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

Therefore, the matrix X can be expressed as,

`X=\begin{bmatrix}{2} & {4} & {} \\ {1} & {-(3)/(2)} & {} \\ {-5} & {1} & {}\end{bmatrix}`