Chapter 4.6, Problem 19PPS

To determine

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

Answer to Problem 19PPS

The inverse exists.

Explanation of Solution

Given information:

  $\left[\begin{array}{cc}3& 0\\ 5& 1\end{array}\right]$ .

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,

  $P=\left[\begin{array}{cc}3& 0\\ 0& 2\end{array}\right]$

Step 1.

  $=\left|\begin{array}{cc}3& 0\\ 5& 1\end{array}\right|=\left(3\times 1\right)-\left(0\right)$

  $\begin{array}{l}=3-0\\ =3\end{array}$

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

Step 2.

  $P^{-1}=\frac{1}{(3)(1)-\left(0\right)}\left[\begin{array}{cc}1& 0\\ -5& 3\end{array}\right]$

  $\begin{array}{l}=\frac{1}{3-0}\left[\begin{array}{cc}1& 0\\ -5& 3\end{array}\right]\\ =\frac{1}{3}\left[\begin{array}{cc}1& 0\\ -5& 3\end{array}\right]\end{array}$

  $a=1,b=0,c=-5,d=3$

Step3.

Thus, inverse exits.

  $P^{-1}=\left[\begin{array}{cc}\frac{1}{3}& 0\\ -\frac{5}{3}& 1\end{array}\right]$