Let for A=[[|gathered 1 2 3 3 1 1 1 2 gathered]]|,|A|=2.

Question

Let for $A=[[|\begin{gathered} 1 \quad  2 \quad  3 \ \alpha  \quad  3 \quad  1 \ 1 \quad  1 \quad  2 \end{gathered}]]|,|A|=2$. If $|2\, adj\, (2\, adj\, (2A))|={32}^{n}$, then $3n+\alpha $ is equal to

Accepted Answer

$\begin{gathered} A=[[|\begin{array}{ccc}1 \quad  2 \quad  3 \ \alpha  \quad  3 \quad  1 \ 1 \quad  1 \quad  2\end{array}]]| \ |A|=2 \end{gathered}$

$\begin{gathered} \Rightarrow 1(6-1)-2(2\alpha -1)+3(\alpha -3)=2 \ \ \Rightarrow 5-4\alpha +2+3\alpha -9=2 \ \ \Rightarrow -\alpha -4=0 \ \ \Rightarrow \alpha =-4 $

Now, $|2\, adj\, (2\, adj\, (2A))|={32}^{n}$

$\Rightarrow$ ${2}^{3}|adj(2adj(2A))|$ = ${32}^{n}$

$ \Rightarrow 8|Adj(2Adj(2A))|={32}^{n} \ \ \Rightarrow 8|Adj(2	imes {2}^{2}Adj(A))|={32}^{n} \ \ \Rightarrow 8|Adj({2}^{3}AdjA)|={32}^{n} \ \ \Rightarrow 8|{2}^{6}Adj(AdjA)|={32}^{n} \ \ \Rightarrow {2}^{3}{({2}^{6})}^{3}|Adj(Adj)|={32}^{n} \ \ \Rightarrow {2}^{3}\cdot {2}^{18}|A{|}^{4}={32}^{n} \ \ \Rightarrow {2}^{21}\cdot {2}^{4}={32}^{n} \ \ \Rightarrow {2}^{25}={32}^{n} \ \ \Rightarrow {({2}^{5})}^{5}={32}^{n} \ \ \Rightarrow (32{)}^{5}={32}^{n} \end{gathered}$

$\begin{gathered} 	herefore n=5 \ \Rightarrow 3n+\alpha =15-4=11 \end{gathered}$

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