If $A = \begin{bmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix}$, then $A^{-1}$ exists, if

None of these

The correct answer is Option (4) → None of these ##

We have,

$A = \begin{bmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix}$

On expanding along $R_1$, we get

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

We know that, $A^{-1}$ exists, if $A$ is non-singular matrix i.e., $|A| \neq 0$.

$∴2 + 5\lambda + 6 \neq 0$

$\Rightarrow 5\lambda \neq -8$

$∴\lambda \neq -\frac{8}{5}$

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