Let A be a non-singular matrix of order 3.

Question

Let $A$ be a non-singular matrix of order 3. If $\det (3adj(2adj((\det A)A)))={3}^{-13}\cdot {2}^{-10}$ and $\det (3adj(2A))={2}^{m}\cdot {3}^{n}$, then $|3m+2n|$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 14

Accepted Answer

$|adj(2adj(|A|A))|={3}^{-13}\cdot {2}^{-10}$

Let $|A|A=B\Rightarrow |B|=\parallel A|A|=|A{|}^{3}|A|=|A{|}^{4}$

$\begin{gathered} \Rightarrow \, adj(|A|A)=(adjB) \ \Rightarrow \, 2adj(|A|A)=(2adjB)=C \ |3adj(C)|={3}^{3}\cdot |C{|}^{2} \end{gathered}$

$\begin{gathered} |C|=|(2adjB)|={2}^{3}|B{|}^{2}={2}^{3}\cdot {|{A}^{4}|}^{2}={2}^{3}\cdot |A{|}^{8} \ \Rightarrow |3adjC|={3}^{3}\cdot {({2}^{3}|A{|}^{8})}^{2}={3}^{-13}\cdot {2}^{-10} \ \, ={2}^{6}|A{|}^{16}={3}^{-16}\cdot {2}^{-10} \ \Rightarrow |A{|}^{16}=(3\cdot 2{)}^{-16}={(\frac{1}{6})}^{16} \ \Rightarrow |A|=\pm \frac{1}{6} \end{gathered}$

$\begin{gathered} |{2A|={3}^{3}|2A|}^{2}={3}^{3}\cdot {({2}^{3}|A|)}^{2}={3}^{3}\cdot {2}^{6}|A{|}^{2} \ ={3}^{3}\cdot {2}^{6}\cdot \frac{1}{36}=\frac{27	imes 64}{36}=48 \end{gathered}$

$\begin{gathered} \Rightarrow {2}^{m}\cdot {3}^{n}={2}^{4}\cdot {3}^{1}\Rightarrow m=4 \ \, n=1 \ \Rightarrow |3	imes 4+2	imes 1|=14 \end{gathered}$

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