Chapter 16.6, Problem 4ST

To determine

To Find: The value matrix A:

  $\left[\begin{array}{cc}2& 1\\ -1& 4\end{array}\right]A=\left[\begin{array}{cc}3& -1\\ -6& 5\end{array}\right]$

Answer to Problem 4ST

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

Explanation of Solution

Given: $\left[\begin{array}{cc}2& 1\\ -1& 4\end{array}\right]A=\left[\begin{array}{cc}3& -1\\ -6& 5\end{array}\right]$

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:

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

  $\begin{array}{l}\left[\begin{array}{cc}2& 1\\ -1& 4\end{array}\right]A=\left[\begin{array}{cc}3& -1\\ -6& 5\end{array}\right]\\ A={\left[\begin{array}{cc}2& 1\\ -1& 4\end{array}\right]}^{-1}\left[\begin{array}{cc}3& -1\\ -6& 5\end{array}\right]\end{array}$

Finding the determinant:

  $\left|\begin{array}{cc}2& 1\\ -1& 4\end{array}\right|=(2)(4)-(-1)(1)=9$

Finding the inverse:

  ${\left[\begin{array}{cc}2& 1\\ -1& 4\end{array}\right]}^{-1}=\frac{1}{9}\left[\begin{array}{cc}4& -1\\ 1& 2\end{array}\right]$

  $\begin{array}{l}A={\left[\begin{array}{cc}2& 1\\ -1& 4\end{array}\right]}^{-1}\left[\begin{array}{cc}3& -1\\ -6& 5\end{array}\right]\\ \Rightarrow A=\frac{1}{9}\left[\begin{array}{cc}4& -1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}3& -1\\ -6& 5\end{array}\right]\\ \Rightarrow A=\frac{1}{9}\left[\begin{array}{cc}4& -1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}3& -1\\ -6& 5\end{array}\right]\\ \Rightarrow A=\frac{1}{9}\left[\begin{array}{cc}18& -9\\ -9& 9\end{array}\right]=\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]\end{array}$