Chapter 12.3, Problem 3LC

Solution

To find: The inverse of the matrix $\left[\begin{array}{cc}1& 2\\ 3& 5\end{array}\right]$ .

The inverse of the matrix is ${\left[\begin{array}{cc}1& 2\\ 3& 5\end{array}\right]}^{-1}=\left[\begin{array}{cc}-5& 2\\ 3& -1\end{array}\right]$ .

Given information: Matrix is $\left[\begin{array}{cc}1& 2\\ 3& 5\end{array}\right]$ .

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,

  $\left[\begin{array}{cc}1& 2\\ 3& 5\end{array}\right]$

Then the determinant of the matrix is,

  $\begin{array}{c}\left|\begin{array}{cc}1& 2\\ 3& 5\end{array}\right|=1\times 5-2\times 3\\ =5-6\\ =-1\end{array}$

Then inverse of the matrix is,

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

Therefore, the inverse of the matrix is ${\left[\begin{array}{cc}1& 2\\ 3& 5\end{array}\right]}^{-1}=\left[\begin{array}{cc}-5& 2\\ 3& -1\end{array}\right]$ .