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Each column of $a 2\times 2$ matrix can be read as the coordinates of a vector in the 2D plane. For the matrix $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$, the first column $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ represents the vector from the origin to the point (3,1), and the second column $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ represents the vector from the origin to the point (1,2).

Conditions: The matrix is $2\times 2$.; Working in standard Cartesian coordinates.

Each column of $a 2\times 2$ matrix can be read as the coordinates of a vector in the 2D plane. For the matrix $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$, the first column $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ represents the vector from the origin to the point (3,1), and the second column $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ represents the vector from the origin to the point (1,2).

Conditions: The matrix is $2\times 2$.; Working in standard Cartesian coordinates.

The determinant represents the signed area scaling factor of the linear transformation defined by the matrix. Geometrically, the columns of the matrix form two vectors that span a parallelogram.

Conditions: A is $a 2\times 2$ matrix.; The columns of A are interpreted as two vectors in the plane.; Use ordinary Euclidean area in standard orthonormal coordinates.

The determinant represents the signed area scaling factor of the linear transformation defined by the matrix. Geometrically, the columns of the matrix form two vectors that span a parallelogram.

Conditions: A is $a 2\times 2$ matrix.; The columns of A are interpreted as two vectors in the plane.; Use ordinary Euclidean area in standard orthonormal coordinates.

We multiply by 5 because the absolute value of the determinant of the transformation matrix $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$ is 5. The determinant represents the area scaling factor of the linear transformation.

Conditions: The transformation matrix is $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$.; The original area is known.; Use ordinary Euclidean area in standard orthonormal coordinates.

We multiply by 5 because the absolute value of the determinant of the transformation matrix $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$ is 5. The determinant represents the area scaling factor of the linear transformation.

Conditions: The transformation matrix is $A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}$.; The original area is known.; Use ordinary Euclidean area in standard orthonormal coordinates.

A 3D linear transformation is fully determined by tracking where the standard basis vectors ($\hat{i}$, $\hat{j}$, and $\hat{k}$) land. The coordinates of these three transformed vectors are recorded as column vectors to form a 3x3 matrix.

Conditions: Working in three-dimensional Cartesian space; Using the standard basis vectors aligned with x, y, and z axes; The transformation is linear (preserves grid lines parallel/evenly spaced and fixes origin)

A 3D linear transformation is fully determined by tracking where the standard basis vectors ($\hat{i}$, $\hat{j}$, and $\hat{k}$) land. The coordinates of these three transformed vectors are recorded as column vectors to form a 3x3 matrix.

Conditions: Working in three-dimensional Cartesian space; Using the standard basis vectors aligned with x, y, and z axes; The transformation is linear (preserves grid lines parallel/evenly spaced and fixes origin)

Starting from $Av = \lambda v$, we rewrite the right side as $(\lambda I)v$ and move all terms to one side to get $(A - \lambda I)v = 0$. Since we seek non-zero solutions for $v$, the matrix $(A - \lambda I)$ must squash space into a lower dimension (have a non-trivial null space).

Conditions: $v$ is a non-zero eigenvector; $A$ is a square matrix; $I$ is the identity matrix

Starting from $Av = \lambda v$, we rewrite the right side as $(\lambda I)v$ and move all terms to one side to get $(A - \lambda I)v = 0$. Since we seek non-zero solutions for $v$, the matrix $(A - \lambda I)$ must squash space into a lower dimension (have a non-trivial null space).

Conditions: $v$ is a non-zero eigenvector; $A$ is a square matrix; $I$ is the identity matrix