Question

I'll send a picture of the prob.

Answer 1

Given:

$\begin{gathered} A=\begin{bmatrix}{-7} & {-8} & {} \\ {-2} & {6} & {} \\ {4} & {4} & {}\end{bmatrix} \\ B=\begin{bmatrix}{-3} & {0} & {} \\ {0} & {3} & {} \\ {-6} & {6} & {}\end{bmatrix} \end{gathered}$

The matrix equation is,

$2X+A=B$

Rewriting the equation,

$\begin{gathered} 2X=B-A \\ X=\frac{1}{2}(B-A) \end{gathered}$

Now, let's find B-A.

$\begin{gathered} B-A=\begin{bmatrix}{-3} & {0} & {} \\ {0} & {3} & {} \\ {-6} & {6} & {}\end{bmatrix}-\begin{bmatrix}{-7} & {-8} & {} \\ {-2} & {6} & {} \\ {4} & {4} & {}\end{bmatrix} \\ =\begin{bmatrix}{-3-(-7)} & {0-(-8)} & {} \\ {0-(-2)} & {3-6} & {} \\ {-6-4} & {6-4} & {}\end{bmatrix} \\ =\begin{bmatrix}{4} & {8} & {} \\ {2} & {-3} & {} \\ {-10} & {2} & {}\end{bmatrix} \end{gathered}$

Now,

$\begin{gathered} X=\frac{1}{2}(B-A) \\ =\frac{1}{2}\begin{bmatrix}{4} & {8} & {} \\ {2} & {-3} & {} \\ {-10} & {2} & {}\end{bmatrix} \\ =\begin{bmatrix}{\frac{4}{2}} & {\frac{8}{2}} & {} \\ {\frac{2}{2}} & {\frac{-3}{2}} & {} \\ {\frac{-10}{2}} & {\frac{2}{2}} & {}\end{bmatrix} \\ =\begin{bmatrix}{2} & {4} & {} \\ {1} & {-\frac{3}{2}} & {} \\ {-5} & {1} & {}\end{bmatrix} \end{gathered}$

Therefore, the matrix X can be expressed as,

$X=\begin{bmatrix}{2} & {4} & {} \\ {1} & {-\frac{3}{2}} & {} \\ {-5} & {1} & {}\end{bmatrix}$