If $A = $\frac{1}{5!6!7!}$ \begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$, then…

If $A = $\frac{1}{5!6!7!}$ \begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$, then $| \text{adj}(\text{adj}(2A)) |$ is equal to
  1. 220
  2. 28
  3. 212
  4. 216

Solution

Given, $A = \frac{1}{5!6!7!}$ $\begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$ $\Rightarrow |A| = \frac{1}{5!6!7!}\begin{vmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{vmatrix}$ $\Rightarrow |A| = \frac{1}{5!6!7!}5!6!7!\begin{vmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{vmatrix}$ $\Rightarrow |A| = 2$ Now using the property of adjoint of matrix we get, $|adj(adj(A))| = |A|^{(n-1)^2}$ and $|kA| = k^n|A|$, where $n$ is order of square matrix, Now using the above two formula we get, $|adj(adj(2A))| = |2A|^4 = 2^{12}|A|^4$ $\Rightarrow |adj(adj(2A))| = 2^{12} \cdot 2^4$ $\Rightarrow |adj(adj(2A))| = 2^{16}$

Asked in: JEE Main 2023 (10 Apr Shift 2)

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