College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 38

Solve the matrix equation $\displaystyle \frac{1}{2} (X - 2B) = A$ for the unknown matrix, $X$ or show that no solution exists, where

$\displaystyle A = \left[ \begin{array}{cc}
2 & 1 \\
3 & 2
\end{array} \right] \qquad B = \left[ \begin{array}{cc}
1 & -2 \\
-2 & 4
\end{array} \right] \qquad C = \left[ \begin{array}{ccc}
0 & 1 & 3 \\
-2 & 4 & 0
\end{array} \right]$


$
\begin{equation}
\begin{aligned}

\frac{1}{2} (X - 2B) =& A
&& \text{Given equation}
\\
\\
X- 2B =& 2A
&& \text{Multiply both sides by scalar } 2
\\
\\
X =& 2A + 2B
&& \text{Add matrix } 2B
\\
\\
X =& 2 \left[ \begin{array}{cc}
2 & 1 \\
3 & 2
\end{array} \right] + 2 \left[ \begin{array}{cc}
1 & -2 \\
-2 & 4
\end{array} \right]
&&
\\
\\
X =& \left[ \begin{array}{cc}
4 & 2 \\
6 & 4
\end{array} \right] + \left[ \begin{array}{cc}
2 & -4 \\
-4 & 8
\end{array} \right]
&&
\\
\\
X =& \left[ \begin{array}{cc}
6 & -2 \\
2 & 12
\end{array} \right]
&&

\end{aligned}
\end{equation}
$