Let the determinant of a square matrix A of order m be m - n , where m and n satisfy 4 m + n = 22 and 17 m + 4 n = 93 . If det(n?adj(adj(mA))) = 3 a ? 5 b ? 6 c , then a + b + c is equal to [2023]

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Solution

Q.
Let the determinant of a square matrix $A$ of order $m$ be $m - n$ , where $m$ and $n$ satisfy $4 m + n = 22 and 17 m + 4 n = 93 .$ If $det(n adj(adj(mA))) = 3^{a} 5^{b} 6^{c},$ then $a + b + c$ is equal to [2023]

1 109

2 101

3 84

4 96

Ans.
(4)

Given $4 m + n = 22$

$17 m + 4 n = 93$

Solving the above two equations, we get $m = 5$ and $n = 2$ .

$∴$ $A is a square matrix of order 5 and$ $| A |$ $= 5 - 2 = 3$

Now, we know that $adj (k A) = k^{n - 1} adj (A),$ where $A$ is a matrix of order $n$

$∴$ $adj (m A) = adj (5 A) = 5^{5 - 1} (adj A) = 5^{4} (adj A)$

Again, $adj (5^{4} adj A) = {(5^{4})}^{4} adj A (adj A) = 5^{16} | A |^{5 - 2} \cdot A = 5^{16} 3^{3} A$

Now, $\det (n adj (adj m A)) = \det (2 \cdot 5^{16} \cdot 3^{3} \cdot A)$

$= (2 \cdot 5^{16} \cdot 3^{5}) \det A = 2^{5} \cdot 5^{80} \cdot 3^{15} \cdot 3 = 2^{5} \cdot 5^{80} \cdot 3^{16} = 6^{5} \cdot 5^{80} \cdot 3^{11}$

$∴$ $a = 11, b = 80, c = 5$

$\Rightarrow a + b + c = 80 + 11 + 5 = 96$

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