Let A be a matrix such that |adj (adj (adj A))| = 81. If , then is equal to [2025]

Question

Let A be a $3 \times 3$ matrix such that $| a d j (a d j (a d j A)) |$ = 81. If $S = {n \in Z : (| adj {(adj A) |)}^{\frac{{(n - 1)}^{2}}{2}} = | A |^{(3 n^{2} - 5 n - 4)}}$ , then $\sum_{n \in S}$ $| A^{(n^{2} + n)} |$ is equal to [2025]

Smart Achievers · Accepted Answer

(4)

We have, $| a d j (a d j (a d j A)) |$ = 81

$\Rightarrow | adj A |^{4} = 81 \Rightarrow | adj A | = 3$

$\Rightarrow | A |^{2} = 3 \Rightarrow | A | = \sqrt{3}$

Now, ${(| A |^{4})}^{\frac{{(n - 1)}^{2}}{2}} =$ $| A |^{(3 n^{2} - 5 n - 4)}$

$\Rightarrow 2 {(n - 1)}^{2} = 3 n^{2} - 5 n - 4$

$\Rightarrow 2 n^{2} - 4 n + 2 = 3 n^{2} - 5 n - 4$

$\Rightarrow n^{2} - n - 6 = 0 \Rightarrow (n - 3) (n + 2) = 0$

$\Rightarrow n = 3, - 2$

So, $\sum_{n \in S}$ $| A^{(n^{2} + n)} |$ $=$ $| A^{2} |$ $+$ $| A^{12} |$ $= 3 + 729 = 732$ .

Solution

Q.
Let A be a $3 \times 3$ matrix such that $| a d j (a d j (a d j A)) |$ = 81. If $S = {n \in Z : (| adj {(adj A) |)}^{\frac{{(n - 1)}^{2}}{2}} = | A |^{(3 n^{2} - 5 n - 4)}}$ , then $\sum_{n \in S}$ $| A^{(n^{2} + n)} |$ is equal to [2025]

1 820

2 750

3 866

4 732

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