Chapter 4, Problem 15PT

To determine

To calculate: The inverse of the matrix $\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]$ .

Answer to Problem 15PT

The inverse of the matrix $\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]$ is $\left[\begin{array}{cc}\frac{1}{5}& 0\\ 0& 1\end{array}\right]$ .

Explanation of Solution

Given information:

The matrix is given as:

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

Formula used:

For the matrix of order $2\times 2$ ,

  $A=\left[\begin{array}{ll}p\hfill & q\hfill \\ r\hfill & s\hfill \end{array}\right]$

The determinant is given as $\left|A\right|=ps-qr$ .

And its inverse is $A^{-1}=\frac{1}{\left|A\right|}\left[\begin{array}{cc}s& -q\\ -r& p\end{array}\right]$

Calculation:

Consider the matrix,

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

Recall that for the matrix of order $2\times 2$ ,

  $A=\left[\begin{array}{ll}p\hfill & q\hfill \\ r\hfill & s\hfill \end{array}\right]$

The determinant is given as $\left|A\right|=ps-qr$ .

And its inverse is $A^{-1}=\frac{1}{\left|A\right|}\left[\begin{array}{cc}s& -q\\ -r& p\end{array}\right]$

Therefore, determinant of the matrix is given by,

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

Since, determinant is non zero so its inverse exists.

And inverse is,

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

Thus, the inverse of the matrix $\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]$ is $\left[\begin{array}{cc}\frac{1}{5}& 0\\ 0& 1\end{array}\right]$ .