College Algebra, Chapter 7, 7.2, Section 7.2, Problem 18

Solve the matrix equation $5(X - C) = D$ for the unknown matrix $X$, where

We solve for $X$


$
\begin{equation}
\begin{aligned}

5(X - C) =& D
&& \text{Given equation}
\\
\\
5X - 5C =& D
&& \text{Distributive Properties of Scalar Multiplication}
\\
\\
5X =& D + 5C
&& \text{Add the matrix $5C$ to each side}
\\
\\
X =& \frac{1}{5} (D + 5C)
&& \text{Multiply each side by the scalar } \frac{1}{5}

\end{aligned}
\end{equation}
$


So,


$
\begin{equation}
\begin{aligned}

X =& \frac{1}{5} \left( \left[ \begin{array}{cc}
10 & 20 \\
30 & 20 \\
10 & 0
\end{array} \right] + 5 \left[ \begin{array}{cc}
2 & 3 \\
1 & 0 \\
0 & 2
\end{array} \right] \right)

&& \text{Substitute the matrices $D$ and $C$}
\\
\\
X =& \frac{1}{5} \left( \left[ \begin{array}{cc}
10 & 20 \\
30 & 20 \\
10 & 0
\end{array} \right] + \left[ \begin{array}{cc}
10 & 15 \\
5 & 0 \\
0 & 10
\end{array} \right] \right)
&& \text{Simplify matrix } C
\\
\\
X =& \frac{1}{5} \left[ \begin{array}{cc}
20 & 35 \\
35 & 20 \\
10 & 10
\end{array} \right]
&& \text{Add matrices}
\\
\\
X =& \left[ \begin{array}{cc}
\displaystyle \frac{20}{5} & \displaystyle \frac{35}{5} \\
\displaystyle \frac{35}{5} & \displaystyle \frac{20}{5} \\
\displaystyle \frac{10}{5} & \displaystyle \frac{10}{5}
\end{array} \right]
&& \text{Multiply the scalar } \frac{1}{5}
\\
\\
X =& \left[ \begin{array}{cc}
4 & 7 \\
7 & 4 \\
2 & 2
\end{array} \right]
&&

\end{aligned}
\end{equation}
$