Chapter 10.5, Problem 20E

To determine

To calculate: The inverse ofmatrix if exist.

Answer to Problem 20E

Inverse of matrix is $A=\left[\begin{array}{ccc}2& 1& 0\\ 1& 1& 4\\ 2& 1& 2\end{array}\right]$ is $A^{-1}=\frac{1}{2}\left|\begin{array}{ccc}-2& -2& 4\\ 4& 4& -8\\ 0& 0& 1\end{array}\right|$

Explanation of Solution

Given information:

  $\left[\begin{array}{ccc}2& 1& 0\\ 1& 1& 4\\ 2& 1& 2\end{array}\right]$

Calculation:

Let call the given matrix as $A$ as,

  $A=\left[\begin{array}{ccc}2& 1& 0\\ 1& 1& 4\\ 2& 1& 2\end{array}\right]$

  $\begin{array}{l}\left|A\right|=2\left|\begin{array}{cc}1& 4\\ 1& 2\end{array}\right|-1\left|\begin{array}{cc}1& 4\\ 2& 2\end{array}\right|\\ \left|A\right|=2\left|2-4\right|-1\left|2-8\right|\\ \left|A\right|=2\left|-2\right|-1\left|-6\right|\\ \left|A\right|=-4+6\\ \left|A\right|=2\end{array}$

  $\begin{array}{l}{C}_{11}=\left|\begin{array}{cc}1& 4\\ 1& 2\end{array}\right|=-2\\ C_{12}=-\left|\begin{array}{cc}1& 4\\ 2& 2\end{array}\right|=-\left(-2-6\right)=4\\ C_{13}=0\\ C_{21}=-\left|\begin{array}{cc}1& 0\\ 1& 2\end{array}\right|=-2\\ C_{22}=\left|\begin{array}{cc}2& 0\\ 2& 2\end{array}\right|=4\\ C_{23}=-\left|\begin{array}{cc}2& 1\\ 2& 1\end{array}\right|=-\left(2-2\right)=0\\ C_{31}=\left|\begin{array}{cc}1& 0\\ 1& 4\end{array}\right|=4\\ C_{32}=-\left|\begin{array}{cc}2& 0\\ 1& 4\end{array}\right|=-8\\ C_{33}=\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|=2-1=1\end{array}$

Now,

  $\begin{array}{l}adjA={\left|\begin{array}{ccc}-2& 4& 0\\ -2& 4& 0\\ 4& -8& 1\end{array}\right|}^r\\ \\ A^{-1}=\frac{1}{\left|A\right|}\left|\begin{array}{ccc}-2& -2& 4\\ 4& 4& -8\\ 0& 0& 1\end{array}\right|\\ =\frac{1}{2}\left|\begin{array}{ccc}-2& -2& 4\\ 4& 4& -8\\ 0& 0& 1\end{array}\right|\end{array}$

Inverse of matrix is $A$ is $A^{-1}=\frac{1}{2}\left|\begin{array}{ccc}-2& -2& 4\\ 4& 4& -8\\ 0& 0& 1\end{array}\right|$