Matrices — Class 12 Maths Solution

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

In order to use elementary row operations we may write $A = IA$.

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

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

$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}1&3\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\{5/22}&{1/22}\end{array}} \right]A$

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

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

$\Rightarrow$ $I = BA$,

where B is the inverse of A.
$\therefore$ $B = \frac{1}{{22}}\left[ {\begin{array}{llllllllllllllllllll}7&{ - 3}\\5&{ - 1}\end{array}} \right]$

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

In order to use elementary row operations,

we write $A = IA$
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 3}\\{ - 2}&6\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}1&0\\0&1\end{array}} \right]A$

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

Since, we obtain all zeroes in a row of the matrix A on LHS,

so ${A^{ - 1}}$ does not exist.