Chapter 11.4, Problem 69AYU

To determine

To find: The inverse of a matrix using a graphing utility. The answers to be rounded to two decimal places.

Answer to Problem 69AYU

Solution:

$\left[\begin{array}{cc}\begin{array}{cc}0.02& -0.04\\ -0.02& 0.05\end{array}& \begin{array}{cc}-0.01& 0.01\\ 0.03& -0.03\end{array}\\ \begin{array}{cc}0.02& 0.01\\ -0.02& 0.06\end{array}& \begin{array}{cc}-0.04& 0.00\\ 0.07& 0.06\end{array}\end{array}\right]$

Explanation of Solution

Given:

$\text{A }=\left[\begin{array}{cc}\begin{array}{cc}44& 21\\ -2& 10\end{array}& \begin{array}{cc}18& 6\\ 15& 5\end{array}\\ \begin{array}{cc}21& 12\\ -8& -16\end{array}& \begin{array}{cc}-12& 4\\ 4& 9\end{array}\end{array}\right]$

Calculation:

The solution for the problem using free numerical and graphing utility is given below:

The inverse of matrix $\text{A }=\left[\begin{array}{cc}\begin{array}{cc}44& 21\\ -2& 10\end{array}& \begin{array}{cc}18& 6\\ 15& 5\end{array}\\ \begin{array}{cc}21& 12\\ -8& -16\end{array}& \begin{array}{cc}-12& 4\\ 4& 9\end{array}\end{array}\right]$ is $\left[\begin{array}{cc}\begin{array}{cc}0.02& -0.04\\ -0.02& 0.05\end{array}& \begin{array}{cc}-0.01& 0.01\\ 0.03& -0.03\end{array}\\ \begin{array}{cc}0.02& 0.01\\ -0.02& 0.06\end{array}& \begin{array}{cc}-0.04& 0.00\\ 0.07& 0.06\end{array}\end{array}\right]$.