If $A = \begin{bmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix}$, then $A^{-1}$ exists, if
Answer & explanation
Correct answer: option 4
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) x (-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 $\lambda \neq -\frac{8}{5}$.