Chapter 10, Problem 71RE

To determine

The determinant and inverse matrix of $\left[\begin{array}{ccc}3& 0& 1\\ 2& -3& 0\\ 4& -2& 1\end{array}\right]$ is to be estimated.

Answer to Problem 71RE

The inverse matrix of $\left[\begin{array}{ccc}3& 0& 1\\ 2& -3& 0\\ 4& -2& 1\end{array}\right]$ is $\left[\begin{array}{ccc}3& 2& -3\\ 2& 1& -2\\ -8& -6& 9\end{array}\right]$

Explanation of Solution

Given:

  $\left[\begin{array}{ccc}3& 0& 1\\ 2& -3& 0\\ 4& -2& 1\end{array}\right]$

Calculations:

The given matrix is defined as matrix $A$

  $A=\left[\begin{array}{ccc}3& 0& 1\\ 2& -3& 0\\ 4& -2& 1\end{array}\right]$

The determinant of the given matrix is calculated as:

  $\begin{array}{l}A=\left[\begin{array}{ccc}3& 0& 1\\ 2& -3& 0\\ 4& -2& 1\end{array}\right]\\ A=3\left(-3\right)+0+1\left(-4+12\right)\\ A=-9+0+8\\ A=1\end{array}$

Know that:

  $A\times A^{-1}=I$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

  $\begin{array}{l}{R}_1\to R_1-3R_3,R_2\to R_2-2R_3\\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& \frac{1}{9}\end{array}\right]=\left[\begin{array}{ccc}3& 2& -3\\ 2& 1& -2\\ -\frac{8}{9}& -\frac{2}{3}& 1\end{array}\right]\\ R_3\to \left(9\right)R_3\\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}3& 2& -3\\ 2& 1& -2\\ -8& -6& 9\end{array}\right]\end{array}$

So, the inverse matrix is

  $A^{-1}=\left[\begin{array}{ccc}3& 2& -3\\ 2& 1& -2\\ -8& -6& 9\end{array}\right]$

Conclusion:

Hence, $A^{-1}=\left[\begin{array}{ccc}3& 2& -3\\ 2& 1& -2\\ -8& -6& 9\end{array}\right]$