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}$