To find the images of the vectors $\mathbf{u} = \begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix}$ and $\mathbf{v} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$ under the linear transformation $T$ defined by the matrix $A = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -3 \end{bmatrix}$, we perform matrix multiplication.

Step 1: Compute $T(\mathbf{u})$

$T(\mathbf{u}) = A \mathbf{u} = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix}$

Calculating each component:

• First component: $-2 \cdot 1 + 0 \cdot 4 + 0 \cdot (-2) = -2$

• Second component: $0 \cdot 1 + 2 \cdot 4 + 0 \cdot (-2) = 8$

• Third component: $0 \cdot 1 + 0 \cdot 4 + (-3) \cdot (-2) = 6$

Thus, $T(\mathbf{u}) = \begin{bmatrix} -2 \\ 8 \\ 6 \end{bmatrix}$

Step 2: Compute $T(\mathbf{v})$

$T(\mathbf{v}) = A \mathbf{v} = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix}$

Calculating each component:

• First component: $-2a$

• Second component: $2b$

• Third component: $-3c$

Thus, $T(\mathbf{v}) = \begin{bmatrix} -2a \\ 2b \\ -3c \end{bmatrix}$

Final Results

$T(\mathbf{u}) = \begin{bmatrix} -2 \\ 8 \\ 6 \end{bmatrix}$

$T(\mathbf{v}) = \begin{bmatrix} -2a \\ 2b \\ -3c \end{bmatrix}$

If you have any questions or need further details, feel free to ask! Here are some related questions:

1. What are the properties of the matrix $A$?
2. How does this transformation affect different types of vectors?
3. Can you find the image of another vector under the same transformation?
4. What is the geometric interpretation of this linear transformation?
5. How can we find the inverse of a transformation defined by a matrix like $A$?

Tip: When dealing with linear transformations, always verify if the matrix is diagonal, as it simplifies calculations significantly.