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:

Step1: Subtract the matrices

Subtract the corresponding elements of the two matrices inside the parentheses.
For the first row, first column: $x^2 - (-4y) = x^2 + 4y$
First row, second column: $4 - 1 = 3$
Second row, first column: $-5y - (y + x) = -5y - y - x = -6y - x$
Second row, second column: $-3x - (-3x) = -3x + 3x = 0$
Third row, first column: $6y - x$
Third row, second column: $-4 - 5 = -9$
So the resulting matrix after subtraction is:

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

Step2: Multiply by 3

Multiply each element of the resulting matrix by 3.
First row, first column: $3(x^2 + 4y) = 3x^2 + 12y$
First row, second column: $3\times3 = 9$
Second row, first column: $3(-6y - x) = -18y - 3x$
Second row, second column: $3\times0 = 0$
Third row, first column: $3(6y - x) = 18y - 3x$
Third row, second column: $3\times(-9) = -27$

Answer:

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