Let A be a square matrix of order 3 such that (A)=-2 and (3adj(-6adj(3A)))=2^m+n 3^mn,mn.

Question

Let $A$ be a square matrix of order 3 such that $\det (A)=-2$ and $\det (3adj(-6adj(3A)))={2}^{m+n}\cdot {3}^{mn},m>n$. Then $4m+2n$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 34

Accepted Answer

$\begin{gathered} AadjA=|A|I,\det (\lambda A)={\lambda }^{n}\det A \ \det (3adj(-6adj(3A)))={3}^{3}\det (adj(-6adj(3A))) \ ={3}^{3}(-6adj(3A){)}^{2} \ ={3}^{3}(-6{)}^{6}|3A{|}^{4} \ ={3}^{9}{2}^{6}\cdot {3}^{12}\cdot (-2{)}^{4} \ ={3}^{21}\cdot {2}^{10} \end{gathered}$

Now comparing with given condition

$\begin{gathered} {2}^{m+n}{3}^{mn}={2}^{10}\cdot {3}^{21} \ m+n=10,mn=21 \ \Rightarrow m=7,n=3(m>n) \ 	herefore 4m+2n=28+6=34 \end{gathered}$

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