

Domain

Row operation on the augmented matrix.

[[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]]

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The augmented matrix shown at the start is [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]].

Symbol

[[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]]

Meaning

Augmented coefficient matrix for a three-variable linear system before the next row reduction step.

Domain

Entries are real numbers; rows correspond to equations and columns correspond to variables plus constants.

-2R_1+R2R_2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The arrow label above the first transformation is -2R_1 + R2R_2.

  2. Audio
    Observation

    The speaker says, 'It's called a matrix row operation.'

Symbol

-2R_1+R2R_2

Meaning

Replace row 2 by the sum of -2 times row 1 and row 2.

Domain

Applies to the current augmented matrix; only row 2 changes.

R2+R3R_2+R_3

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The second arrow label is R2+R3R_2+R_3.

  2. Audio
    Observation

    The speaker says, 'So it's going to be R two plus R three.'

Symbol

R2+R3R_2+R_3

Meaning

Replace row 3 by the sum of row 2 and row 3.

Domain

Applies to the intermediate matrix after the first row operation; only row 3 changes.

R1R_1

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    R1R_1 appears in the row-operation labels and in the spoken explanation of entries.

Symbol

R1R_1

Meaning

First row of the current augmented matrix.

Domain

Used as a source row in elementary row operations.

R2R_2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    R2R_2 appears in both row-operation labels and in the spoken entry-by-entry calculations.

Symbol

R2R_2

Meaning

Second row of the current augmented matrix.

Domain

First used as the target row in -2R_1+R2R_2, then as a source row in R2+R3R_2+R_3.

R3R_3

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    R3R_3 appears in the second row-operation label and in the spoken calculation of the new third row.

Symbol

R3R_3

Meaning

Third row of the current augmented matrix.

Domain

Target row in the second elementary row operation.

[ [1, 1, -1 | -2], [0, -3, 3 | 9], [0, 0, 4 | 8] ]

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    A3×4A 3\times 4 augmented matrix is written on the left at the start and later transformed into a3×4a 3\times 4 upper-triangular augmented matrix on the right.

Symbol

[ [1, 1, -1 | -2], [0, -3, 3 | 9], [0, 0, 4 | 8] ]

Meaning

Augmented coefficient matrix for a three-variable linear system before normalization to row echelon form.

Domain

Entries are real numbers; the vertical bar separates coefficient columns from the constants column.

Knowledge points · 22

System of three linear equations in three variables

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator introduces a system of equations with three variables.

  2. Formula
    Observation

    Three equations are written: x+y−z=−2x + y - z = -2, 2x−y+z=52x - y + z = 5, and -x + 2y+2z=12y + 2z = 1.

Definition
Explanation

The video begins with a concrete linear system containing three unknowns x, y, and z. Each equation is linear, with coefficients multiplying the variables and constants on the right-hand side.

Formula
{x+y−z=−22x−y+z=5−x+2y+2z=1\begin{cases} x + y - z = -2 \\ 2x - y + z = 5 \\ -x + 2y + 2z = 1 \end{cases}
Conditions
  1. The system has three equations and three variables.

  2. The video does not state a domain for the variables.

Constructing the augmented matrix from a linear system

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator says the first thing to do is convert the system to an augmented matrix.

  2. Formula
    Observation

    The matrix entries are filled row by row from the coefficients and constants of the three equations.

Method
Explanation

Each row of the augmented matrix corresponds to one equation. The entries before the vertical bar are the coefficients of x, y, and z in order; the entry after the vertical bar is the constant term from the right-hand side of that equation.

Formula
[11−1−22−115−1221]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ -1 & 2 & 2 & 1 \end{array}\right]
Conditions
  1. The variables must be placed in a consistent order across all equations.

  2. Missing variables would require zero coefficients, but this example has all three variables present in every equation.

Prerequisites
  1. System of three linear equations in three variables

Target pattern for row echelon form in this example

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator says he wants to convert the matrix into row echelon form.

  2. Diagram
    Observation

    Red circles mark the diagonal positions intended to become 1, and blue circles mark positions below the first pivot intended to become 0.

Uncertainties
  1. The video visually indicates the target pattern but does not give a full formal definition of row echelon form.

Definition
Explanation

In the demonstrated setup, the goal is to make the leading diagonal entries equal to 1 and the entries below the first pivot equal to 0. The visual marking identifies which positions are being targeted before the row operation is performed.

Formula
Conditions
  1. This is the specific target shown for the current 3-by-3 augmented matrix.

  2. The video does not explicitly state all general row-echelon-form conditions.

Prerequisites
  1. Constructing the augmented matrix from a linear system

Elementary row operation R1+R3→R3R_1 + R_3 \to R_3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator says he will add row one and row three together and apply that change to row three.

  2. Formula
    Observation

    The operation is written as R1+R3R_1 + R_3 above the transformation arrow.

Method
Explanation

The operation replaces row 3 with the entrywise sum of row 1 and row 3. Rows 1 and 2 are copied unchanged, while each entry in row 3 is computed by adding the corresponding entries from rows 1 and 3.

Formula
R3←R1+R3R_3 \leftarrow R_1 + R_3
Conditions
  1. The operation is applied to the augmented matrix.

  2. The vertical bar is preserved, so the constant column is transformed together with the coefficient columns.

Prerequisites
  1. Constructing the augmented matrix from a linear system
  2. Target pattern for row echelon form in this example

Elementary row operation used to eliminate a leading entry

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker names the process a matrix row operation and explains that one row is multiplied by -2 and added to another row.

  2. Formula
    Observation

    The written operation is -2R_1+R2R_2.

Method
Explanation

The video demonstrates replacing row 2 with -2 times row 1 plus row 2 so that the first-column entry in row 2 becomes 0. The method is applied column by column while rows 1 and 3 remain unchanged.

Formula
R2←−2R1+R2R_2 \leftarrow -2R_1+R_2
Conditions
  1. Applied to an augmented matrix.

  2. Only the target row changes.

  3. The multiplier is chosen to cancel the pivot-column entry in the target row.

Second elimination step toward row echelon form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, 'Now, we need this number to be a zero,' referring to the (3,2) entry, then applies R2+R3R_2+R_3.

  2. Formula
    Observation

    The second operation is written as R2+R3R_2+R_3 and produces a third row [0,0,4|8].

Method
Explanation

After the first elimination, the matrix has a nonzero entry 3 below the second pivot position. The video adds row 2 to row 3 to make the (3,2) entry zero, advancing the matrix toward upper-triangular form.

Formula
R3←R2+R3R_3 \leftarrow R_2+R_3
Conditions
  1. Applied to the intermediate matrix after the first row operation.

  2. Only row 3 changes.

  3. The goal is to zero the entry below the second pivot position.

Prerequisites
  1. Elementary row operation used to eliminate a leading entry

Goal of converting to row echelon form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says he wants to convert the matrix to row echelon form and turn the leading diagonal entries into ones.

  2. Diagram
    Observation

    Circles mark the diagonal entries 1, -3, and 4 in the starting matrix.

Method
Explanation

The video treats row echelon form as the target shape obtained by making the leading nonzero entry in each pivot row equal to 1. The first pivot is already 1, so only rows 2 and 3 need scaling.

Formula
Conditions
  1. The matrix is already in upper-triangular shape with zeros below the main diagonal.

  2. The intended next step is normalization of pivots to 1.

Scaling row 2 by -1/3

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The arrow label shows -1/3 R2R_2.

  2. Audio
    Observation

    The speaker explains multiplying row two by negative one third, equivalently dividing by negative three.

  3. Diagram
    Observation

    The resulting second row is written as 0, 1, -1 | -3.

Method
Explanation

To normalize the second pivot, every entry in row 2 is multiplied by -1/3. This changes -3 to 1, 3 to -1, and 9 to -3 while leaving the leading zero unchanged.

Formula
R2←−13R2R_2 \leftarrow -\frac{1}{3}R_2
Conditions
  1. Applied to the whole second row of the augmented matrix.

  2. Used after the matrix is already upper triangular.

Prerequisites
  1. Goal of converting to row echelon form

Scaling row 3 by 1/41/4

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The arrow label shows 1/4R31/4 R_3.

  2. Audio
    Observation

    The speaker says row three is multiplied by one fourth.

  3. Diagram
    Observation

    The resulting third row is written as 0, 0, 1 | 2.

Method
Explanation

To normalize the third pivot, every entry in row 3 is multiplied by 1/41/4. This changes 4 to 1 and 8 to 2 while preserving the zeros in the first two positions.

Formula
R3←14R3R_3 \leftarrow \frac{1}{4}R_3
Conditions
  1. Applied to the whole third row of the augmented matrix.

  2. Used after the matrix is already upper triangular.

Prerequisites
  1. Goal of converting to row echelon form

Back substitution after row echelon form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says to plug z into the second equation to get y, then plug y and z into the first equation to get x.

  2. Diagram
    Observation

    Equations are rewritten sequentially from bottom to top until x=1x = 1 is boxed.

Method
Explanation

Once the augmented matrix is converted back into equations, the bottom equation gives z directly. That value is substituted upward into the second equation to solve for y, and then both y and z are substituted into the first equation to solve for x.

Formula
Conditions
  1. The system has been reduced to an upper-triangular augmented matrix with unit pivots.

  2. Variables are read from left to right as x, y, z.

Prerequisites
  1. Goal of converting to row echelon form
  2. Matrix columns correspond to variables

Matrix columns correspond to variables

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    Red x, y, z labels are placed above the first three matrix columns before the rows are rewritten as equations.

  2. Audio
    Observation

    The speaker corrects an initial x=2x = 2 reading to z=2z = 2 by matching the last row to the z column.

Method
Explanation

Each coefficient column keeps the same variable identity: the first column is x, the second is y, and the third is z. Reading a row as an equation requires this column-to-variable mapping.

Formula
Conditions
  1. The matrix columns retain the order x, y, z.

Row echelon form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Speaker says, 'this form is row echelon form where you have a diagonal of ones and zeros beneath it.'

  2. Diagram
    Observation

    The right-hand matrix shown is [[1,1,-1|-2],[0,1,-1|-3],[0,0,1|2]], with leading 1s on the diagonal and zeros below.

Definition
Explanation

In this clip, row echelon form is described as a matrix shape with a diagonal of 1s and zeros beneath that diagonal. The speaker also states that the entries above the diagonal do not matter for this description.

Formula
Conditions
  1. Applies to the coefficient part of the augmented matrix shown.

  2. The video emphasizes leading 1s on the diagonal and zeros below them.

Claims and conditions · 9

Naming of the transformation

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, 'It's called a matrix row operation.'

Proposition
Statement

The transformation being applied to the augmented matrix is called a matrix row operation.

Hypotheses
  1. The speaker is referring to the written operation -2R_1+R2R_2.

Quantifiers

No explicit quantifier beyond the demonstrated operation.

Purpose of the second row operation

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, 'Now, we need this number to be a zero. So all we got to do is add rows two and three.'

  2. Formula
    Observation

    The circled target entry is the 3 in position (3,2), and the next written operation is R2+R3R_2+R_3.

Proposition
Statement

To make the (3,2) entry zero, the video adds row 2 to row 3 and records the result in row 3.

Hypotheses
  1. The current matrix is [[1,1,-1|-2],[0,-3,3|9],[0,3,1|-1]].

Quantifiers

For the displayed matrix at this stage.

Pivots must be normalized to 1 for the displayed row echelon form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says he wants to convert these three numbers into a one and notes the first is already one.

  2. Diagram
    Observation

    The diagonal entries 1, -3, 4 are circled.

Proposition
Statement

In the worked example, the entries on the main diagonal of the upper-triangular augmented matrix are made equal to 1 by scaling the corresponding rows.

Hypotheses
  1. The matrix is already upper triangular.

  2. Each displayed pivot is nonzero.

Quantifiers

For each pivot row shown in the example.

Column order determines variable order

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says column one is the coefficients for x, this is for y, and z.

  2. Diagram
    Observation

    Red labels x, y, z are placed above the first three columns.

Proposition
Statement

In the augmented matrix, the first, second, and third coefficient columns correspond respectively to the unknowns x, y, and z.

Hypotheses
  1. The augmented matrix represents a three-variable linear system.

  2. Columns are ordered consistently from left to right.

Quantifiers

For the three coefficient columns in this example.

Informal characterization of row echelon form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    'this form is row echelon form where you have a diagonal of ones and zeros beneath it. It really doesn't matter what's here.'

  2. Diagram
    Observation

    Blue marks highlight the diagonal 1s and the zero region below them in [[1,1,-1|-2],[0,1,-1|-3],[0,0,1|2]].

Uncertainties
  1. The statement is given informally rather than as a full formal definition with all standard row-echelon conditions.

Proposition
Statement

For the displayed 3×33\times 3 coefficient block, row echelon form is characterized by having a diagonal of 1s and zeros beneath that diagonal, while the entries above the diagonal are irrelevant to this description.

Hypotheses
  1. The matrix is in the displayed upper-triangular pattern with leading 1s on the diagonal.

Quantifiers

Stated for the example matrix shown on screen.

Result of applying -3R_1+2R22R_2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board lists -3(2)+2(3)=02(3)=0, -3(1)+2(2)=+12(2)=+1, -3(-1)+2(1)=52(1)=5, and -3(1)+2(10)=172(10)=17.

  2. Diagram
    Observation

    The transformed matrix on the right has second row 0, 1, 5, 17.

Proposition
Statement

Applying -3R_1+2R22R_2 to [21−113211002−3−5]\left[\begin{array}{ccc|c}2&1&-1&1\\3&2&1&10\\0&2&-3&-5\end{array}\right] gives [21−110151702−3−5]\left[\begin{array}{ccc|c}2&1&-1&1\\0&1&5&17\\0&2&-3&-5\end{array}\right].

Hypotheses
  1. The starting matrix is the one shown on the board.

  2. The operation replaces row 2 only.

Quantifiers

For each column j=1j=1,2,3,4, the new (2,j) entry equals -3 times the old (1,j) entry plus 2 times the old (2,j) entry.

First entry after applying -2R_2+R3R_3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says for row 2 and 3 column 1 it is 0 and 0, so negative 2 times 0 plus 0 is 0.

  2. Diagram
    Observation

    The third matrix begins with first column 2, 0, 0.

Proposition
Statement

In the intermediate matrix, applying -2R_2+R3R_3 leaves the first entry of row 3 equal to 0 because -2\cdot 0+0=00+0=0.

Hypotheses
  1. The current matrix is the one after the first row operation.

  2. The operation targets row 3.

Quantifiers

Specifically for column 1 of row 3.

Choosing -2R_2+R3R_3 creates a zero in column 2 of row 3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, "So negative two plus two, that will give us the zero that we wanted."

  2. Formula
    Observation

    The computation -2(1)+2=02=0 is written on screen.

Proposition
Statement

For the displayed matrix, applying -2R_2+R3R_3 makes the (3,2) entry equal to 0 because -2(1)+2=02=0.

Hypotheses
  1. The current (2,2) entry is 1.

  2. The current (3,2) entry is 2.

  3. The operation replaces row 3 by -2 times row 2 plus row 3.

Quantifiers

For the specific matrix shown in this example.

Upper-triangular form is enough to solve by back substitution in this example

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, "Now, once you have these three zeros, you can go ahead and get all the answers without putting it in row echelon form, which is what we're going to do in this example."

Uncertainties
  1. The speaker's wording contrasts this stage with further reduction to row echelon form; the clip does not define row echelon form formally.

Proposition
Statement

After obtaining the three zeros below the pivots, the system can be solved directly by translating the triangular matrix into equations and using back substitution.

Hypotheses
  1. The augmented matrix has been reduced to upper-triangular form.

  2. The coefficient columns correspond to variables x, y, z.

Quantifiers

For the worked example shown in the clip.

Derivations and proofs · 18

Derivation of the matrix after applying R1+R3→R3R_1 + R_3 \to R_3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator computes each entry of the new third row aloud: 1 plus negative 1, 1 plus 2, negative 1 plus 2, and negative 2 plus 1.

  2. Formula
    Observation

    The resulting matrix is written with third row 0, 3, 1, -1.

Proof
Steps
  1. Expression
    [11−1−22−115−1221]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ -1 & 2 & 2 & 1 \end{array}\right]
    Explanation

    Start from the augmented matrix built directly from the system.

    Justification

    Definition of the augmented matrix from the given equations.

    Shown in the video
  2. Expression
    R3←R1+R3R_3 \leftarrow R_1 + R_3
    Explanation

    Apply the elementary row operation that replaces row 3 by the sum of row 1 and row 3.

    Justification

    Stated by the narrator and written above the transformation arrow.

    Shown in the video
  3. Expression
    1+(−1)=01 + (-1) = 0
    Explanation

    Compute the first entry of the new row 3.

    Justification

    Entrywise addition of the first entries of rows 1 and 3.

    Shown in the video
  4. Expression
    1+2=31 + 2 = 3
    Explanation

    Compute the second entry of the new row 3.

    Justification

    Entrywise addition of the second entries of rows 1 and 3.

    Shown in the video
  5. Expression
    −1+2=1-1 + 2 = 1
    Explanation

    Compute the third entry of the new row 3.

    Justification

    Entrywise addition of the third entries of rows 1 and 3.

    Shown in the video
  6. Expression
    −2+1=−1-2 + 1 = -1
    Explanation

    Compute the augmented constant entry of the new row 3.

    Justification

    Entrywise addition of the right-side entries of rows 1 and 3.

    Shown in the video
  7. Expression
    [11−1−22−115031−1]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ 0 & 3 & 1 & -1 \end{array}\right]
    Explanation

    Write the transformed augmented matrix with rows 1 and 2 unchanged and row 3 replaced by the computed sum.

    Justification

    Result of the elementary row operation.

    Shown in the video
Conclusion

After applying R1+R3→R3R_1 + R_3 \to R_3, the augmented matrix becomes [11−1−22−115031−1]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ 0 & 3 & 1 & -1 \end{array}\right].

Entry-by-entry derivation of -2R_1+R2R_2

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker computes each entry of the new row 2 aloud, column by column.

  2. Formula
    Observation

    The written calculations include -2(1)+2=02=0, -2(1)+(-1)=-3, -2(-1)+1=31=3, and -2(-2)+5=95=9.

Proof
Steps
  1. Expression
    R2←−2R1+R2R_2 \leftarrow -2R_1+R_2
    Explanation

    The operation replaces row 2 by -2 times row 1 plus row 2.

    Justification

    Stated verbally and written on screen as the matrix row operation.

    Shown in the video
  2. Expression
    −2(1)+2=0-2(1)+2=0
    Explanation

    Column 1 of the new row 2 is computed from row 1 entry 1 and row 2 entry 2.

    Justification

    Direct substitution into the written operation.

    Shown in the video
  3. Expression
    −2(1)+(−1)=−3-2(1)+(-1)=-3
    Explanation

    Column 2 of the new row 2 is computed from row 1 entry 1 and row 2 entry -1.

    Justification

    Direct substitution into the written operation.

    Shown in the video
  4. Expression
    −2(−1)+1=3-2(-1)+1=3
    Explanation

    Column 3 of the new row 2 is computed from row 1 entry -1 and row 2 entry 1.

    Justification

    Direct substitution into the written operation.

    Shown in the video
  5. Expression
    −2(−2)+5=9-2(-2)+5=9
    Explanation

    Column 4 of the new row 2 is computed from row 1 entry -2 and row 2 entry 5.

    Justification

    Direct substitution into the written operation.

    Shown in the video
  6. Expression
    [11−1−20−339031−1]\begin{bmatrix}1&1&-1&-2\\0&-3&3&9\\0&3&1&-1\end{bmatrix}
    Explanation

    Rows 1 and 3 stay the same, and the newly computed row 2 is inserted.

    Justification

    The speaker says everything else will be the same and rewrites the other numbers.

    Shown in the video
Conclusion

Applying -2R_1+R2R_2 transforms [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]] into [[1,1,-1|-2],[0,-3,3|9],[0,3,1|-1]].

Entry-by-entry derivation of R2+R3R_2+R_3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, 'So if we add R two plus R three from column one, that's zero plus zero... And for column two, R two plus R three is going to be negative three plus three, which is zero... And then for column three, it's three plus one... And then for the fourth column, nine plus negative one is eight.'

  2. Formula
    Observation

    The written operation is R2+R3R_2+R_3 and the resulting third row is [0,0,4|8].

Proof
Steps
  1. Expression
    R3←R2+R3R_3 \leftarrow R_2+R_3
    Explanation

    The operation replaces row 3 by the sum of row 2 and row 3.

    Justification

    Written on the arrow and stated verbally.

    Shown in the video
  2. Expression
    0+0=00+0=0
    Explanation

    Column 1 of the new row 3 is obtained by adding the first entries of rows 2 and 3.

    Justification

    Direct substitution from the intermediate matrix.

    Shown in the video
  3. Expression
    −3+3=0-3+3=0
    Explanation

    Column 2 of the new row 3 is obtained by adding the second entries of rows 2 and 3.

    Justification

    Direct substitution from the intermediate matrix.

    Shown in the video
  4. Expression
    3+1=43+1=4
    Explanation

    Column 3 of the new row 3 is obtained by adding the third entries of rows 2 and 3.

    Justification

    Direct substitution from the intermediate matrix.

    Shown in the video
  5. Expression
    9+(−1)=89+(-1)=8
    Explanation

    Column 4 of the new row 3 is obtained by adding the constant entries of rows 2 and 3.

    Justification

    Direct substitution from the intermediate matrix.

    Shown in the video
  6. Expression
    [11−1−20−3390048]\begin{bmatrix}1&1&-1&-2\\0&-3&3&9\\0&0&4&8\end{bmatrix}
    Explanation

    Rows 1 and 2 remain unchanged, and the computed row is written as the new row 3.

    Justification

    The speaker states that row ones and row two will stay the same, then writes the final matrix.

    Shown in the video
Conclusion

Applying R2+R3R_2+R_3 transforms [[1,1,-1|-2],[0,-3,3|9],[0,3,1|-1]] into [[1,1,-1|-2],[0,-3,3|9],[0,0,4|8]].

Derivation of the row-echelon-form matrix

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Two stacked operation labels appear between the matrices: -1/3 R2R_2 and 1/4R31/4 R_3.

  2. Diagram
    Observation

    The left matrix is [1 1 -1 | -2; 0 -3 3 | 9; 0 0 4 | 8] and the right matrix becomes [1 1 -1 | 2; 0 1 -1 | -3; 0 0 1 | 2].

  3. Audio
    Observation

    The speaker verbally computes each scaled entry.

Uncertainties
  1. The first-row constant is initially written as 2 on the right-hand matrix, but the speaker later corrects it to -2.

Proof
Steps
  1. Expression
    [11−1∣−20−33∣9004∣8]\begin{bmatrix}1&1&-1&|&-2\\0&-3&3&|&9\\0&0&4&|&8\end{bmatrix}
    Explanation

    Start from the given upper-triangular augmented matrix.

    Justification

    Observed directly on screen.

    Shown in the video
  2. Expression
    R2←−13R2R_2 \leftarrow -\frac{1}{3}R_2
    Explanation

    Scale the entire second row by -1/3 to make its pivot equal to 1.

    Justification

    Stated by the speaker and written above the arrow.

    Shown in the video
  3. Expression
    [01−1∣−3]\begin{bmatrix}0&1&-1&|&-3\end{bmatrix}
    Explanation

    The second row becomes 0, 1, -1 | -3 because -3(-1/3)=1, 3(−1/3)=−13(-1/3)=-1, and 9(−1/3)=−39(-1/3)=-3.

    Justification

    Arithmetic explicitly spoken and shown.

    Shown in the video
  4. Expression
    R3←14R3R_3 \leftarrow \frac{1}{4}R_3
    Explanation

    Scale the entire third row by 1/41/4 to make its pivot equal to 1.

    Justification

    Stated by the speaker and written below the arrow.

    Shown in the video
  5. Expression
    [001∣2]\begin{bmatrix}0&0&1&|&2\end{bmatrix}
    Explanation

    The third row becomes 0, 0, 1 | 2 because 4(1/4)=14(1/4)=1 and 8(1/4)=28(1/4)=2.

    Justification

    Arithmetic explicitly spoken and shown.

    Shown in the video
  6. Expression
    [11−1∣−201−1∣−3001∣2]\begin{bmatrix}1&1&-1&|&-2\\0&1&-1&|&-3\\0&0&1&|&2\end{bmatrix}
    Explanation

    After correcting the sign transfer in the first row, the final augmented matrix is in the displayed row echelon form.

    Justification

    Combines the visible result with the later spoken correction that the first-row constant should be -2.

    Derived from the video
Conclusion

The original augmented matrix is converted to a unit-pivot upper-triangular augmented matrix suitable for back substitution.

Solving the system by back substitution

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The equations x+y−z=−2x + y - z = -2, y−z=−3y - z = -3, z=2z = 2 are written, then solved step by step.

  2. Audio
    Observation

    The speaker narrates substituting z into the second equation and then y and z into the first.

Proof
Steps
  1. Expression
    z=2z = 2
    Explanation

    The third equation directly gives the value of z.

    Justification

    Read from the last row of the echelon-form matrix.

    Shown in the video
  2. Expression
    y−z=−3y - z = -3
    Explanation

    Use the second equation before substitution.

    Justification

    Read from the second row of the echelon-form matrix.

    Shown in the video
  3. Expression
    y−2=−3y - 2 = -3
    Explanation

    Substitute z=2z = 2 into the second equation.

    Justification

    Explicitly stated by the speaker.

    Shown in the video
  4. Expression
    y=−1y = -1
    Explanation

    Add 2 to both sides: -3 + 2=−12 = -1.

    Justification

    Algebraic simplification shown and spoken.

    Shown in the video
  5. Expression
    x+y−z=−2x + y - z = -2
    Explanation

    Use the first equation before substitution.

    Justification

    Read from the first row of the echelon-form matrix after sign correction.

    Shown in the video
  6. Expression
    x+(−1)−(2)=−2x + (-1) - (2) = -2
    Explanation

    Substitute y=−1y = -1 and z=2z = 2 into the first equation.

    Justification

    Explicitly written on screen.

    Shown in the video
  7. Expression
    x−3=−2x - 3 = -2
    Explanation

    Combine constants: -1 - 2=−32 = -3.

    Justification

    Spoken arithmetic and visible simplification.

    Shown in the video
  8. Expression
    x=1x = 1
    Explanation

    Add 3 to both sides: -2 + 3=13 = 1.

    Justification

    Final algebraic step shown and boxed.

    Shown in the video
Conclusion

The solution of the system is x=1x = 1, y=−1y = -1, z=2z = 2.

Back substitution for the first example

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Visible equations include y−z=−3y - z = -3, z=2z = 2, and x+(−1)−(2)=−2x + (-1) - (2) = -2.

  2. Diagram
    Observation

    Colored boxes show y=−1y = -1, z=2z = 2, x=1x = 1, and the final triple (1,-1,2).

  3. Audio
    Observation

    'Now we have the final answer... in the form of x, y, z, it's going to be 1, negative 1, comma 2.'

Proof
Steps
  1. Expression
    z=2z = 2
    Explanation

    The bottom equation gives z directly.

    Justification

    Read from the displayed equation z=2z = 2.

    Shown in the video
  2. Expression
    y−z=−3y - z = -3
    Explanation

    Substitute z=2z = 2 into the second equation to solve for y.

    Justification

    Displayed equation y−z=−3y - z = -3 together with z=2z = 2.

    Shown in the video
  3. Expression
    y=−1y = -1
    Explanation

    Solving y−2=−3y - 2 = -3 yields y=−1y = -1.

    Justification

    Algebraic simplification of the previous step.

    Derived from the video
  4. Expression
    x+y−z=−2x + y - z = -2
    Explanation

    Substitute y=−1y = -1 and z=2z = 2 into the first equation to solve for x.

    Justification

    Displayed equation x+y−z=−2x + y - z = -2 and previously found values.

    Shown in the video
  5. Expression
    x+(−1)−(2)=−2x + (-1) - (2) = -2
    Explanation

    The substitution is written explicitly on screen.

    Justification

    Directly visible in the worked solution.

    Shown in the video
  6. Expression
    x=1x = 1
    Explanation

    Solving x−3=−2x - 3 = -2 gives x=1x = 1.

    Justification

    Algebraic simplification shown in the board work.

    Derived from the video
  7. Expression
    (1,−1,2)(1, -1, 2)
    Explanation

    The ordered triple records the solution in the order x, y, z.

    Justification

    Speaker states the final answer in this form and the board shows the boxed values.

    Shown in the video
Conclusion

The first system has solution (x,y,z)=(1,-1,2).

First elimination step in the second example

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Initial augmented matrix is [[2,1,-1|1],[3,2,1|10],[2,-1,2|6]].

  2. Formula
    Observation

    Operation label R3−R1R_3 - R_1 is written next to the arrow.

  3. Audio
    Observation

    Speaker computes 2−2=02-2=0, 1−(−1)=21-(-1)=2, -1-2=-3, and 1−6=−51-6=-5.

  4. Diagram
    Observation

    Resulting third row is written as [0, 2, -3 | -5].

Proof
Steps
  1. Expression
    [21−1∣1321∣102−12∣6]\begin{bmatrix}2&1&-1&|&1\\3&2&1&|&10\\2&-1&2&|&6\end{bmatrix}
    Explanation

    Start from the augmented matrix of the new system.

    Justification

    Matrix is written directly from the equations.

    Shown in the video
  2. Expression
    R3←R3−R1R_3 \leftarrow R_3 - R_1
    Explanation

    Replace row 3 by row 3 minus row 1 to eliminate the first entry in row 3.

    Justification

    Explicit operation label and spoken explanation.

    Shown in the video
  3. Expression
    2−2=02-2=0
    Explanation

    First entry of the new row 3.

    Justification

    Column-wise subtraction shown and spoken.

    Shown in the video
  4. Expression
    1−(−1)=21-(-1)=2
    Explanation

    Second entry of the new row 3.

    Justification

    Column-wise subtraction shown and spoken.

    Shown in the video
  5. Expression
    −1−2=−3-1-2=-3
    Explanation

    Third entry of the new row 3.

    Justification

    Column-wise subtraction shown and spoken.

    Shown in the video
  6. Expression
    1−6=−51-6=-5
    Explanation

    Augmented constant entry of the new row 3.

    Justification

    Column-wise subtraction shown and spoken.

    Shown in the video
  7. Expression
    [21−1∣1321∣1002−3∣−5]\begin{bmatrix}2&1&-1&|&1\\3&2&1&|&10\\0&2&-3&|&-5\end{bmatrix}
    Explanation

    Rows 1 and 2 remain unchanged while row 3 becomes [0,2,-3|-5].

    Justification

    Speaker says everything else stays the same and the board shows the updated matrix.

    Shown in the video
Conclusion

After applying R3−R1R_3 - R_1, the augmented matrix becomes [[2,1,-1|1],[3,2,1|10],[0,2,-3|-5]].

Why the next elimination must use row 1, not row 3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Speaker says he needs to make the remaining lower-left entry zero and warns that subtracting the wrong pair would destroy the existing zero.

  2. Diagram
    Observation

    Circles are drawn around the 3 in row 2 column 1 and the 0 in row 3 column 1.

Uncertainties
  1. The exact next operation is not completed within this clip.

Intuitive argument
Steps
  1. Expression
    a31=0a_{31}=0
    Explanation

    The first entry of row 3 has already been made zero.

    Justification

    Result of the previous step R3−R1R_3 - R_1.

    Derived from the video
  2. Expression
    a21=3a_{21}=3
    Explanation

    The first entry of row 2 still needs to be eliminated.

    Justification

    Visible in the current matrix.

    Shown in the video
  3. Expression
    Usingrow3toeliminaterow2wouldspoila31=0Using row 3 to eliminate row 2 would spoil a_{31}=0
    Explanation

    If one subtracted multiples involving the already-zeroed row 3 in the wrong way, the zero in row 3 could be lost.

    Justification

    Spoken warning in the clip.

    Shown in the video
Conclusion

The elimination order matters: after creating a zero below a pivot, later steps should preserve that zero rather than undo it.

Column-by-column derivation of the new row 2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Four separate arithmetic lines are written for the four columns of the new row 2.

  2. Audio
    Observation

    The speaker narrates each column calculation in order.

Proof
Steps
  1. Expression
    −3(2)+2(3)=0-3(2)+2(3)=0
    Explanation

    Column 1 uses the entries 2 from row 1 and 3 from row 2.

    Justification

    Direct substitution into -3R_1+2R22R_2.

    Shown in the video
  2. Expression
    −3(1)+2(2)=+1-3(1)+2(2)=+1
    Explanation

    Column 2 uses the entries 1 from row 1 and 2 from row 2.

    Justification

    Direct substitution into -3R_1+2R22R_2.

    Shown in the video
  3. Expression
    −3(−1)+2(1)=5-3(-1)+2(1)=5
    Explanation

    Column 3 uses the entries -1 from row 1 and 1 from row 2.

    Justification

    Direct substitution into -3R_1+2R22R_2.

    Shown in the video
  4. Expression
    −3(1)+2(10)=17-3(1)+2(10)=17
    Explanation

    Column 4 uses the constants 1 from row 1 and 10 from row 2.

    Justification

    Direct substitution into -3R_1+2R22R_2.

    Shown in the video
Conclusion

The second row becomes [0, 1, 5, 17], producing the intermediate augmented matrix shown on the board.

Beginning of the derivation for -2R_2+R3R_3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker states the next goal is to make the (3,2) entry zero and chooses -2R_2+R3R_3.

  2. Formula
    Observation

    Only the first entry of the new row 3 is computed before the clip ends.

Uncertainties
  1. Columns 2, 3, and 4 of the final row 3 are not completed within this 180-second segment.

Proof
Steps
  1. Expression
    −2R2+R3-2R_2+R_3
    Explanation

    The presenter announces the next row operation after obtaining a 1 in the (2,2) position.

    Justification

    Chosen to eliminate the 2 in position (3,2).

    Shown in the video
  2. Expression
    −2(0)+0=0-2(0)+0=0
    Explanation

    For column 1, both the relevant entries in row 2 and row 3 are 0.

    Justification

    Substitution into the announced operation.

    Shown in the video
Conclusion

The first entry of the new row 3 remains 0; the remaining entries are not finished in this clip.

Derivation of the new (3,2) entry under -2R_2+R3R_3

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes -2(1)+2=02=0.

  2. Audio
    Observation

    The speaker identifies column 2 row 2 as 1 and column 2 row 3 as 2, then says the result is the desired zero.

Proof
Steps
  1. Expression
    −2(1)+2=0-2(1)+2=0
    Explanation

    Use the current row-2 and row-3 entries in column 2 to compute the replacement value for row 3.

    Justification

    Direct application of the elementary row operation R3←−2R2+R3R_3 \leftarrow -2R_2+R_3.

    Shown in the video
Conclusion

The (3,2) entry becomes 0 after the row operation.

Derivation of the new (3,3) entry under -2R_2+R3R_3

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes -2(5)+(-3)=-13.

  2. Audio
    Observation

    The speaker says, "negative two times five is negative ten plus negative three, that's going to be negative thirteen."

Proof
Steps
  1. Expression
    −2(5)+(−3)=−13-2(5)+(-3)=-13
    Explanation

    Apply the same row operation to column 3 using the row-2 entry 5 and the row-3 entry -3.

    Justification

    Elementary row operation arithmetic on corresponding entries.

    Shown in the video
Conclusion

The (3,3) entry becomes -13 after the row operation.

Worked examples · 9

Beginning a Gaussian elimination solution of a 3-variable linear system

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator frames the clip as using Gaussian elimination to solve a system of equations with three variables.

  2. Formula
    Observation

    The full worked start of the example is visible: system, augmented matrix, row operation label, and transformed matrix.

Uncertainties
  1. The clip ends before the full solution of the system is obtained.

Problem

Use Gaussian elimination and row echelon form to solve the system x+y−z=−2x + y - z = -2, 2x−y+z=52x - y + z = 5, -x + 2y+2z=12y + 2z = 1.

Given
  1. Equation 1: x+y−z=−2x + y - z = -2.

  2. Equation 2: 2x−y+z=52x - y + z = 5.

  3. Equation 3: -x + 2y+2z=12y + 2z = 1.

Goal

Convert the system to an augmented matrix and begin transforming it toward row echelon form.

Steps
  1. Expression
    {x+y−z=−22x−y+z=5−x+2y+2z=1\begin{cases} x + y - z = -2 \\ 2x - y + z = 5 \\ -x + 2y + 2z = 1 \end{cases}
    Explanation

    Write the given linear system.

    Justification

    Problem statement shown on the board.

    Shown in the video
  2. Expression
    [11−1−22−115−1221]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ -1 & 2 & 2 & 1 \end{array}\right]
    Explanation

    Convert the system into an augmented matrix by placing coefficients of x, y, z before the vertical bar and constants after it.

    Justification

    Method for forming an augmented matrix from a linear system.

    Shown in the video
  3. Expression
    R3←R1+R3R_3 \leftarrow R_1 + R_3
    Explanation

    Choose an elementary row operation to eliminate the -1 in the first column of row 3.

    Justification

    The narrator states that row 1 and row 3 will be added and the change applied to row 3.

    Shown in the video
  4. Expression
    [11−1−22−115031−1]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ 0 & 3 & 1 & -1 \end{array}\right]
    Explanation

    Compute the new row 3 entrywise while leaving rows 1 and 2 unchanged.

    Justification

    Arithmetic shown in the derivation: 1+(−1)=01+(-1)=0, 1+2=31+2=3, -1+2=12=1, -2+1=−11=-1.

    Shown in the video
Answer

The clip reaches the intermediate augmented matrix [11−1−22−115031−1]\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ 0 & 3 & 1 & -1 \end{array}\right]; the final solution is not shown within this segment.

Verification

The displayed arithmetic for the new third row matches entrywise addition of the original first and third rows.

Worked Gaussian elimination step on a 3x4 augmented matrix

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The full worked example starts from [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]] and ends at [[1,1,-1|-2],[0,-3,3|9],[0,0,4|8]].

  2. Audio
    Observation

    The speaker narrates the arithmetic for both row operations.

Problem

Reduce the augmented matrix [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]] by eliminating the entries below the pivots in columns 1 and 2.

Given
  1. Initial augmented matrix [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]].

  2. First operation -2R_1+R2R_2.

  3. Second operation R2+R3R_2+R_3.

Goal

Obtain an upper-triangular augmented matrix suitable for back substitution.

Steps
  1. Expression
    R2←−2R1+R2R_2 \leftarrow -2R_1+R_2
    Explanation

    Eliminate the 2 in position (2,1).

    Justification

    Chosen because -2 times the pivot 1 cancels the entry 2.

    Shown in the video
  2. Expression
    [11−1−20−339031−1]\begin{bmatrix}1&1&-1&-2\\0&-3&3&9\\0&3&1&-1\end{bmatrix}
    Explanation

    Write the intermediate matrix after recomputing row 2.

    Justification

    Rows 1 and 3 are unchanged; row 2 is replaced by the computed values.

    Shown in the video
  3. Expression
    R3←R2+R3R_3 \leftarrow R_2+R_3
    Explanation

    Eliminate the 3 in position (3,2).

    Justification

    Adding row 2 to row 3 makes the second-column entry zero because -3+3=03=0.

    Shown in the video
  4. Expression
    [11−1−20−3390048]\begin{bmatrix}1&1&-1&-2\\0&-3&3&9\\0&0&4&8\end{bmatrix}
    Explanation

    Write the final matrix after recomputing row 3.

    Justification

    Rows 1 and 2 are unchanged; row 3 is replaced by the computed values.

    Shown in the video
Answer

[[1,1,-1|-2],[0,-3,3|9],[0,0,4|8]]

Verification

Each displayed arithmetic line matches the corresponding entry in the final matrix.

Worked example: solve a 3-variable system from an upper-triangular augmented matrix

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The full worked example starts from an augmented matrix, converts it to row echelon form, rewrites it as equations, and solves for x, y, z.

  2. Audio
    Observation

    The speaker narrates each transformation and substitution step.

Uncertainties
  1. The initial transcription of the first-row constant on the right-hand matrix omits the minus sign until the speaker corrects it later.

Problem

Given the augmented matrix [[1, 1, -1 | -2], [0, -3, 3 | 9], [0, 0, 4 | 8]], convert it to row echelon form and solve the corresponding linear system.

Given
  1. The augmented matrix is already upper triangular.

  2. The variables correspond to columns 1, 2, 3 as x, y, z.

Goal

Find the values of x, y, and z.

Steps
  1. Expression
    R2←−13R2,R3←14R3R_2 \leftarrow -\frac{1}{3}R_2,\quad R_3 \leftarrow \frac{1}{4}R_3
    Explanation

    Normalize the second and third pivots to 1.

    Justification

    Elementary row scaling preserves the solution set and produces the displayed echelon form.

    Shown in the video
  2. Expression
    [11−1∣−201−1∣−3001∣2]\begin{bmatrix}1&1&-1&|&-2\\0&1&-1&|&-3\\0&0&1&|&2\end{bmatrix}
    Explanation

    Write the resulting augmented matrix after the two scalings and the sign correction in the first row.

    Justification

    Visible final matrix after correction.

    Derived from the video
  3. Expression
    x+y−z=−2,y−z=−3,z=2x+y-z=-2,\quad y-z=-3,\quad z=2
    Explanation

    Translate each row back into a linear equation using the column labels x, y, z.

    Justification

    Standard correspondence between augmented matrix rows and equations.

    Shown in the video
  4. Expression
    z=2z=2
    Explanation

    Solve the bottom equation immediately.

    Justification

    Direct reading from the third row.

    Shown in the video
  5. Expression
    y−2=−3⇒y=−1y-2=-3 \Rightarrow y=-1
    Explanation

    Substitute z into the second equation and solve for y.

    Justification

    Back substitution.

    Shown in the video
  6. Expression
    x+(−1)−2=−2⇒x−3=−2⇒x=1x+(-1)-2=-2 \Rightarrow x-3=-2 \Rightarrow x=1
    Explanation

    Substitute y and z into the first equation and solve for x.

    Justification

    Back substitution and elementary algebra.

    Shown in the video
Answer

x=1x = 1, y=−1y = -1, z=2z = 2

Verification

The final values are obtained by direct substitution into the three echelon-form equations shown on screen.

Completed back-substitution example

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Board shows x+y−z=−2x + y - z = -2, y−z=−3y - z = -3, z=2z = 2, and the computed values x=1x=1, y=−1y=-1, z=2z=2.

  2. Audio
    Observation

    'So this is the solution.'

Uncertainties
  1. The original full system before row reduction is not restated in this clip; only the reduced equations and final back substitution are visible.

Problem

Use the already reduced equations to find the solution triple (x,y,z).

Given
  1. z=2z = 2

  2. y−z=−3y - z = -3

  3. x+y−z=−2x + y - z = -2

Goal

Find x, y, and z and write the solution as an ordered triple.

Steps
  1. Expression
    z=2z = 2
    Explanation

    Take the value of z from the bottom equation.

    Justification

    Directly given on the board.

    Shown in the video
  2. Expression
    y−2=−3⇒y=−1y - 2 = -3 \Rightarrow y = -1
    Explanation

    Substitute z into the middle equation and solve for y.

    Justification

    Algebraic substitution.

    Derived from the video
  3. Expression
    x+(−1)−2=−2⇒x=1x + (-1) - 2 = -2 \Rightarrow x = 1
    Explanation

    Substitute y and z into the top equation and solve for x.

    Justification

    Algebraic substitution shown on the board.

    Shown in the video
  4. Expression
    (1,−1,2)(1,-1,2)
    Explanation

    Collect the results in the order x, y, z.

    Justification

    Speaker states the final answer in this form.

    Shown in the video
Answer

(1,-1,2)

Verification

The board displays the boxed intermediate values x=1x=1, y=−1y=-1, z=2z=2 and then the circled triple (1,-1,2).

New Gaussian elimination example

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Equations written are 2x+y−z=12x + y - z = 1, 3x+2y+z=103x + 2y + z = 10, 2x−y+2z=62x - y + 2z = 6.

  2. Audio
    Observation

    'use the Gaussian elimination with back substitution to solve this system of equations.'

  3. Formula
    Observation

    Augmented matrix written as [[2,1,-1|1],[3,2,1|10],[2,-1,2|6]] and then partially reduced to [[2,1,-1|1],[3,2,1|10],[0,2,-3|-5]].

Uncertainties
  1. The clip ends before the second example is fully solved.

  2. At the end, the speaker begins discussing changes to the second row but does not complete the operation on screen.

Problem

Solve the system 2x+y−z=12x + y - z = 1, 3x+2y+z=103x + 2y + z = 10, 2x−y+2z=62x - y + 2z = 6 using Gaussian elimination with back substitution.

Given
  1. 2x+y−z=12x + y - z = 1

  2. 3x+2y+z=103x + 2y + z = 10

  3. 2x−y+2z=62x - y + 2z = 6

Goal

Convert the system to an augmented matrix and begin elimination toward a triangular form suitable for back substitution.

Steps
  1. Expression
    {2x+y−z=13x+2y+z=102x−y+2z=6\begin{cases}2x+y-z=1\\3x+2y+z=10\\2x-y+2z=6\end{cases}
    Explanation

    Write the system of three linear equations.

    Justification

    Equations are spoken and written on the board.

    Shown in the video
  2. Expression
    [21−1∣1321∣102−12∣6]\begin{bmatrix}2&1&-1&|&1\\3&2&1&|&10\\2&-1&2&|&6\end{bmatrix}
    Explanation

    Convert the system into an augmented matrix by taking coefficients and constants.

    Justification

    Speaker explicitly says to convert it into an augmented matrix and writes each row.

    Shown in the video
  3. Expression
    R3←R3−R1R_3 \leftarrow R_3 - R_1
    Explanation

    Eliminate the first entry of row 3 using row 1.

    Justification

    Operation label and spoken explanation.

    Shown in the video
  4. Expression
    [21−1∣1321∣1002−3∣−5]\begin{bmatrix}2&1&-1&|&1\\3&2&1&|&10\\0&2&-3&|&-5\end{bmatrix}
    Explanation

    Update the matrix after the first elimination step.

    Justification

    Computed entrywise on the board.

    Shown in the video
  5. ExpressionNext target: eliminate the 3 in row 2 column 1 while preserving the 0 in row 3 column 1
    Explanation

    The speaker identifies the remaining lower-left entry to clear and warns against using the wrong row pairing.

    Justification

    Audio plus circles around the relevant entries.

    Shown in the video
Answer

Not completed within the clip; the matrix has been reduced to [[2,1,-1|1],[3,2,1|10],[0,2,-3|-5]] and the next elimination step is only introduced.

Verification

The visible board state confirms the first elimination step and the speaker’s stated plan for the next step.

Worked Gaussian elimination step 1

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The full worked example starts from a specific 3×43\times 4 augmented matrix and transforms it step by step.

  2. Audio
    Observation

    The speaker explains the choice of multipliers and computes each entry aloud.

Problem

Start from [21−113211002−3−5]\left[\begin{array}{ccc|c}2&1&-1&1\\3&2&1&10\\0&2&-3&-5\end{array}\right] and eliminate the (2,1) entry using a row operation involving rows 1 and 2.

Given
  1. Initial augmented matrix as shown.

  2. Target: make the entry in row 2, column 1 equal to 0.

Goal

Compute the new row 2 and write the resulting intermediate matrix.

Steps
  1. Expression
    Choose −3R1+2R2\text{Choose }-3R_1+2R_2
    Explanation

    Multiply row 1 by -3 and row 2 by 2 so the first-column contributions cancel.

    Justification

    Because -3\cdot 2+2⋅3=02+2\cdot 3=0.

    Shown in the video
  2. Expression
    Copy rows 1 and 3 unchanged\text{Copy rows 1 and 3 unchanged}
    Explanation

    Rows not being replaced are rewritten exactly as before.

    Justification

    Elementary row replacement affects only the target row.

    Shown in the video
  3. Expression
    New row 2 =[0,1,5,17]\text{New row 2 }=[0,1,5,17]
    Explanation

    Compute each column separately using the entries from rows 1 and 2.

    Justification

    Column-wise substitution into -3R_1+2R22R_2.

    Shown in the video
Answer

[21−110151702−3−5]\left[\begin{array}{ccc|c}2&1&-1&1\\0&1&5&17\\0&2&-3&-5\end{array}\right]

Verification

The displayed intermediate matrix matches the four computed column values written above the arrow.

Worked Gaussian elimination step 2 (partial)

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says the next thing needed is to make this a zero and applies changes to row 3.

  2. Formula
    Observation

    The board writes -2R_2+R3R_3 and begins constructing the third matrix.

Uncertainties
  1. The example is incomplete within the supplied segment because only the first entry of the new row 3 is computed before time runs out.

Problem

From [21−110151702−3−5]\left[\begin{array}{ccc|c}2&1&-1&1\\0&1&5&17\\0&2&-3&-5\end{array}\right], eliminate the (3,2) entry using rows 2 and 3.

Given
  1. Intermediate augmented matrix after step 1.

  2. Target: make the entry in row 3, column 2 equal to 0.

Goal

Set up and begin computing the new row 3.

Steps
  1. Expression
    Choose −2R2+R3\text{Choose }-2R_2+R_3
    Explanation

    Multiply row 2 by -2 and add row 3 so the second-column contributions cancel.

    Justification

    Because -2\cdot 1+2=01+2=0.

    Shown in the video
  2. Expression
    Copy rows 1 and 2 unchanged\text{Copy rows 1 and 2 unchanged}
    Explanation

    The presenter starts the new matrix by rewriting the first two rows.

    Justification

    Only row 3 is the target row.

    Shown in the video
  3. Expression
    −2(0)+0=0-2(0)+0=0
    Explanation

    Compute the first entry of the new row 3.

    Justification

    Substitution into -2R_2+R3R_3.

    Shown in the video
Answer

Partial result: the new row 3 begins with 0 in column 1; the rest is not shown in this clip.

Verification

The visible board content confirms only the first entry of the transformed row 3.

Worked example: finish Gaussian elimination and solve by back substitution

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The clip works from an intermediate augmented matrix to an upper-triangular matrix and then solves the resulting system.

  2. Formula
    Observation

    Visible matrices include [[2,1,-1|1],[0,1,5|17],[0,2,-3|-5]] and the final [[2,1,-1|1],[0,1,5|17],[0,0,-13|-39]].

  3. Audio
    Observation

    The speaker narrates the row operation, the conversion to equations, and the back substitution steps.

Uncertainties
  1. The original full 3x3 system before any row operations is not shown in this clip.

  2. The last simplification from 2x=22x=2 to x=1x=1 is not fully visible before the segment ends.

Problem

Given the intermediate augmented matrix after one prior row operation, apply -2R_2+R3R_3 to obtain an upper-triangular matrix and solve the corresponding linear system.

Given
  1. Intermediate augmented matrix: [21−1∣1015∣1702−3∣−5]\begin{bmatrix}2&1&-1&|&1\\0&1&5&|&17\\0&2&-3&|&-5\end{bmatrix}.

  2. Row operation to apply: -2R_2+R3R_3.

  3. Variables correspond to columns 1, 2, 3 as x, y, z.

Goal

Find the values of x, y, and z.

Steps
  1. Expression
    R3←−2R2+R3R_3 \leftarrow -2R_2+R_3
    Explanation

    Replace row 3 by -2 times row 2 plus row 3.

    Justification

    Elementary row operation used to create zeros below the pivot in column 2.

    Shown in the video
  2. Expression
    [21−1∣1015∣1700−13∣−39]\begin{bmatrix}2&1&-1&|&1\\0&1&5&|&17\\0&0&-13&|&-39\end{bmatrix}
    Explanation

    The updated matrix is upper triangular.

    Justification

    Column-by-column computation gives (3,2)=0, (3,3)=-13, and (3,4)=-39.

    Shown in the video
  3. Expression
    2x+y−z=1,  y+5z=17,  −13z=−392x+y-z=1,\; y+5z=17,\; -13z=-39
    Explanation

    Translate each row of the triangular matrix into an equation.

    Justification

    Columns 1, 2, 3 correspond to x, y, z and the last column is the constant term.

    Shown in the video
  4. Expression
    z=3z=3
    Explanation

    Solve the bottom equation first.

    Justification

    Divide -13z=-39 by -13; two negatives yield a positive quotient.

    Shown in the video
  5. Expression
    y=2y=2
    Explanation

    Substitute z=3z=3 into y+5z=17y+5z=17.

    Justification

    Back substitution from the second equation after simplifying y+15=17y+15=17.

    Shown in the video
  6. Expression
    2x=22x=2
    Explanation

    Substitute y=2y=2 and z=3z=3 into 2x+y−z=12x+y-z=1.

    Justification

    Back substitution gives 2x+2−3=12x+2-3=1, hence 2x−1=12x-1=1 and then 2x=22x=2.

    Shown in the video
Answer

z=3z=3, y=2y=2, and the visible work leads to 2x=22x=2, indicating x=1x=1.

Verification

Substituting x=1x=1, y=2y=2, z=3z=3 into the displayed equations gives 2(1)+2−3=12(1)+2-3=1, 2+5(3)=172+5(3)=17, and -13(3)=-39.

Solving a Partially Given System

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The system of equations 2x+y−z=12x + y - z = 1 and y+5z=17y + 5z = 17 is shown, along with the given z=3z = 3.

Problem

Given the system of equations 2x+y−z=12x + y - z = 1 and y+5z=17y + 5z = 17, and knowing that z=3z = 3, find the values of x and y.

Given
  1. 2x+y−z=12x + y - z = 1

  2. y+5z=17y + 5z = 17

  3. z=3z = 3

Goal

Find the values of x and y.

Steps
  1. Expression
    y+5(3)=17⇒y+15=17⇒y=2y + 5(3) = 17 \Rightarrow y + 15 = 17 \Rightarrow y = 2
    Explanation

    Substitute z=3z = 3 into the second equation and solve for y.

    Justification

    Substitution and basic algebra.

    Shown in the video
  2. Expression
    2x+2−(3)=1⇒2x−1=1⇒2x=2⇒x=12x + 2 - (3) = 1 \Rightarrow 2x - 1 = 1 \Rightarrow 2x = 2 \Rightarrow x = 1
    Explanation

    Substitute y=2y = 2 and z=3z = 3 into the first equation and solve for x.

    Justification

    Substitution and basic algebra.

    Shown in the video
Answer

x=1x = 1, y=2y = 2, z=3z = 3

Verification

The values satisfy both original equations: 2(1)+2−3=12(1) + 2 - 3 = 1 and 2+5(3)=172 + 5(3) = 17.

Visual events · 21

Writing the three-equation system

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    White handwritten equations appear sequentially on a black background.

Objects
  1. Equation 1: x+y−z=−2x + y - z = -2

  2. Equation 2: 2x−y+z=52x - y + z = 5

  3. Equation 3: -x + 2y+2z=12y + 2z = 1

Changes
  1. The first equation is written, then the second, then the third.

Invariants
  1. All three equations remain visible after being written.

Interpretation

The visual sequence establishes the linear system that will be converted into matrix form.

Building the augmented matrix

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    A large bracket and vertical bar are drawn, then matrix entries are filled row by row.

Objects
  1. Left bracket and right bracket

  2. Vertical separator bar

  3. Coefficient entries 1, 1, -1; 2, -1, 1; -1, 2, 2

  4. Constant entries -2, 5, 1

Changes
  1. The matrix frame is drawn first.

  2. Row 1 entries are filled, then row 2, then row 3.

Invariants
  1. The original system remains visible to the left of the matrix.

Interpretation

The animation shows the direct correspondence between each equation and a row of the augmented matrix.

Marking target entries for row echelon form

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    Red circles mark diagonal entries, and blue circles mark entries below the first pivot.

Uncertainties
  1. The exact color assignment of every circled entry is visually clear for the main targets, but the clip does not label the circles with text.

Objects
  1. Red circles around diagonal target positions

  2. Blue circles around lower-left target positions

Changes
  1. Target positions are highlighted before any row operation is performed.

Invariants
  1. The underlying matrix entries are not changed by the circles.

Interpretation

The colored marks distinguish entries intended to become pivots or zeros during Gaussian elimination.

Applying and displaying R1+R3→R3R_1 + R_3 \to R_3

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    R1+R3R_1 + R_3 is written above the arrow, and a new matrix is drawn below.

  2. Animation
    Observation

    The new third-row entries 0, 3, 1, -1 are written after the unchanged first two rows.

Objects
  1. Original augmented matrix

  2. Transformation arrow labeled R1+R3R_1 + R_3

  3. New augmented matrix

Changes
  1. The operation label is added.

  2. Rows 1 and 2 are copied unchanged.

  3. Row 3 is replaced by the computed sum.

Invariants
  1. The vertical bar separating coefficients from constants remains in the same position.

Interpretation

The visual transformation demonstrates one Gaussian elimination step that creates a zero below the first pivot.

Visual tracking of the first row operation

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    During the first operation, individual entries in rows 1 and 2 are circled as the speaker computes the new row 2.

  2. Audio
    Observation

    The speaker refers to 'column one', 'column two', 'column three', and 'the fourth column' while circling entries.

Objects
  1. Matrix entries in row 1

  2. Matrix entries in row 2

  3. Circles around selected entries

Changes
  1. Circles move from one column pair to the next as each new entry of row 2 is calculated.

Invariants
  1. Rows 1 and 3 are not altered during this visual sequence.

  2. The operation label -2R_1+R2R_2 remains fixed.

Interpretation

The circling shows that the new row 2 is formed entrywise from corresponding entries of rows 1 and 2.

Highlighting the next pivot-elimination target

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The entry 3 in position (3,2) is circled before the second operation is introduced.

  2. Audio
    Observation

    The speaker says, 'Now, we need this number to be a zero.'

Objects
  1. Entry 3 in row 3, column 2

  2. Arrow for the next operation

Changes
  1. Attention shifts from the completed first elimination to the remaining nonzero entry below the second pivot.

Invariants
  1. The intermediate matrix remains visible while the target is identified.

Interpretation

The circled 3 marks the quantity that the next row operation is designed to eliminate.

Construction of the final reduced matrix

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    A new matrix is written above the previous work, with rows 1 and 2 copied and row 3 filled in as [0,0,4|8].

  2. Formula
    Observation

    The final displayed matrix is [[1,1,-1|-2],[0,-3,3|9],[0,0,4|8]].

Objects
  1. Intermediate matrix

  2. Final matrix written above it

Changes
  1. Row 3 is replaced by the sum of rows 2 and 3, producing zeros below the second pivot.

Invariants
  1. Rows 1 and 2 remain the same from the intermediate matrix to the final matrix.

Interpretation

The visual progression shows the matrix becoming upper triangular, which is the structural goal of this Gaussian elimination stage.

Highlighting the pivot entries

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    White circles are drawn around the diagonal entries 1, -3, and 4 of the left matrix.

Objects
  1. Left augmented matrix

  2. Diagonal entries 1, -3, 4

  3. White circles

Changes
  1. Circles appear sequentially around the three diagonal entries.

Invariants
  1. The matrix entries themselves do not change during the circling.

Interpretation

The visual emphasis identifies which entries must be turned into 1 to reach the displayed row echelon form.

Displaying simultaneous row scalings

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    A long rightward arrow is drawn between the two matrices, with -1/3 R2R_2 above it and 1/4R31/4 R_3 below it.

Objects
  1. Original matrix

  2. Transformed matrix

  3. Rightward arrow

  4. Operation labels -1/3 R2R_2 and 1/4R31/4 R_3

Changes
  1. The arrow and operation labels are added.

  2. The right-hand matrix is written entry by entry.

Invariants
  1. Row 1 remains structurally the same except for the later sign correction in its constant term.

Interpretation

The animation presents the echelon-form conversion as two independent scaling operations applied to rows 2 and 3.

Labeling columns with variables

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    Red letters x, y, z are written above the first three columns of the transformed matrix.

Objects
  1. Transformed augmented matrix

  2. Red labels x, y, z

Changes
  1. Column labels are added above the matrix.

Invariants
  1. The numeric entries of the matrix remain unchanged while the labels are added.

Interpretation

The labels establish the mapping from matrix columns to unknowns before rewriting the rows as equations.

Marking solved variables and correcting a sign

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The speaker boxes z=2z = 2 in blue, corrects the first-row constant to -2, boxes y=−1y = -1 in red, and finally boxes x=1x = 1 in green.

Objects
  1. Equation list

  2. Blue box around z=2z = 2

  3. Red box around y=−1y = -1

  4. Green box around x=1x = 1

  5. Corrected -2 in the first equation

Changes
  1. Boxes are drawn around solved values.

  2. The first-row constant is corrected from 2 to -2.

Invariants
  1. The lower two equations remain y−z=−3y - z = -3 and z=2z = 2 once established.

Interpretation

The color-coded boxes track the back-substitution order, and the correction fixes a transcription error in the first equation.

Highlighting the row-echelon pattern

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    Blue circle/oval and diagonal stroke mark the matrix [[1,1,-1|-2],[0,1,-1|-3],[0,0,1|2]].

  2. Audio
    Observation

    Speaker explains the diagonal of ones and zeros beneath it.

Objects
  1. 3×43\times 4 augmented matrix

  2. blue circle

  3. blue diagonal line

  4. blue marks over lower-left zeros

Changes
  1. A blue oval is drawn around the whole matrix.

  2. A blue diagonal stroke is added through the leading 1s.

  3. Additional blue marks emphasize the zero region below the diagonal.

Invariants
  1. The numerical entries of the matrix do not change during this highlighting.

Interpretation

The visual annotation isolates the structural pattern the speaker calls row echelon form: leading 1s on the diagonal and zeros below.

Misconceptions · 11

Row echelon form is not just “make the diagonal 1”

Approximate timing
Supplementary explanation
Evidence
  1. Diagram
    Observation

    The video highlights diagonal entries to become 1 and some lower entries to become 0, but does not verbally define all row-echelon-form rules.

Uncertainties
  1. This is an analyst-added caution because the video itself does not explicitly state the misconception.

Misconception

A viewer might infer from the circled diagonal that row echelon form only requires 1s on the diagonal.

Clarification

The clip only shows the target pattern for this example. A full row-echelon-form definition also involves leading entries, zero rows, and the staircase placement of pivots, which are not fully stated here.

In R1+R3→R3R_1 + R_3 \to R_3, only row 3 is replaced

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator explicitly says the change is applied to row three.

  2. Formula
    Observation

    Rows 1 and 2 are copied unchanged in the new matrix.

Misconception

A viewer might think adding row 1 and row 3 changes both rows or changes row 1.

Clarification

The operation replaces row 3 by the sum R1+R3R_1 + R_3 while leaving row 1 and row 2 unchanged, as shown in the transformed matrix.

Unchanged rows during an elementary row replacement

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, 'Everything else will be the same. So let's rewrite the other numbers.'

  2. Formula
    Observation

    Rows 1 and 3 are copied unchanged into the next matrix.

Misconception

One might think a row operation changes the whole matrix at once.

Clarification

In the demonstrated replacement operation, only the target row is rewritten; the other rows are copied exactly.

Why the multiplier is chosen

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, 'By negative two, because this will become negative two. And then add that to row two.'

  2. Formula
    Observation

    The written operation is -2R_1+R2R_2, and the (2,1) entry becomes 0.

Misconception

The scalar multiplier may look arbitrary.

Clarification

Here -2 is chosen specifically so that -2 times the pivot entry 1 cancels the 2 in row 2, column 1.

Misreading the last row as solving for x

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, 'Now, it shouldn't be x. It should be z because this is the column for z.'

  2. Diagram
    Observation

    The mistaken line x=2x = 2 is replaced by z=2z = 2.

Misconception

One may look at the final nonzero entry and incorrectly assign it to x.

Clarification

The variable is determined by the column position: the third coefficient column corresponds to z, so the last row gives z=2z = 2, not x=2x = 2.

Dropping a minus sign when copying the augmented column

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, 'this is negative two. I forgot to transfer the sign here, so that should be negative two as well.'

  2. Diagram
    Observation

    The first-row constant on the right-hand matrix is corrected to -2.

Misconception

When rewriting the matrix or equations, the sign of a constant can be accidentally omitted.

Clarification

The first-row constant must remain -2; losing the minus sign changes the equation and therefore the final solution.

Thinking row echelon form requires zeros above the diagonal

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Speaker says, 'It really doesn't matter what's here,' while indicating the upper part of the matrix.

  2. Diagram
    Observation

    The displayed row-echelon example still has nonzero entries above the diagonal.

Misconception

Some learners may expect all off-diagonal entries to be zero once they see leading 1s on the diagonal.

Clarification

In this clip, row echelon form is presented as requiring a diagonal of 1s and zeros beneath it; entries above the diagonal may remain nonzero.

Believing any later row subtraction is safe after creating a zero

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Speaker warns that if he subtracts the wrong pair, 'this will no longer be zero.'

  2. Diagram
    Observation

    Circles distinguish the remaining 3 in row 2 column 1 from the newly created 0 in row 3 column 1.

Misconception

After one entry below the pivot has been zeroed, a learner might think any subsequent subtraction among nearby rows is harmless.

Clarification

The clip stresses that the next elimination must be chosen so it does not destroy the zero already produced in row 3 column 1.

Do not alter unrelated rows during a row replacement

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker explicitly says rows 1 and 3 will not change in the first operation, and later says row 1 is the same and row 2 does not change in the second operation.

  2. Diagram
    Observation

    Unchanged rows are copied verbatim before the new row is computed.

Misconception

One might think a row operation changes every row of the matrix.

Clarification

In the demonstrated elementary row replacement, only the target row is recomputed; all other rows are copied exactly from the previous matrix.

The augmented column must also be transformed

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The fourth column is included in the arithmetic: -3(1)+2(10)=172(10)=17.

  2. Audio
    Observation

    The speaker says “Now for column 4” and computes it like the others.

Misconception

One might forget to apply the row operation to the constants column.

Clarification

The clip treats column 4 exactly like the coefficient columns, showing that the augmented entry changes under the same linear combination.

Mistaking a quotient of two negatives for a negative number

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker first says, "Negative thirty nine divided by negative thirteen is negative three," then immediately corrects: "actually not negative three, I take that back. That is positive three."

Misconception

One may think (-39)/(-13) equals -3.

Clarification

Dividing two negative numbers gives a positive result, so (-39)/(-13)=3.

Concept relations · 20

System of three linear equations in three variables → Constructing the augmented matrix from a linear system

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator says the first thing to do is convert the system to an augmented matrix.

  2. Formula
    Observation

    The matrix entries correspond directly to the coefficients and constants of the written system.

Application
Explanation

The augmented-matrix method is applied to the given three-variable linear system.

Constructing the augmented matrix from a linear system → Target pattern for row echelon form in this example

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    After the matrix is complete, the narrator says he will convert this matrix into row echelon form.

  2. Diagram
    Observation

    Target entries in the matrix are circled.

Prerequisite
Explanation

The augmented matrix must be constructed before its entries can be targeted for row echelon form.

Target pattern for row echelon form in this example → Elementary row operation R1+R3→R3R_1 + R_3 \to R_3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator says to make the blue-circled numbers zero first, then performs R1+R3R_1 + R_3.

  2. Formula
    Observation

    The operation produces a 0 in the first entry of row 3.

Application
Explanation

The elementary row operation is used to achieve the zero target identified for row echelon form.

Elementary row operation R1+R3→R3R_1 + R_3 \to R_3 → Derivation of the matrix after applying R1+R3→R3R_1 + R_3 \to R_3

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Each new row-3 entry is computed aloud.

  2. Formula
    Observation

    The transformed matrix displays the computed values.

Proof dependency
Explanation

The derivation depends on the definition and application of the elementary row operation R3←R1+R3R_3 \leftarrow R_1 + R_3.

Elementary row operation used to eliminate a leading entry → Second elimination step toward row echelon form

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Two successive row replacements produce zeros below the pivots in columns 1 and 2.

  2. Audio
    Observation

    The speaker explicitly aims to make a displayed entry zero before applying the next operation.

Application
Explanation

The general elementary row-operation method is applied repeatedly to move the augmented matrix toward row echelon form.

Worked Gaussian elimination step on a 3x4 augmented matrix → Elementary row operation used to eliminate a leading entry

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The worked example contains both displayed operations and their resulting matrices.

Application
Explanation

The example concretely demonstrates the row-replacement method on a specific 3x4 augmented matrix.

Entry-by-entry derivation of R2+R3R_2+R_3 → Entry-by-entry derivation of -2R_1+R2R_2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The second operation is applied to the matrix produced by the first operation.

Proof dependency
Explanation

The derivation of the final matrix depends on the intermediate matrix obtained from the first elimination step.

Goal of converting to row echelon form → Back substitution after row echelon form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker first converts to row echelon form and then says what needs to be done is convert back into a system and use back substitution.

  2. Diagram
    Observation

    The transformed matrix is immediately rewritten as three equations and solved from bottom to top.

Application
Explanation

Normalizing the pivots to 1 produces the triangular equation form that makes back substitution straightforward.

Scaling row 2 by -1/3 → Goal of converting to row echelon form

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The two displayed operations are exactly the row scalings used to obtain the echelon-form matrix.

Proof dependency
Explanation

The row-echelon goal in this example is achieved by applying the displayed scalings to rows 2 and 3.

Matrix columns correspond to variables → Back substitution after row echelon form

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    Red x, y, z labels are placed above the columns just before the equations are written below.

Prerequisite
Explanation

Correctly matching columns to variables is required before the matrix rows can be translated into the equations used for back substitution.

Row echelon form → Back substitution for the first example

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The solved example uses equations z=2z=2, y−z=−3y-z=-3, x+y−z=−2x+y-z=-2 coming from an upper-triangular matrix.

  2. Audio
    Observation

    Speaker first gives the solution by back substitution and then names the matrix shape as row echelon form.

Application
Explanation

The row-echelon-shaped matrix makes back substitution possible because the bottom variable can be read directly and then substituted upward.

Gaussian elimination as used here → Row echelon form

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    Speaker says he will not convert all the way to row echelon form and only needs enough zeros to solve by elimination.

Contrast
Explanation

The clip contrasts doing just enough elimination for back substitution with fully producing the row-echelon pattern discussed earlier.

Find an answer · 27

How do you convert a system of three linear equations into an augmented matrix?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator explains converting the system to an augmented matrix.

  2. Formula
    Observation

    The augmented matrix is filled from the system coefficients and constants.

Knowledge points
  1. System of three linear equations in three variables
  2. Constructing the augmented matrix from a linear system

Which entries are targeted to become 1 or 0 when preparing a matrix for row echelon form in this example?

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    Red and blue circles mark target entries in the matrix.

  2. Audio
    Observation

    The narrator says he wants certain numbers to be 1 and others to be 0.

Knowledge points
  1. Target pattern for row echelon form in this example
  2. Marking target entries for row echelon form

Why does the video use R1+R3→R3R_1 + R_3 \to R_3 in Gaussian elimination?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narrator says he will add row one and row three and apply the change to row three.

  2. Formula
    Observation

    The first entry of the new row 3 becomes 0.

Knowledge points
  1. Elementary row operation R1+R3→R3R_1 + R_3 \to R_3
  2. Derivation of the matrix after applying R1+R3→R3R_1 + R_3 \to R_3

What is the augmented matrix after applying the first row operation R1+R3R_1 + R_3?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The transformed matrix is written below the original matrix.

Knowledge points
  1. Derivation of the matrix after applying R1+R3→R3R_1 + R_3 \to R_3
  2. Beginning a Gaussian elimination solution of a 3-variable linear system

Why does the video multiply row 1 by -2 before adding it to row 2?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker explains choosing -2 so the entry becomes -2 and then adds it to row 2.

Knowledge points
  1. Elementary row operation used to eliminate a leading entry
  2. Why the multiplier is chosen

How is each entry of the new second row calculated in -2R_1+R2R_2?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The four arithmetic lines for the new row 2 are written explicitly.

Knowledge points
  1. Elementary row operation used to eliminate a leading entry
  2. Entry-by-entry derivation of -2R_1+R2R_2

Why does the next step add row 2 to row 3?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says the needed number must be zero and then adds rows two and three.

Knowledge points
  1. Second elimination step toward row echelon form
  2. Purpose of the second row operation
  3. Entry-by-entry derivation of R2+R3R_2+R_3

What is the resulting augmented matrix after these two Gaussian elimination steps?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The final matrix shown is [[1,1,-1|-2],[0,-3,3|9],[0,0,4|8]].

Knowledge points
  1. Worked Gaussian elimination step on a 3x4 augmented matrix
  2. Entry-by-entry derivation of R2+R3R_2+R_3

Why does the video multiply row 2 by -1/3 and row 3 by 1/41/4?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker explains converting the diagonal entries into ones.

  2. Formula
    Observation

    The operations -1/3 R2R_2 and 1/4R31/4 R_3 are shown.

Knowledge points
  1. Goal of converting to row echelon form
  2. Scaling row 2 by -1/3
  3. Scaling row 3 by 1/41/4

Why does the last row give z=2z = 2 rather than x=2x = 2?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker corrects x=2x = 2 to z=2z = 2 based on the column for z.

Knowledge points
  1. Matrix columns correspond to variables
  2. Misreading the last row as solving for x

How is back substitution performed after converting the augmented matrix to row echelon form?

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The equations are solved from bottom to top and the results are boxed in sequence.

Knowledge points
  1. Back substitution after row echelon form
  2. Solving the system by back substitution

Where does the sign error occur and how is it corrected?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says he forgot to transfer the sign and corrects the first-row constant to -2.

Knowledge points
  1. Dropping a minus sign when copying the augmented column
  2. Derivation of the row-echelon-form matrix
Coverage and review notes

Covered · Black screen with introductory audio about using Gaussian elimination to solve a system with three variables; no mathematical object is visible yet.

Covered · The three linear equations are written and read aloud.

Covered · The narrator asks how matrices can be used and states that the first step is conversion to an augmented matrix.

Covered · The augmented matrix is drawn and filled row by row from the system.

Covered · The narrator states the row-echelon-form goal and circles target entries.

Covered · The narrator chooses to make the blue-circled entries zero and announces R1+R3R_1 + R_3 applied to row 3.

Covered · The transformed matrix is constructed entry by entry using the row operation.

Covered · The narrator begins describing the next step, saying he wants to convert the 2 into a 0 and will apply changes to row 2, but the actual next operation is not completed within the provided clip. Adjacent contiguous segment resolves this boundary.

Covered · The first elimination step -2R_1+R2R_2 is fully shown and explained entry by entry.

Covered · The second elimination step R2+R3R_2+R_3 is fully shown and the final upper-triangular augmented matrix is written.

Covered · Opening display of the original augmented matrix and statement that the goal is conversion to row echelon form.

Covered · Pivot entries are identified as the numbers to turn into 1.

Covered · The two row-scaling operations are written and the echelon-form matrix is constructed.

Covered · Columns are labeled x, y, z and the matrix is rewritten as a system of equations.

Covered · The speaker corrects the mistaken x=2x = 2 to z=2z = 2 and fixes the missing minus sign in the first-row constant.

Covered · Back substitution solves z=2z = 2, then y=−1y = -1, then x=1x = 1.

Covered · Final back substitution and solution triple for the first example.

Covered · Informal explanation and visual highlighting of row echelon form.

Covered · Brief transition with no new mathematical content beyond moving to the next example.

Covered · New system is written and the instructor explains the elimination goal without fully reaching row echelon form.

Covered · Conversion of the new system into an augmented matrix.

Covered · Application of R3−R1R_3 - R_1 and writing the updated matrix.

Covered · Warning about preserving the newly created zero while targeting the remaining lower-left entry.

Covered · The speaker begins discussing changes to the second row, but no completed new operation is shown before the clip ends.

Covered · Introduces the initial augmented matrix, circles the entries used for elimination, and explains that rows 1 and 3 will be copied unchanged while row 2 is targeted.

Covered · Writes -3R_1+2R22R_2, computes all four column entries of the new row 2, and displays the intermediate matrix [2 1 -1 | 1; 0 1 5 | 17; 0 2 -3 | -5].

Covered · Begins the second elimination step with -2R_2+R3R_3 and computes only the first entry of the new row 3; the remaining entries are not reached within the supplied 180-second segment. Adjacent contiguous segment resolves this boundary.

Covered · The clip applies -2R_2+R3R_3 column by column and fills in the new third-row entries 0, -13, and -39.

Covered · The completed upper-triangular matrix remains on screen as the setup for the next step.

Covered · The speaker explains that the triangular form is sufficient, labels the columns x, y, z, and rewrites the rows as equations.

Covered · The bottom equation is solved for z, including a spoken sign mistake and immediate correction.

Covered · The value z=3z=3 is substituted into the second equation to obtain y=2y=2.

Covered · The values y=2y=2 and z=3z=3 are substituted into the first equation; the visible work reaches 2x=22x=2 before the clip ends.

Covered · The instructor solves for x and y using back-substitution and presents the final solution.

Covered · The screen is completely black with no audio or visual content.

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  • Gaussian elimination ExplanationAt 0:10
    Why this connection?

    Reviewed current material from 10 seconds converts a three-variable system to an augmented matrix, applies elementary row replacements to create echelon form, and completes back substitution; transient board errors are explicitly corrected in the reviewed notes.