Question

Question: In Exercise 12, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.

12. $\left(\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{3}\\ -\mathbf{2}& \mathbf{2}& \mathbf{1}\\ \mathbf{0}& \mathbf{1}& \mathbf{1}\end{array}\right)$

Short answer

The adjugate matrix is $\left(\begin{array}{ccc}1& 2& -5\\ 2& 1& -7\\ -2& -1& 4\end{array}\right)$, and the inverse matrix is

Step-by-step solution

First, find the determinant

Let $A=\left(\begin{array}{ccc}1& 1& 3\\ -2& 2& 1\\ 0& 1& 1\end{array}\right)$. Then,

$\begin{array}{c}detA=\left|\begin{array}{ccc}1& 1& 3\\ -2& 2& 1\\ 0& 1& 1\end{array}\right|\\ =0-1\left|\begin{array}{cc}1& 3\\ -2& 1\end{array}\right|+1\left|\begin{array}{cc}1& 1\\ -2& 2\end{array}\right|\\ =-\left(7\right)+4\\ detA=-3\ne 0\end{array}$

Here, $detA\ne 0$. Hence, the inverse of A exists.

Compute the adjugate matrix  

The nine cofactorsare:

$\begin{array}{c}{C}_{11}={\left(-1\right)}^2\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|\\ =1\end{array}$

$\begin{array}{c}{C}_{12}={\left(-1\right)}^3\left|\begin{array}{cc}-2& 1\\ 0& 1\end{array}\right|\\ =-\left(-2\right)\\ =2\end{array}$

$\begin{array}{c}{C}_{13}={\left(-1\right)}^4\left|\begin{array}{cc}-2& 2\\ 0& 1\end{array}\right|\\ =-2\end{array}$

$\begin{array}{c}{C}_{21}={\left(-1\right)}^3\left|\begin{array}{cc}1& 3\\ 1& 1\end{array}\right|\\ =-\left(-2\right)\\ =2\end{array}$

$\begin{array}{c}{C}_{22}={\left(-1\right)}^4\left|\begin{array}{cc}1& 3\\ 0& 1\end{array}\right|\\ =1\end{array}$

$\begin{array}{c}{C}_{23}={\left(-1\right)}^5\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|\\ =-1\end{array}$

$\begin{array}{c}{C}_{31}={\left(-1\right)}^4\left|\begin{array}{cc}1& 3\\ 2& 1\end{array}\right|\\ =-5\end{array}$

$\begin{array}{c}{C}_{32}={\left(-1\right)}^5\left|\begin{array}{cc}1& 3\\ -2& 1\end{array}\right|\\ =-7\end{array}$

$\begin{array}{c}{C}_{33}={\left(-1\right)}^6\left|\begin{array}{cc}1& 1\\ -2& 2\end{array}\right|\\ =4\end{array}$

The adjugate matrix is the transpose of the matrix of cofactors. Hence,

$\begin{array}{c}\mathrm{a}\mathrm{d}\mathrm{j}A=\left(\begin{array}{ccc}{C}_{11}& C_{21}& C_{31}\\ C_{12}& C_{22}& C_{32}\\ C_{13}& C_{23}& C_{33}\end{array}\right)\\ =\left(\begin{array}{ccc}1& 2& -5\\ 2& 1& -7\\ -2& -1& 4\end{array}\right)\end{array}$

Use Theorem 8 to find $A^{-\mathbf{1}}$

By Theorem 8,