Let $B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}$, $\alpha > 2$ be…

Let $B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}$, $\alpha > 2$ be the adjoint of a matrix $A$ and $|A| = 2$. Then $\begin{bmatrix} \alpha & -2\alpha & \alpha \end{bmatrix} B \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}$ is equal to
  1. 0
  2. 16
  3. -16
  4. 32

Solution

We have been given that $B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}$, $\alpha > 2$ And $\text{adj}(A) = B$, $|A| = 2$ $\Rightarrow | \text{adj}(A) | = |B|$ We know that $|\text{adj} A| = |A|^{n-1}$ $\Rightarrow 2^2 = (8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(-\alpha)$ $\Rightarrow \alpha^2 - 6\alpha + 8 = 0$ $\Rightarrow (\alpha - 4)(\alpha - 2) = 0$ $\alpha = 4, 2$ but $\alpha > 2$ so $\boxed{\alpha = 4}$ Now $[\alpha, -2\alpha, \alpha]B[\alpha, -2\alpha, \alpha] = [4, -8, 4]\begin{bmatrix} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{bmatrix}[4, -8, 4]$ $= [12, 12, 8][4, -8, 4] = 48 - 96 + 32 = -16$ Hence this is the correct option.

Asked in: JEE Main 2023 (13 Apr Shift 1)

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