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

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { As } A \operatorname{adj} A=|A| l, \operatorname{det}(\lambda A)=\lambda^n \operatorname{det} A \ & \operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=3^3 \operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3 A))) \ & =3^3(-6 \operatorname{adj}(3 A))^2 \ & =3^3(-6)^6|3 A|^4 \ & =3^9 2^6 \cdot 3^{12} \cdot(-2)^4 \ & =3^{21} \cdot 2^{10} \end{aligned}$ Now comparing with given condition $\begin{aligned} & 2^{m+n} 3^{m n}=2^{10} \cdot 3^{21} \ & m+n=10, m n=21 \ & \Rightarrow \quad m=7, n=3(m>n) \ & 	herefore \quad 4 m+2 n=28+6=34 \end{aligned}$

Let $A$ be a square matrix of order 3 such that $\operatorname{det}(A)=-2$ and $\operatorname{det}(3…

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