Chapter 16.6, Problem 20WE

To determine

To Find: The value of matrix $X$ :

  $\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]X-\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]$

Answer to Problem 20WE

  $\left[\begin{array}{cc}1& -2\\ -1& 5\end{array}\right]$

Explanation of Solution

Given:

  $\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]X-\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]$

Formula Used:

1. Determinant formula:
$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-cb$
2. Inverse formula:
${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
3. Matrix Multiplication formula:

  $\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{cc}E& F\\ G& H\end{array}\right]=\left[\begin{array}{cc}AE+BG& AF+BH\\ CE+DG& CF+DH\end{array}\right]$

  $\begin{array}{l}\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]X-\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]X=\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]+\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]X=\left[\begin{array}{cc}2& 5\\ 1& 4\end{array}\right]\\ \Rightarrow X={\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]}^{-1}\left[\begin{array}{cc}2& 5\\ 1& 4\end{array}\right]\\ \Rightarrow X=\frac{1}{1}\left[\begin{array}{cc}2& -3\\ -3& 5\end{array}\right]\left[\begin{array}{cc}2& 5\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}1& -2\\ -1& 5\end{array}\right]\end{array}$