Let A be a non-singular matrix of order 3. If det ( 3 ? adj( 2 ? adj(( det A ) A ) ) ) = 3 - 13 · 2 - 10 and det ( 3 ? adj ( 2 A ) ) = 2 m · 3 n , then | 3 m + 2 n | is equal to ______. [2024]

Question

Let $A$ be a non-singular matrix of order 3. If $\det (3 adj(2 adj((\det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $\det (3 adj (2 A)) = 2^{m} \cdot 3^{n},$ then $| 3 m + 2 n |$ is equal to ______. [2024]

Smart Achievers · Accepted Answer

(14)

We have, $\det (3 adj(2 adj((\det A) A))) = 3^{- 13} \cdot 2^{- 10}$ $= 3^{- 13} \cdot 2^{- 10}$

Let $| A |$ $A = B$

$\Rightarrow | B | = | A | | A | = | A |^{3} | A | = | A |^{4}$ ...(i)

$\Rightarrow a d j (| A | A) = (adj B)$

$\Rightarrow 2 a d j (| A | A) = (2 adj B) = C (say)$

Now, $| 3 adj (C) |$ $= 3^{3} | C |^{2}$ ...(ii)

$∴$ $| C |$ $=$ $| (2 adj B) |$ $= 2^{3} | B |^{2} = 2^{3} {(| A |^{4})}^{2}$ (using (i))

$= 2^{3} \cdot | A |^{8}$ ...(iii)

From (ii) and (iii), we get

$| 3 adj C |$ $= 3^{3} \cdot {(2^{3} | A |^{8})}^{2} = 3^{3} 2^{6} | A |^{16} = 3^{- 13} 2^{- 10}$

$\Rightarrow | A |^{16} = {(3 \cdot 2)}^{- 16} = {(\frac{1}{6})}^{16} \Rightarrow | A | = \pm \frac{1}{6}$

So, $| 3 adj 2 A |$ $= 3^{3} | 2 A |^{2} = 3^{3} \cdot {(2^{3} | A |)}^{2} = 3^{3} \cdot 2^{6} | A |^{2}$

$3^{3} \cdot 2^{6} \cdot \frac{1}{36} = 2^{m} \cdot 3^{n}$ $(∵ Given)$

$\Rightarrow 2^{m} \cdot 3^{n} = 48$

$\Rightarrow 2^{m} \cdot 3^{n} = 2^{4} \cdot 3^{1} \Rightarrow m = 4$ and $n = 1$

So, $3 m + 2 n = 12 + 2 = 14$

Solution

Q.
Let $A$ be a non-singular matrix of order 3. If $\det (3 adj(2 adj((\det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $\det (3 adj (2 A)) = 2^{m} \cdot 3^{n},$ then $| 3 m + 2 n |$ is equal to ______. [2024]

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Solution

Q. Let A be a non-singular matrix of order 3. If det(3 adj(2 adj((detA)A)))=3-13·2-10 and det(3 adj(2A))=2m·3n, then |3m+2n| is equal to ______. [2024]

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Q.
Let $A$ be a non-singular matrix of order 3. If $\det (3 adj(2 adj((\det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $\det (3 adj (2 A)) = 2^{m} \cdot 3^{n},$ then $| 3 m + 2 n |$ is equal to ______. [2024]