Chapter 4.6, Problem 18PPS

To determine

To find: the inverse of the matrix, if it exists.

Answer to Problem 18PPS

The inverse exists.

Explanation of Solution

Given information:

  $\left[\begin{array}{cc}2& 3\\ 3& 2\end{array}\right]$ .

Formula used:

Inverse of a $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ is, where $ad-bc\ne 0$ .

The value of a second order determinant is the difference of the products of the products of the two diagonals.

Calculation:

Consider,

  $P=\left[\begin{array}{cc}2& 3\\ 3& 2\end{array}\right]$

Step 1.

  $=\left|\begin{array}{cc}2& 3\\ 3& 2\end{array}\right|=\left(2\times 2\right)-\left(3\times 3\right)$

  $\begin{array}{l}=4-9\\ =-5\end{array}$

Since the determinant does not equal to 0, $P^{-1}$ exists.

Step 2.

  $P^{-1}=\frac{1}{(2)(2)-(3)(3)}\left[\begin{array}{cc}2& -3\\ -3& 2\end{array}\right]$

  $\begin{array}{l}=\frac{1}{4-9}\left[\begin{array}{cc}2& -3\\ -3& 2\end{array}\right]\\ =-\frac{1}{5}\left[\begin{array}{cc}2& -3\\ -3& 2\end{array}\right]\end{array}$

  $a=2,b=-3,c=-3,d=2$

Step3.

Thus, inverse exits.

  $P^{-1}=\left[\begin{array}{cc}-\frac{2}{5}& \frac{3}{5}\\ \frac{3}{5}& -\frac{2}{5}\end{array}\right]$