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) $\lambda \neq 2$
(c) $\lambda \neq-2$
(d) none of these
Solution:
Given: $A=\left[\begin{array}{ccc}2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3\end{array}\right]$
$A^{-1}$ exists only if $|A| \neq 0$
$\left|\begin{array}{ccc}2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3\end{array}\right| \neq 0$
$\Rightarrow 2(6-5)+1(5 \lambda+6) \neq 0$
$\Rightarrow 2(1)+5 \lambda+6 \neq 0$
$\Rightarrow 2+5 \lambda+6 \neq 0$
$\Rightarrow 5 \lambda+8 \neq 0$
$\Rightarrow 5 \lambda \neq-8$
$\Rightarrow \lambda \neq \frac{-8}{5}$
Thus, $A^{-1}$ exists if $\lambda \in R-\left\{-\frac{8}{5}\right\}$
Hence, the correct option is (a).