Let and A be a matrix of order such that , where is the identity matrix of order . If det((a + 1) adj((a – 1A)) is , then m + n is equal to : [2025]

Question

Let $a \in R$ and A be a matrix of order $3 \times 3$ such that $\det (A) = - 4 and A + I = [\begin{matrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{matrix}]$ , where $I$ is the identity matrix of order $3 \times 3$ . If det((a + 1) adj((a – 1A)) is $2^{m} 3^{n},$ $m, n \in {0, 1, 2, . . ., 20}$ , then m + n is equal to : [2025]

Smart Achievers · Accepted Answer

(1)

We have, $A = [\begin{matrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{matrix}] - I = [\begin{matrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{matrix}]$

Also, $3 \times 3$ $= - 4 \Rightarrow [\begin{matrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{matrix}] = - 4 \Rightarrow - 2 (a - 1) = - 4 \Rightarrow a = 3$

Now, det((a + 1) adj((a – 1)A)) = |4 adj(2A)|

$= 4^{3}$ $| a d j (2 A) |$ $= 4^{3}$ $| 2 A |^{3 - 1} = 4^{3} | 2 A |^{2} = 4^{3} {(2)}^{6} | A |^{2}$

$= 4^{3} \times 2^{6} \times {(- 4)}^{2} = 4^{5} \times 2^{6} = {(2^{2})}^{5} \times 2^{6} = 2^{16} = 2^{m} \times 3^{n}$

$\Rightarrow$ m = 16 and n = 0

$∴$ m + n = 16 + 0 = 16.

Solution

1 16

2 17

3 15

4 14

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