Let $A$ be a non-singular matrix of order 3. If $$\operatorname{det}(3 \operatorname{adj}(2 \operatorname{adj}((\operatorname{det} A) A)))=3^{-13} \cdot 2^{-10}$$ and $$\operatorname{det}(3\operatorname{adj}(2 \mathrm{A}))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}$$, then $|3 \mathrm{~m}+2 \mathrm{n}|$ is equal to _________.
Answer (integer)
14
Solution
<p>$|\operatorname{adj}(2 \operatorname{adj}(|A| A))|=3^{-13} \cdot 2^{-10}$</p>
<p>Let $|A| A=B \Rightarrow|B|=\| A|A|=|A|^3|A|=|A|^4$</p>
<p>$$\begin{aligned}
\Rightarrow \quad & \operatorname{adj}(|A| A)=(\operatorname{adj} B) \\
\Rightarrow \quad & 2 \operatorname{adj}(|A| A)=(2 \operatorname{adj} B)=C \text { (say) } \\
& |\operatorname{3adj}(C)|=3^3 \cdot|C|^2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& |C|=|(2 \operatorname{adj} B)|=2^3|B|^2=2^3 \cdot\left|A^4\right|^2=2^3 \cdot|A|^8 \\
& \Rightarrow|\operatorname{3adj} C|=3^3 \cdot\left(2^3|A|^8\right)^2=3^{-13} \cdot 2^{-10} \\
& \quad=2^6|A|^{16}=3^{-16} \cdot 2^{-10} \\
& \Rightarrow|A|^{16}=(3 \cdot 2)^{-16}=\left(\frac{1}{6}\right)^{16} \\
& \Rightarrow|A|= \pm \frac{1}{6}
\end{aligned}$$</p>
<p>$$\begin{array}{r}
\mid \text { 3adj }\left.2 A\left|=3^3\right| 2 A\right|^2=3^3 \cdot\left(2^3|A|\right)^2=3^3 \cdot 2^6|A|^2 \\
=3^3 \cdot 2^6 \cdot \frac{1}{36}=\frac{27 \times 64}{36}=48
\end{array}$$</p>
<p>$$
\begin{aligned}
& \Rightarrow 2^m \cdot 3^n=2^4 \cdot 3^1 \Rightarrow m=4 \\
& \qquad n=1 \\
& \Rightarrow|3 \times 4+2 \times 1|=14 \\
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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