Let $A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ and $B = \frac{1}{3} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & \lambda \end{bmatrix}$. If $AB = I$, then the value of $\lambda$ is

$-2$

The correct answer is Option (2) → $-2$ ##

$A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$

$B = \frac{1}{3} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & \lambda \end{bmatrix}$

Here, $AB = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & \lambda \end{bmatrix}\times \frac{1}{3}$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -2-9+12 & 0-2+2 & 1+3+2\lambda \\ 18-18 & 0+4-3 & 0-6-3\lambda \\ -6-18+24 & 0-4+4 & 3+6+4\lambda \end{bmatrix}\times \frac{1}{3}$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 4+2\lambda \\ 0 & 1 & -6-3\lambda \\ 0 & 0 & 9+4\lambda \end{bmatrix}\times \frac{1}{3}$

After equating: $0 = \frac{4+2\lambda}{3}$

$⇒\lambda = -2$