If $A = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&\lambda &{ - 3}\\0&2&5\\1&1&3\end{array}} \right|$, then ${A^{ - 1}}$ exists, if

We have,
$A = \left| {\begin{array}{cccccccccccccccccccc}2&\lambda &{ - 3}\\0&2&5\\1&1&3\end{array}} \right|$

Expanding along ${R_1}$,

$|A| = 2(6 - 5) - \lambda ( - 5) - 3( - 2) = 2 + 5\lambda + 6$

We know that, ${A^{ - 1}}$ exists, if $A$ is non-singular matrix i.e.,

$|A| \ne 0$.
$\therefore 2 + 5\lambda + 6 \ne 0$

$\Rightarrow$ $5\lambda \ne - 8$

$\therefore \lambda \ne \frac{{ - 8}}{5}$

So, ${A^{ - 1}}$ exists if and only if $\lambda \ne \frac{{ - 8}}{5}$.