Let $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$. If $|adj(adj(adj^2(A)))| =…

Let $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$. If $|adj(adj(adj^2(A)))| = 16^n$, then $n$ is equal to.
  1. 8
  2. 10
  3. 9
  4. 12

Solution

Given that $A=\begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$ and $|\text{adj}(\text{adj}(\text{adj}^2 A))|=16^n$. We know that $|\text{adj} A|=|A|^{n-1}$ and $|\text{adj}(\text{adj} A)|=|A|^{(n-1)^2}$. Similarly, $|\text{adj}(\text{adj}(\text{adj} A))|=|A|^{(n-1)^3}$ where $n$ is the order of the square matrix. Let us find $|A|$. $|A|=2(2 \times 2 - (-1)(-1)) - 1(1 \times 2 - 0(-1)) + 0(1 \times (-1) - 0 \times 2)$ $=2(3) - 2 + 0 = 4$ $\Rightarrow (2A)^{(n-1)^3}=(2A)^{(3-1)^3}$ $=(2A)^8=(2^8)^3|A|^8$ $\Rightarrow 2^{24} \times 4^8 = 16^n$ $\Rightarrow 16^6 \times 16^4 = 16^n$ $\Rightarrow 16^{10} = 16^n$ $\Rightarrow n = 10$ Therefore, the value of $n$ is $10$.

Asked in: JEE Main 2023 (08 Apr Shift 1)

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