Question

Find inverse, by elementary row operations (if possible), of the given matrix

(i) $\left[\begin{array}{cc}1& 3\\ -5& 7\end{array}\right]$

(ii) $\left[\begin{array}{cc}1& -3\\ -2& 6\end{array}\right]$

Answer 1

(i)

Let $A=\left[\begin{array}{cc}1& 3\\ -5& 7\end{array}\right]$

We can write the given matrix as

$A=IA$

$\therefore \left[\begin{array}{cc}1& 3\\ -5& 7\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$

$[\text{Applying}{R}_2\to R_2+5R_1]$

$\Rightarrow \left[\begin{array}{cc}1& 3\\ 0& 22\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]A$
$[\text{Applying}{R}_2\to \frac{1}{22}{R}_2]$

$\Rightarrow \left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ \frac{5}{22}& \frac{1}{22}\end{array}\right]A$
$[\text{Applying}{R}_1\to R_1+3R_2]$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{7}{22}& -\frac{3}{22}\\ \frac{5}{22}& \frac{1}{22}\end{array}\right]A$

$\Rightarrow I=BA$

Where $B$ is the inverse of $A$

$\therefore B=\frac{1}{22}\left[\begin{array}{cc}7& -3\\ 5& 1\end{array}\right]$

Hence, inverse of given matrix is
$\frac{1}{22}\left[\begin{array}{cc}7& -3\\ 5& 1\end{array}\right]$

(ii)
Let $A=\left[\begin{array}{cc}1& -3\\ -2& 6\end{array}\right]$

We can write the given matrix as

$A=IA$

$\Rightarrow \left[\begin{array}{cc}1& -3\\ -2& 6\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$

Using $R_2\to R_2+2R_1$

$\Rightarrow \left[\begin{array}{cc}1& -3\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]A$ ...(1)

Since all the elements are zero in row number 2 in the matrix in L.H.S of the above eq. (1)

Hence, $A^{-1}$ does not exist.