Let |A|=6, where A is a 3 3 matrix.

Question

Let $|A|=6$, where A is a $3	imes 3$ matrix. If $|adj(3adj({A}^{2}\cdot adj(2A)))|={2}^{m}\cdot {3}^{n},m,n\in N$, then $m+n$ is equal to

$____$ .

Answer: 62

Accepted Answer

Let

$X={A}^{2}\cdot adj(2A)$

so the given expression is

$|adj(3adj(X))|.$

Since $A$ is a $3	imes 3$ matrix and $|A|=6$:

Find $|{A}^{2}|$: $|{A}^{2}|=|A{|}^{2}={6}^{2}=36$

Find $|adj(2A)|$: First,

$|2A|={2}^{3}|A|=8\cdot 6=48$

For a $3	imes 3$ matrix $M$,

$|adj(M)|=|M{|}^{3-1}=|M{|}^{2}$

So,

$|adj(2A)|=|2A{|}^{2}={48}^{2}$

Find $|X|$: $|X|=|{A}^{2}|\cdot |adj(2A)|=36\cdot {48}^{2}$

Now,

$36={2}^{2}\cdot {3}^{2},\, 48={2}^{4}\cdot 3$

So,

${48}^{2}={2}^{8}\cdot {3}^{2}$

Hence,

$|X|=({2}^{2}\cdot {3}^{2})({2}^{8}\cdot {3}^{2})={2}^{10}\cdot {3}^{4}$

Find $|3adj(X)|$: First,

$|adj(X)|=|X{|}^{2}=({2}^{10}\cdot {3}^{4}{)}^{2}={2}^{20}\cdot {3}^{8}$

Since $3adj(X)$ is a $3	imes 3$ matrix,

$|3adj(X)|={3}^{3}\cdot |adj(X)|$

Therefore,

$|3adj(X)|=27\cdot {2}^{20}\cdot {3}^{8}={2}^{20}\cdot {3}^{11}$

Find the required determinant: Again using

$|adj(M)|=|M{|}^{2}$

we get

$|adj(3adj(X))|={({2}^{20}\cdot {3}^{11})}^{2}={2}^{40}\cdot {3}^{22}$

So,

$m=40,\, n=22$

and

$m+n=40+22=62$

$62$

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