Chapter 12.3, Problem 31PPSE

Solution

Whether the inverse of the matrix $\left[\begin{array}{cc}1& 3\\ 2& 0\end{array}\right]$ exists or not. If exists, then find it.

The inverse of the matrix is ${\left[\begin{array}{cc}1& 3\\ 2& 0\end{array}\right]}^{-1}=\left[\begin{array}{cc}0& \frac{1}{2}\\ \frac{1}{3}& -\frac{1}{6}\end{array}\right]$ .

Given information: Matrix is $\left[\begin{array}{cc}1& 3\\ 2& 0\end{array}\right]$ .

Formula used: The determinant of the matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is,

  $\begin{array}{c}\det A=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\\ =ad-bc\end{array}$ .

If the determinant of the matrix is not zero, then inverse exists.

Inverse of the matrix is,

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

Calculation:

Consider the matrix as,

  $\left[\begin{array}{cc}1& 3\\ 2& 0\end{array}\right]$ .

The determinant of the matrix is,

  $\begin{array}{c}\left|\begin{array}{cc}1& 3\\ 2& 0\end{array}\right|=1\times 0-3\times 2\\ =0-6\\ =-6\end{array}$

Since, determinant is non-zero.

Hence, inverse exists.

Then inverse of the matrix is,

  $\begin{array}{c}{\left[\begin{array}{cc}1& 3\\ 2& 0\end{array}\right]}^{-1}=\frac{1}{\left(-6\right)}\left[\begin{array}{cc}0& -3\\ -2& 1\end{array}\right]\\ =\left[\begin{array}{cc}\frac{0}{\left(-6\right)}& \frac{-3}{-6}\\ \frac{-2}{-6}& \frac{1}{-6}\end{array}\right]\\ =\left[\begin{array}{cc}0& \frac{1}{2}\\ \frac{1}{3}& -\frac{1}{6}\end{array}\right]\end{array}$

Therefore, the inverse of the matrix is ${\left[\begin{array}{cc}1& 3\\ 2& 0\end{array}\right]}^{-1}=\left[\begin{array}{cc}0& \frac{1}{2}\\ \frac{1}{3}& -\frac{1}{6}\end{array}\right]$ .