Question

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

(a) $\lambda =2$
(b) $\lambda \ne 2$
(c) $\lambda \ne -2$
(d) None of these

Answer 1

(d) We have,
$A=\left|\begin{array}{ccc}2& \lambda & -3\\ 0& 2& 5\\ 1& 1& 3\end{array}\right|$
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 $\lambda \ne \frac{-8}{5}$