Let $A$ be a square matrix of order 3 such that $\det(A)=-2$ and $\det(3\,\mathrm{adj}(-6\,\mathrm{adj}(3A)))=2^{m+n}\cdot3^{mn}$, $mn$. Then $4m+2n$ is equal to _______

Question

Let $A$ be a square matrix of order 3 such that $\det(A)=-2$ and $\det(3\,\mathrm{adj}(-6\,\mathrm{adj}(3A)))=2^{m+n}\cdot3^{mn}$, $m>n$. Then $4m+2n$ is equal to _______

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] For order 3: $\mathrm{adj}(kM)=k^2\,\mathrm{adj}M$, $\mathrm{adj}(\mathrm{adj}M)=(\det M)M$ and $\det(3A)=-54$. So $\mathrm{adj}(-6\,\mathrm{adj}(3A))=36\cdot(-54)\cdot3A=-5832A$, and $\det(3\cdot(-5832A))=(-17496)^3(-2)=2\cdot(2^3\cdot3^7)^3=2^{10}\cdot3^{21}$. So $m+n=10$ and $mn=21$, giving $m=7$, $n=3$ and $4m+2n=34$. Answer: 34.

Q. Let $A$ be a square matrix of order 3 such that $\det(A)=-2$ and $\det(3\,\mathrm{adj}(-6\,\mathrm{adj}(3A)))=2^{m+n}\cdot3^{mn}$, $m>n$. Then $4m+2n$ is equal to _______

Explanation

Reviewed by Dr. Abubacker

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