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:

$$
LATEXBLOCK2
=
LATEXBLOCK3
=
LATEXBLOCK4
$$

Step 2: Multiply the resulting matrix by 3

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

$$ 3 \times
LATEXBLOCK5
=
LATEXBLOCK6
=
LATEXBLOCK7
$$

Answer:

$$ \boxed{
LATEXBLOCK0
} $$