# Given the matrices A=begin{bmatrix}-1 & 3 2 & -1 3&1 end{bmatrix} text{ and } B=begin{bmatrix}0 & -2 1 & 3 4 & -3 end{bmatrix} find the 3 times 2 matrix X that is a solution of the equation. 8X+A=B

Given the matrices
find the $3×2$ matrix X that is a solution of the equation. 8X+A=B
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Talisha
Step 1
Given that :
The matrices ,

The equation is 8x + A = B
Step 2
By using,
Subtraction of matrix is possible by subtracting the element of another matrix if they have the same order.
So, the difference between two matrices is obtained by subtracting the corresponding elements of the given matrices.
In scalar multiple, each entry in the matrix is multiplied by the given scalar.
Step 3
To find matrix X:
consider ,
8X+A=B
Solve for X
8X=B-A
$X=\frac{1}{8}\left[B-A\right]$
then,
Step 4
$X=\frac{1}{8}\left[\left[\begin{array}{cc}0& -2\\ 1& 3\\ 4& -3\end{array}\right]-\left[\begin{array}{cc}-1& 3\\ 2& -1\\ 3& 1\end{array}\right]\right]$
$=\frac{1}{8}\left[\begin{array}{cc}0-\left(-1\right)& -2-3\\ 1-2& 3-\left(-1\right)\\ 4-3& -3-1\end{array}\right]$
$=\frac{1}{8}\left[\begin{array}{cc}1& -5\\ -1& 4\\ 1& -4\end{array}\right]$
$=\left[\begin{array}{cc}\frac{1}{8}& -\frac{5}{8}\\ -\frac{1}{8}& \frac{4}{8}\\ \frac{1}{8}& -\frac{4}{8}\end{array}\right]$
Step 5
$=\left[\begin{array}{cc}\frac{1}{8}& -\frac{5}{8}\\ -\frac{1}{8}& \frac{4}{8}\\ \frac{1}{8}& -\frac{4}{8}\end{array}\right]$
$⇒X=\left[\begin{array}{cc}\frac{1}{8}& -\frac{5}{8}\\ -\frac{1}{8}& \frac{4}{8}\\ \frac{1}{8}& -\frac{4}{8}\end{array}\right]$
Step 6
Therefore,
The $3×2$ matrix X is the solution of the equation 8X+A=B is, $X=\left[\begin{array}{cc}\frac{1}{8}& -\frac{5}{8}\\ -\frac{1}{8}& \frac{4}{8}\\ \frac{1}{8}& -\frac{4}{8}\end{array}\right]$
Jeffrey Jordon