Question

Given the matrices A and B shown below , solve for X in the equation -frac{1}{3}X+frac{1}{2}A=B A=begin{bmatrix}-10 & 4 8 & 8 end{bmatrix}, B=begin{bmatrix}9 & -1 4& 2 end{bmatrix}

Matrices
ANSWERED
asked 2020-12-09
Given the matrices A and B shown below , solve for X in the equation \(-\frac{1}{3}X+\frac{1}{2}A=B\)
\(A=\begin{bmatrix}-10 & 4 \\8 & 8 \end{bmatrix}, B=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}\)

Answers (1)

2020-12-10
Step 1
the given matrices are:
\(A=\begin{bmatrix}-10 & 4 \\8 & 8 \end{bmatrix}, B=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}\)
we have to solve for X in the equation \(-\frac{1}{3}X+\frac{1}{2}A=B\)
Step 2
the given matrices are: \(A=\begin{bmatrix}-10 & 4 \\8 & 8 \end{bmatrix} \text{ and }B=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}\)
\(-\frac{1}{3}X+\frac{1}{2}A=B\)
\(\frac{-1}{3}X+\frac{1}{2}\begin{bmatrix}-10 & 4 \\8 & 8 \end{bmatrix}=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}\)
\(\frac{-1}{3}X+\begin{bmatrix}-\frac{10}{2} & \frac{4}{2} \\\frac{8}{2} & \frac{8}{2} \end{bmatrix}=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}\)
\(\frac{-1}{3}X+\begin{bmatrix}-5 & 2 \\4 & 4 \end{bmatrix}=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}\)
\(\frac{-1}{3}X=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}-\begin{bmatrix}-5 & 2 \\4 & 4 \end{bmatrix}\)
\(\frac{-1}{3}X=\begin{bmatrix}9 & -1 \\4& 2 \end{bmatrix}+\begin{bmatrix}5 & -2 \\-4 & -4 \end{bmatrix}\)
\(\frac{-1}{3}X=\begin{bmatrix}9+5 & -1-2 \\4-4& 2-4 \end{bmatrix}\)
\(\frac{-1}{3}X=\begin{bmatrix}14 & -3 \\0& -2 \end{bmatrix}\)
\(X=-3\begin{bmatrix}14 & -3 \\0& -2 \end{bmatrix}\)
\(X=\begin{bmatrix}14\times(-3) & -3\times(-3) \\0\times(-3)& -2\times(-3) \end{bmatrix}\)
\(X=\begin{bmatrix}-42 & 9 \\0 & 6 \end{bmatrix}\)
Step 3
therefore the matrix X is \(\begin{bmatrix}-42 & 9 \\0 & 6 \end{bmatrix}\)
therefore the solution of X for the equation \(-\frac{1}{3}X+\frac{1}{2}A=B \text{ is } X=\begin{bmatrix}-42 & 9 \\0 & 6 \end{bmatrix}\)
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