Chapter 4.6, Problem 23PPS

To determine

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

Answer to Problem 23PPS

The inverse exists.

Explanation of Solution

Given information:

  $\left[\begin{array}{cc}6& -5\\ 4& 9\end{array}\right]$ .

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,

  $P=\left[\begin{array}{cc}6& -5\\ 4& 9\end{array}\right]$

Step 1.

  $=\left|\begin{array}{cc}6& -5\\ 4& 9\end{array}\right|=\left(6\times 9\right)-\left(-5\times 4\right)$

  $\begin{array}{l}=54+20\\ =74\end{array}$

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

Step 2.

  $P^{-1}=\frac{1}{(6)(9)-(-5)(4)}\left[\begin{array}{cc}9& 5\\ -4& 6\end{array}\right]$

  $\begin{array}{l}=\frac{1}{54+20}\left[\begin{array}{cc}9& 5\\ -4& 6\end{array}\right]\\ =\frac{1}{74}\left[\begin{array}{cc}9& 5\\ -4& 6\end{array}\right]\end{array}$

  $a=9,b=5,c=-4,d=6$

Step3.

Thus, inverse exits.

  $P^{-1}=\left[\begin{array}{cc}\frac{9}{74}& \frac{\infty 5}{74}\\ -\frac{2}{37}& \frac{3}{37}\end{array}\right]$