Chapter 2.5, Problem 47E

To determine

To show: expression $A^{-1}$ is the multiplicative inverse of $A$ for any non singular second order matrix.

Answer to Problem 47E

See explanation

Explanation of Solution

Given information:

If $A=\left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]$ and $I=\left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}0\\ 1\end{array}\end{array}\right]$ then proof $A{A}^{-1}=I$

Calculation:

Evaluate $A^{-1}$ as follows:

  $\begin{array}{l}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}\begin{array}{l}d\\ -c\end{array}& \begin{array}{l}-b\\ a\end{array}\end{array}\right]\\ =\left[\begin{array}{cc}\begin{array}{c}\frac{d}{ad-bc}\\ \frac{-c}{ad-bc}\end{array}& \begin{array}{l}\frac{-b}{ad-bc}\\ \frac{a}{ad-bc}\end{array}\end{array}\right]\end{array}$

Solve $A{A}^{-1}$ as follows:

  $\begin{array}{c}A{A}^{-1}=\left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]\left[\begin{array}{cc}\begin{array}{c}\frac{d}{ad-bc}\\ \frac{-c}{ad-bc}\end{array}& \begin{array}{l}\frac{-b}{ad-bc}\\ \frac{a}{ad-bc}\end{array}\end{array}\right]\\ A{A}^{-1}=\left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}0\\ 1\end{array}\end{array}\right]\\ A{A}^{-1}=I\end{array}$

Hence proved