Let $A$ be a square matrix of order 3 such that $\operatorname{det}(A)=-2$ and $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=2^{m+n} \cdot 3^{m n}, m>n$. Then $4 m+2 n$ is equal to __________.
Answer (integer)
34
Solution
<p>$\begin{aligned} & \text { As } A \operatorname{adj} A=|A| I, \operatorname{det}(\lambda A)=\lambda^n \operatorname{det} A \\\\ & \operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=3^3 \operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3 A))) \\\\ & =3^3(-6 \operatorname{adj}(3 A))^2 \\\\ & =3^3(-6)^6|3 A|^4 \\\\ & =3^9 2^6 \cdot 3^{12} \cdot(-2)^4 \\\\ & =3^{21} \cdot 2^{10}\end{aligned}$</p>
<p>Now comparing with given condition</p>
<p>$$ \begin{aligned} & 2^{m+n} 3^{m n}=2^{10} \cdot 3^{21} \\\\ & m+n=10, m n=21 \\\\ & \Rightarrow m=7, n=3(m>n) \\\\ & \therefore 4 m+2 n=28+6=34 \end{aligned} $$</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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