Find a Matrix Inverse Using the Classical Adjugate

1. Formula
   Observation

   The screen displays $A = \begin{bmatrix} 1 & 3 & -2 \\ 2 & 5 & -3 \\ -3 & 2 & -4 \end{bmatrix}$

2. Audio
   Observation

   The narration identifies the symbol in the matrix/inverse exercise.

A

The given3×3 real matrix, whose rows are1,3,−2;2,5,−3;−3,2,−4.

3×3 real matrix

1. Formula
   Observation

   The screen displays $\det(A)$

2. Audio
   Observation

   The narration identifies the symbol in the matrix/inverse exercise.

\det(A)

The determinant of matrix A; in this example, the calculated result is -1

Real numbers

1. Formula
   Observation

   The screen title and problem text display adj A

2. Audio
   Observation

   The narration identifies the symbol in the matrix/inverse exercise.

adj A

The classical adjoint matrix (adjugate matrix) of matrix A, obtained by transposing the cofactor matrix

3×3 real matrix

1. Formula
   Observation

   The screen problem text displays A^{-1}

2. Audio
   Observation

   The narration identifies the symbol in the matrix/inverse exercise.

A^{-1}

The inverse of A. The current0–78second analysis interval has not completed it; the later source does.

3×3 real matrix

1. Formula
   Observation

   The screen displays symbols such as |A_{11}|, |A_{21}|, |A_{31}|, |A_{12}|, |A_{22}|, |A_{32}|, |A_{13}|, |A_{23}|, |A_{33}|

2. Audio
   Observation

   The narration identifies the symbol in the matrix/inverse exercise.

1. The first extraction confused a submatrix/minor with a cofactor. Corrected from the actual screen; this is not attributed to the author.

A_{ij}

Source notation: the2×2 submatrix after deleting row i and column j. Its determinant is |A_ij|; an additional position sign gives the cofactor.

A real2×2 submatrix; i,j belong to{1,2,3}.

1. Formula
   Observation

   For example, to the right of $|A_{11}|$, the screen displays $\begin{vmatrix} 5 & -3 \\ 2 & -4 \end{vmatrix}$

2. Audio
   Observation

   The narration identifies the symbol in the matrix/inverse exercise.

A_{ij}

Deleting row i and column j from A produces a2×2 submatrix. Its determinant is the minor M_ij.

A real2×2 matrix, not a scalar. Its determinant is scalar.

1. Formula
   Observation

   The screen displays $A=\begin{bmatrix}1&3&-2\\2&5&-3\\-3&2&-4\end{bmatrix}$ and uses it repeatedly in subsequent calculations.

A

The $3\times 3$ matrix given in the problem.

Matrix elements are real numbers.

1. Formula
   Observation

   Editorial expansion from verified A; the actual final minor and determinant value support: $\det(A)=1\begin{vmatrix}5&-3\\2&-4\end{vmatrix}-3\begin{vmatrix}2&-3\\-3&-4\end{vmatrix}+(-2)\begin{vmatrix}2&5\\-3&2\end{vmatrix}=-1\neq 0$

1. Editorial mathematical correction follows the verified given matrix and actual final minor determinant: deleting the first row and second column leaves two in its top-left entry. The early expansion glyph cannot currently be reread because ordinary reacquisition was blocked; no author error is asserted.

\det(A)

The determinant of matrix $A$.

Scalar value; calculated here as $-1$.

1. Formula
   Observation

   The screen uses $|A_{11}|,|A_{21}|,|A_{31}|,\dots,|A_{33}|$ to denote multiple $2\times 2$ determinants.

1. The video does not verbally define $A_{ij}$, but from the notation it represents the submatrix obtained by deleting the $i$-th row and $j$-th column.

A_{ij}

The2×2 submatrix after deleting row i and column j from A; bars denote its determinant.

$i,j\in\{1,2,3\}$.

1. Formula
   Observation

   The screen writes $adj A=\begin{bmatrix}|A_{11}|&-|A_{21}|&|A_{31}|\\-|A_{12}|&|A_{22}|&-|A_{32}|\\|A_{13}|&-|A_{23}|&|A_{33}|\end{bmatrix}$.

2. Audio
   Observation

   The narration identifies the adjugate and inverse and explains their role.

adj A

The classical adjugate of A is the transpose of the signed cofactor matrix, not the direct transpose of unsigned minors or a conjugate transpose.

It has the same order as$A$: a$3\times 3$ matrix.

1. Formula
   Observation

   The screen writes $A^{-1}=\frac{1}{-1}\begin{bmatrix}-14&8&1\\17&-10&-1\\19&-11&-1\end{bmatrix}$.

2. Audio
   Observation

   The narration identifies the adjugate and inverse and explains their role.

A^{-1}

The inverse matrix of $A$.

Exists when $\det(A)\neq 0$.

1. Caption evidence
   Observation

   Title "5-1 Finding the Inverse Matrix Using the Classical Adjoint Matrix"

2. Audio
   Observation

   The narration proceeds through determinant checking and minor calculation.

3. Formula
   Observation

   The problem asks to "find adj A, and use adj A to find A^{-1}"

1. The later interval applies the inverse formula; the source does not prove the general theorem.

Check a nonzero determinant, compute minor determinants, apply cofactor signs and transpose, then divide by the determinant. The current0–78second interval is the first half; the later source completes the adjugate and inverse.

1. A is a square matrix

2. To use this method to find the inverse, one must first confirm $\det(A)\neq 0$

1. Formula
   Observation

   The screen lists item by item |A_{11}|, |A_{21}|, |A_{31}|, |A_{12}|, |A_{22}|, |A_{32}|, |A_{13}|, |A_{23}|, |A_{33}| and their 2×2 determinants

2. Audio
   Observation

   The narration proceeds through determinant checking and minor calculation.

1. This is an editorial clarification of actual source notation; original provider receipts are retained.

In the source, A_ij is the deleted-row/column submatrix, and |A_ij| is its determinant M_ij. The cofactor is C_ij=(-1)^(i+j)M_ij. For example M_21=−8 but C_21=8. The later adjugate construction explicitly applies signs;−8 is not the signed cofactor.

1. A is the given3×3 real square matrix.

2. Delete row i and column j; i,j belong to{1,2,3}.

1. Method for finding the inverse matrix using the classical adjoint matrix

1. Formula
   Observation

   The screen gives $adj A=\begin{bmatrix}|A_{11}|&-|A_{21}|&|A_{31}|\\-|A_{12}|&|A_{22}|&-|A_{32}|\\|A_{13}|&-|A_{23}|&|A_{33}|\end{bmatrix}$.

2. Audio
   Observation

   The narration follows the minors, adjugate construction and inverse formula.

First form C_ij=(-1)^(i+j)|A_ij|, then transpose C. The adjugate entry(i,j) is C_ji=(-1)^(i+j)|A_ji|. Transposing minors without applying signs is incorrect.

1. Applicable to $3\times 3$ matrices.

2. Requires calculating each $2\times 2$ minor determinant first.

1. Calculation of Nine Second-Order Minor Determinants

1. Formula
   Observation

   The screen writes $A^{-1}=\frac{1}{-1}\begin{bmatrix}-14&8&1\\17&-10&-1\\19&-11&-1\end{bmatrix}$.

2. Audio
   Observation

   The narration follows the minors, adjugate construction and inverse formula.

The video uses the formula $A^{-1}=\frac{1}{\det(A)}adj A$ to find the inverse matrix. Since $\det(A)=-1$ was calculated earlier, the entire adjoint matrix is multiplied by $-1$ to obtain the final inverse matrix.

1. A is a real square matrix with det(A)≠0.

2. Here det(A)=−1.

1. Construction Formula for the Classical Adjoint Matrix
2. Non-zero Determinant is the Condition for Invertibility

1. Formula
   Observation

   The screen lists item by item: $|A_{11}|=-14,\ |A_{21}|=-8,\ |A_{31}|=1,\ |A_{12}|=-17,\ |A_{22}|=-10,\ |A_{32}|=1,\ |A_{13}|=19,\ |A_{23}|=11,\ |A_{33}|=-1$.

2. Audio
   Observation

   The narration follows the minors, adjugate construction and inverse formula.

A_ij denotes the2×2 submatrix after deleting row i and column j, and M_ij=det(A_ij). The screen also uses |A_ij| for this determinant, which can be negative; these bars do not denote absolute value.

1. In this example, $A$ is a $3\times 3$ matrix.

2. Each $|A_{ij}|$ is a second-order determinant.

1. Formula
   Observation

   The screen writes $\det(A)=\cdots=-1\neq 0$.

At the beginning, the video calculates $\det(A)=-1$ and explicitly writes $\neq 0$. This step provides the prerequisite for using $A^{-1}=\frac{1}{\det(A)}adj A$ later.

1. Used to determine whether the adjoint method can be used to find the inverse matrix.

1. Formula
   Observation

   The screen gives $\det(A)=-1\neq 0$, establishing invertibility in this example.

2. Audio
   Observation

   The narration uses a nonzero determinant to establish invertibility here.

If $\det(A)\neq 0$, then the inverse matrix of A exists. In this example, since $\det(A)=-1\neq 0$, A^{-1} exists.

1. A is a square matrix

2. $\det(A)\neq 0$

Holds for the 3×3 matrix A in this example; the video states this criterion using a single example

1. Formula
   Observation

   The screen first writes $\det(A)=-1\neq 0$, and later writes $A^{-1}=\frac{1}{-1}adj A$.

2. Audio
   Observation

   The narration applies the adjugate inverse formula after checking a nonzero determinant.

1. The video does not state a general theorem separately, but only applies this logic in this specific example.

According to the calculation shown in the video, since $\det(A)=-1\neq 0$, one can use $A^{-1}=\frac{1}{\det(A)}adj A$ to find $A^{-1}$.

1. $A$ is the given $3\times 3$ matrix.

2. It has been calculated that $\det(A)=-1$.

For the specific matrix $A$ in this example.

1. Formula
   Observation

   Editorial expansion from verified A; the actual final minor and determinant value support: $\det(A)=1\begin{vmatrix}5&-3\\2&-4\end{vmatrix}-3\begin{vmatrix}2&-3\\-3&-4\end{vmatrix}+(-2)\begin{vmatrix}2&5\\-3&2\end{vmatrix}=-1\neq 0$

2. Audio
   Observation

   The narration follows the displayed determinant and minor calculations.

1. The intermediate arithmetic for the three 2×2 sub-determinants is not expanded item by item on the screen, only the final combined result -1 is given

2. Editorial mathematical correction follows the verified given matrix and actual final minor determinant: deleting the first row and second column leaves two in its top-left entry. The early expansion glyph cannot currently be reread because ordinary reacquisition was blocked; no author error is asserted.

1. Expression

   $$\det(A)=1\begin{vmatrix}5&-3\\2&-4\end{vmatrix}-3\begin{vmatrix}2&-3\\-3&-4\end{vmatrix}+(-2)\begin{vmatrix}2&5\\-3&2\end{vmatrix}$$
   Explanation

   Expand along the first row1,3,−2, deleting that row and the corresponding column for each minor.

   Justification

   Use the alternating signs of row expansion. Regional row/column terminology is not an author error.

   Supplementary explanation
2. Expression

   $$=-1$$
   Explanation

   Calculate the three 2×2 determinants and combine them to get $\det(A)=-1$.

   Justification

   The screen and narration give the final determinant value−1.

   Shown in the video
3. Expression

   $$-1\neq 0$$
   Explanation

   From $\det(A)=-1$, it is known that the determinant is not equal to zero.

   Justification

   $\det(A)=-1\neq 0$, which the narration uses to conclude invertibility here.

   Shown in the video

$\det(A)=-1\neq 0$, therefore the inverse matrix of A exists in this example.

1. Formula
   Observation

   The screen displays $|A_{11}|=\begin{vmatrix}5&-3\\2&-4\end{vmatrix}=-14, |A_{21}|=\begin{vmatrix}3&-2\\2&-4\end{vmatrix}=-8, |A_{31}|=\begin{vmatrix}3&-2\\5&-3\end{vmatrix}=1, |A_{12}|=\begin{vmatrix}2&-3\\-3&-4\end{vmatrix}=-17$

2. Audio
   Observation

   The narration follows the displayed determinant and minor calculations.

1. The0–78second analysis interval does not narrate every entry; the later source completes them.

1. Expression

   $$M_{11}=\begin{vmatrix}5&-3\\2&-4\end{vmatrix}=-14$$
   Explanation

   Delete row1 and column1, then evaluate the resulting determinant.

   Justification

   The actual board gives a minor determinant. A separate position sign is needed for its cofactor.

   Supplementary explanation
2. Expression

   $$M_{21}=\begin{vmatrix}3&-2\\2&-4\end{vmatrix}=-8$$
   Explanation

   Delete row2 and column1, then evaluate the resulting determinant.

   Justification

   The actual board gives a minor determinant. A separate position sign is needed for its cofactor.

   Supplementary explanation
3. Expression

   $$M_{31}=\begin{vmatrix}3&-2\\5&-3\end{vmatrix}=1$$
   Explanation

   Delete row3 and column1, then evaluate the resulting determinant.

   Justification

   The actual board gives a minor determinant. A separate position sign is needed for its cofactor.

   Supplementary explanation
4. Expression

   $$M_{12}=\begin{vmatrix}2&-3\\-3&-4\end{vmatrix}=-17$$
   Explanation

   Delete row1 and column2, then evaluate the resulting determinant.

   Justification

   The actual board gives a minor determinant. A separate position sign is needed for its cofactor.

   Supplementary explanation

$M_{11}=-14,\quad M_{21}=-8,\quad M_{31}=1,\quad M_{12}=-17$. These are minor determinants. Signs and transposition are still required; the later interval completes the remaining entries and final answer.

1. Formula
   Observation

   The screen fully displays the entire process from $\det(A)$, each $|A_{ij}|$, $adj A$, to $A^{-1}$.

2. Audio
   Observation

   The instructor explains each step in order and substitutes the numerical values into the adjoint and inverse matrix formulas.

1. Editorial mathematical correction follows the verified given matrix and actual final minor determinant: deleting the first row and second column leaves two in its top-left entry. The early expansion glyph cannot currently be reread because ordinary reacquisition was blocked; no author error is asserted.

1. Expression

   $$\det(A)=1\begin{vmatrix}5&-3\\2&-4\end{vmatrix}-3\begin{vmatrix}2&-3\\-3&-4\end{vmatrix}+(-2)\begin{vmatrix}2&5\\-3&2\end{vmatrix}=-1\neq 0$$
   Explanation

   First, expand along the first row to calculate the determinant of $A$, obtaining $-1$, and confirm it is non-zero.

   Justification

   Editorial expansion independently corrected from the verified given matrix and actual final minor and determinant value; this does not claim the early printed glyph was reread.

   Supplementary explanation
2. Expression

   $$|A_{11}|=-14,\ |A_{21}|=-8,\ |A_{31}|=1,\ |A_{12}|=-17,\ |A_{22}|=-10,\ |A_{32}|=1,\ |A_{13}|=19,\ |A_{23}|=11,\ |A_{33}|=-1$$
   Explanation

   Calculate the nine second-order minor determinants by deleting the corresponding rows and columns one by one.

   Justification

   The screen lists these second-order determinants and their values item by item.

   Shown in the video
3. Expression

   $$adj A=\begin{bmatrix}|A_{11}|&-|A_{21}|&|A_{31}|\\-|A_{12}|&|A_{22}|&-|A_{32}|\\|A_{13}|&-|A_{23}|&|A_{33}|\end{bmatrix}$$
   Explanation

   Arrange the calculated minor determinants according to the adjoint matrix formula, noting the alternating signs and transposed positions.

   Justification

   The definition formula for the adjoint matrix given on the screen.

   Shown in the video
4. Expression

   $$adj A=\begin{bmatrix}-14&8&1\\17&-10&-1\\19&-11&-1\end{bmatrix}$$
   Explanation

   Substitute the numerical values of each $|A_{ij}|$ into the adjoint matrix formula to get the specific matrix.

   Justification

   The screen directly displays this result after substitution.

   Shown in the video
5. Expression

   $$A^{-1}=\frac{1}{\det(A)}adj A=\frac{1}{-1}\begin{bmatrix}-14&8&1\\17&-10&-1\\19&-11&-1\end{bmatrix}$$
   Explanation

   Use the inverse matrix formula, substituting $\det(A)=-1$.

   Justification

   The screen writes $A^{-1}=\frac{1}{-1}adj A$.

   Shown in the video
6. Expression

   $$A^{-1}=\begin{bmatrix}14&-8&-1\\-17&10&1\\-19&11&1\end{bmatrix}$$
   Explanation

   Multiply each element of the adjoint matrix by $-1$ to obtain the final inverse matrix.

   Justification

   The screen finally displays this result.

   Shown in the video

In this example, $A^{-1}=\begin{bmatrix}14&-8&-1\\-17&10&1\\-19&11&1\end{bmatrix}$.

1. Formula
   Observation

   The screen problem box writes "Consider matrix A = ... , find adj A, and use adj A to find A^{-1}"

2. Audio
   Observation

   The narration states the exercise goal and works through it.

1. The current0–78second interval is the first half; the later source gives the complete answer.

2. Editorial mathematical correction follows the verified given matrix and actual final minor determinant: deleting the first row and second column leaves two in its top-left entry. The early expansion glyph cannot currently be reread because ordinary reacquisition was blocked; no author error is asserted.

Consider matrix $A=\begin{bmatrix}1&3&-2\\2&5&-3\\-3&2&-4\end{bmatrix}$, find adj A, and use adj A to find A^{-1}.

1. $A=\begin{bmatrix}1&3&-2\\2&5&-3\\-3&2&-4\end{bmatrix}$

2. The goal is to first find the classical adjoint matrix adj A, then find the inverse matrix A^{-1}

Calculate adj A, and accordingly find A^{-1}.

1. Expression

   $$\det(A)=1\begin{vmatrix}5&-3\\2&-4\end{vmatrix}-3\begin{vmatrix}2&-3\\-3&-4\end{vmatrix}+(-2)\begin{vmatrix}2&5\\-3&2\end{vmatrix}=-1\neq 0$$
   Explanation

   Expand along the first row to establish invertibility.

   Justification

   Editorial expansion independently corrected from the verified given matrix and actual final minor and determinant value; this does not claim the early printed glyph was reread.

   Supplementary explanation
2. Expression

   $$M_{11}=-14,\quad M_{21}=-8,\quad M_{31}=1,\quad M_{12}=-17$$
   Explanation

   Evaluate the first four narrated minor determinants; their cofactors require position signs.

   Justification

   Source-verified determinants after deleting the corresponding row and column.

   Supplementary explanation

$\det(A)=-1\ne0,\quad M_{11}=-14,\ M_{21}=-8,\ M_{31}=1,\ M_{12}=-17$. Only the current0–78second interval has not reached the final answer; the later source completes the adjugate and inverse.

Check invertibility and the minor determinants. These values still require position signs to form cofactors. The final answer is independently verified by multiplication in both orders.

1. Formula
   Observation

   The displayed example heading asks for A^{-1} using adj A, followed by the complete calculation.

2. Audio
   Observation

   The instructor explains step by step how to construct the adjoint matrix from minor determinants and then find the inverse matrix.

Consider the matrix $A=\begin{bmatrix}1&3&-2\\2&5&-3\\-3&2&-4\end{bmatrix}$. Find $adj A$, and use $adj A$ to find $A^{-1}$.

1. $A=\begin{bmatrix}1&3&-2\\2&5&-3\\-3&2&-4\end{bmatrix}$.

2. The goal is to first find $adj A$, then find $A^{-1}$.

Find $adj A$ and $A^{-1}$.

1. Expression

   $$\det(A)=-1\neq 0$$
   Explanation

   First calculate the determinant to confirm invertibility.

   Justification

   The first-row expansion formula given on the screen.

   Shown in the video
2. Expression

   $$|A_{11}|=-14,\ |A_{21}|=-8,\ |A_{31}|=1,\ |A_{12}|=-17,\ |A_{22}|=-10,\ |A_{32}|=1,\ |A_{13}|=19,\ |A_{23}|=11,\ |A_{33}|=-1$$
   Explanation

   Calculate all second-order minor determinants.

   Justification

   Listed item by item on the screen.

   Shown in the video
3. Expression

   $$adj A=\begin{bmatrix}-14&8&1\\17&-10&-1\\19&-11&-1\end{bmatrix}$$
   Explanation

   Substitute into the adjoint matrix formula and organize the signs.

   Justification

   The adjoint matrix result given on the screen.

   Shown in the video
4. Expression

   $$A^{-1}=\frac{1}{-1}\begin{bmatrix}-14&8&1\\17&-10&-1\\19&-11&-1\end{bmatrix}$$
   Explanation

   Apply $A^{-1}=\frac{1}{\det(A)}adj A$.

   Justification

   The screen directly writes this expression.

   Shown in the video
5. Expression

   $$A^{-1}=\begin{bmatrix}14&-8&-1\\-17&10&1\\-19&11&1\end{bmatrix}$$
   Explanation

   Multiply the entire matrix by $-1$ to get the final answer.

   Justification

   The result displayed at the end of the screen.

   Shown in the video

$adj A=\begin{bmatrix}-14&8&1\\17&-10&-1\\19&-11&-1\end{bmatrix}$, $A^{-1}=\begin{bmatrix}14&-8&-1\\-17&10&1\\-19&11&1\end{bmatrix}$.

The video does not perform back-substitution verification; the result is derived solely through formula application.

1. Caption evidence
   Observation

   Black background with white text displays "5-1 Finding the Inverse Matrix Using the Classical Adjoint Matrix" and "Recorded by Lin Bing-sen"

1. Title text

2. Recorder's name

1. Switch from black title page to whiteboard teaching screen

1. No mathematical operation content

This segment is only course title and recorder information, containing no mathematical derivation.

1. Diagram
   Observation

   Top of the screen shows the example and matrix A, middle section is the det(A) expansion, bottom section contains multiple sets of |A_{ij}| 2×2 determinants and results

2. Animation
   Observation

   Cursor sequentially points to elements of A, the det(A) expansion, and minor determinants like A_{11},A_{21},A_{31},A_{12}

1. The current0–78second interval does not complete the adjugate; the later source does.

1. Matrix A

2. Expansion of det(A)

3. Nine minor determinants and values

4. Pointer

1. Cursor first points to matrix A in the problem

2. Then moves to the det(A) expansion

3. The pointer moves through the minor determinants |A_11|, |A_21|, |A_31| and |A_12|.

1. Elements of matrix A remain unchanged throughout

2. Written formulas and values remain visible on the screen

The layout checks the determinant, then calculates minors. Later work applies cofactor signs and transposes.

1. Animation
   Observation

   The mouse cursor points sequentially to each second-order determinant, the adjoint matrix formula, and the final inverse matrix result.

2. Audio
   Observation

   The instructor reads out the corresponding values and steps while pointing.

1. Matrix $A$

2. Nine second-order minor determinants

3. Adjoint matrix formula

4. Inverse matrix result

1. First focuses on $\det(A)$ and each $|A_{ij}|$.

2. Then moves to the sign arrangement of $adj A$.

3. Finally moves to the numerical result of $A^{-1}$.

1. The page remains the same handwritten/blackboard-style example page throughout.

2. Matrix $A$ itself does not change.

The visual movement order corresponds to the calculation order: first find minors, then assemble the adjoint matrix, and finally divide by the determinant to get the inverse matrix.

1. Formula
   Observation

   The actual screen gives |A_21|=−8. The later adjugate uses−|A_21|=8.

2. Audio
   Observation

   The narration evaluates determinants of the submatrices. Their distinction from signed minor determinants is clarified editorially.

1. This is an analyst's supplementary reminder, not an error point explicitly mentioned in the video

Taking |A_ij| directly as a cofactor misses the position sign(-1)^(i+j).

Here A_ij is a submatrix, and M_ij=det(A_ij)=|A_ij| is a minor determinant. Its cofactor is C_ij=(-1)^(i+j)M_ij. Thus M_21=−8 but C_21=8. The later board applies the minus sign; this is not an author arithmetic error.

1. Formula
   Observation

   In the screen's $adj A$, $|A_{21}|,|A_{12}|,|A_{32}|,|A_{23}|$ are placed in positions with negative signs, and the overall arrangement is transposed.

A common mistake is to identify $adj A$ at position $(i,j)$ with $(-1)^{i+j}|A_{ij}|$ without transposing.

The adjugate is the transpose of the signed cofactor matrix C. Entry(i,j) is C_ji=(-1)^(i+j)|A_ji|, requiring both signs and transposition.

1. Formula
   Observation

   The screen gives $\det(A)=-1\neq 0$, establishing invertibility in this example.

2. Audio
   Observation

   The narration checks a nonzero determinant before continuing the inverse calculation.

The method for finding the inverse matrix first relies on the existence criterion that the determinant is non-zero, before proceeding to calculate the classical adjoint matrix.

1. Formula
   Observation

   The problem requires finding adj A first, and the screen subsequently calculates A_{ij} item by item

2. Audio
   Observation

   The narration checks a nonzero determinant before continuing the inverse calculation.

1. An unfinished current analysis interval is not missing whole-video coverage.

Compute minors M_ij, form signed cofactors C_ij, then transpose to obtain the adjugate. The current0–78second interval is only the first half; the later source completes the work.

1. Formula
   Observation

   First lists each $|A_{ij}|$, then substitutes them into the matrix expression for $adj A$.

One must first calculate each second-order minor determinant to construct the classical adjoint matrix.

1. Formula
   Observation

   The screen first gives $adj A$, then writes $A^{-1}=\frac{1}{-1}adj A$.

2. Audio
   Observation

   The narration constructs the adjugate before applying the inverse formula.

The classical adjoint matrix is the core component in the formula for finding the inverse matrix.

1. Formula
   Observation

   The screen first writes $\det(A)=-1\neq 0$, and later uses $\frac{1}{-1}$ as the coefficient in the inverse matrix formula.

A non-zero determinant is the prerequisite for using $A^{-1}=\frac{1}{\det(A)}adj A$, and this determinant value enters directly into the final formula.

1. Audio
   Observation

   The narration describes the adjugate inverse method and minor calculations.

2. Formula
   Observation

   The problem asks to "find adj A, and use adj A to find A^{-1}"

1. Method for finding the inverse matrix using the classical adjoint matrix
2. Non-zero determinant implies existence of inverse matrix

1. Formula
   Observation

   The screen gives $\det(A)=-1\neq 0$, establishing invertibility in this example.

2. Audio
   Observation

   The narration describes the adjugate inverse method and minor calculations.

1. Non-zero determinant implies existence of inverse matrix
2. Method for finding the inverse matrix using the classical adjoint matrix

1. Formula
   Observation

   The screen calculates item by item using |A_{ij}| paired with 2×2 determinants

2. Audio
   Observation

   The narration describes the adjugate inverse method and minor calculations.

1. Minor determinants and signed cofactors
2. A_{ij}

1. Formula
   Observation

   The screen displays $\det(A)=...=-1\neq 0$

2. Audio
   Observation

   The narration describes the adjugate inverse method and minor calculations.

1. Expand along the first row
2. Example: Finding A^{-1} using adj A

1. Formula
   Observation

   The screen displays |A_{11}|=-14, |A_{21}|=-8, |A_{31}|=1, |A_{12}|=-17

2. Audio
   Observation

   The narration describes the adjugate inverse method and minor calculations.

1. The first interval has not finished narrating M_12; its value is verified from the actual screen and the later source continues.

1. Minor determinants in the first interval
2. Example: Finding A^{-1} using adj A

1. Formula
   Observation

   The screen gives the specific arrangement formula for $adj A$.

1. Calculation of Nine Second-Order Minor Determinants
2. Construction Formula for the Classical Adjoint Matrix

1. Formula
   Observation

   The screen writes $A^{-1}=\frac{1}{-1}adj A$ and gives the final matrix.

1. Construction Formula for the Classical Adjoint Matrix
2. Finding the Inverse Matrix Using the Classical Adjoint
3. Non-zero Determinant is the Condition for Invertibility

1. Formula
   Observation

   The screen explicitly writes $\det(A)=-1\neq 0$.

1. Non-zero Determinant is the Condition for Invertibility
2. Finding the Inverse Matrix Using the Classical Adjoint