Chapter 7.6, Problem 71E

To determine

To Find: To find the inverse of matrix.

Answer to Problem 71E

The solution is $x=-23\text{and}y=-18.$

Explanation of Solution

Given information:Given a diagonal matrix.

Formula used: $A^{-1}=\frac{1}{\det (A)}{\left[A_{ij}\right]}^T$

Calculation:

Let the matrix of 2 by 2 order is X and the matrix of order 3 by 3 is Y.

Let $\therefore X=\left[\begin{array}{cc}3& 0\\ 0& -3\end{array}\right]\text{and}Y=\left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& -1\end{array}\right]$

Calculation the inverse of matrix X

  $\begin{array}{l}\det (X)=\left|\begin{array}{cc}3& 0\\ 0& -3\end{array}\right|\\ \det (X)=-9-0=-9\end{array}$

Let the cofactors be represented by $C_{ij}$ of matrix X

  $\begin{array}{l}{C}_{11}=-3\\ C_{12}=0\\ C_{21}=0\\ C_{22}=3\end{array}$

Thus $X^{-1}=\frac{1}{-9}\left[\begin{array}{cc}-3& 0\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& 0\\ 0& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}\right]$

Similarly, the inverse of Y

  $\det (Y)=\left|\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& -1\end{array}\right|$

Expanding along 1^st row

  $\det (Y)=-2$

Let the cofactors be represented by $C_{ij}$ of matrix Y

  $\begin{array}{l}{C}_{11}=-2\\ C_{12}=0\\ C_{13}=0\\ C_{21}=0\\ C_{22}=-1\\ C_{23}=0\\ C_{31}=0\\ C_{32}=0\\ C_{33}=2\end{array}$

Thus, $Y^{-1}=\frac{1}{-2}\left[\begin{array}{ccc}-2& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& 0\\ 0& 0& -1\end{array}\right]$

Conjecture about the inverse of matrix A

  $A^{-1}=\left[\begin{array}{ccccc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a_{11}$}\right.& 0& 0& \dots & 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a_{22}$}\right.& 0& \dots & 0\\ 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a_{33}$}\right.& \dots & 0\\ ⋮& ⋮& ⋮& \cdots & ⋮\\ 0& 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a_{nn}$}\right.\end{array}\right]$