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.
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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