Question

1. $3\left(\

$$\begin{bmatrix}x^2 & 4 \\\\ -5y & -3x \\\\ 6y & -4\\end{bmatrix}$$

-\

$$\begin{bmatrix}-4y & 1 \\\\ y + x & -3x \\\\ x & 5\\end{bmatrix}$$

\
ight)$

Explanation:

Step 1: Subtract the matrices

To subtract two matrices, we subtract their corresponding elements. Let's denote the first matrix as \( A =

$$\begin{bmatrix} x^2 & 4 \\ -5y & -3x \\ 6y & -4 \end{bmatrix}$$

\) and the second matrix as \( B =

$$\begin{bmatrix} -4y & 1 \\ y + x & -3x \\ x & 5 \end{bmatrix}$$

\). Then \( A - B \) is:

\[

$$\begin{bmatrix} x^2 - (-4y) & 4 - 1 \\ -5y - (y + x) & -3x - (-3x) \\ 6y - x & -4 - 5 \end{bmatrix}$$

=

$$\begin{bmatrix} x^2 + 4y & 3 \\ -5y - y - x & -3x + 3x \\ 6y - x & -9 \end{bmatrix}$$

=

$$\begin{bmatrix} x^2 + 4y & 3 \\ -6y - x & 0 \\ 6y - x & -9 \end{bmatrix}$$

\]

Step 2: Multiply the resulting matrix by 3

Now we multiply each element of the matrix \( A - B \) by 3:

\[
3 \times

$$\begin{bmatrix} x^2 + 4y & 3 \\ -6y - x & 0 \\ 6y - x & -9 \end{bmatrix}$$

=

$$\begin{bmatrix} 3(x^2 + 4y) & 3 \times 3 \\ 3(-6y - x) & 3 \times 0 \\ 3(6y - x) & 3 \times (-9) \end{bmatrix}$$

=

$$\begin{bmatrix} 3x^2 + 12y & 9 \\ -18y - 3x & 0 \\ 18y - 3x & -27 \end{bmatrix}$$

\]

Answer:

\[
\boxed{

$$\begin{bmatrix} 3x^2 + 12y & 9 \\ -3x - 18y & 0 \\ -3x + 18y & -27 \end{bmatrix}$$

}
\]