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

Question

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

Accepted Answer

$$
\begin{aligned}
& A=\left[\[\begin{array}{lll} 1 & 2 & 3 \ \alpha & 3 & 1 \ 1 & 1 & 2 \end{array}\end{aligned}$$\right] \\
\& |A|=2

\]

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

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

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

$$
\begin{aligned}
\Rightarrow & 8|\operatorname{Adj}(2 \operatorname{Adj}(2 \mathrm{~A}))| = {32^n}\\
\Rightarrow &  8\left|\operatorname{Adj}\left(2 \times 2^2 \operatorname{Adj}(\mathrm{A})\right)\right| = {32^n} \\
\Rightarrow & 8\left|\operatorname{Adj}\left(2^3 \operatorname{AdjA}\right)\right| = {32^n} \\
\Rightarrow & 8\left|2^6 \operatorname{Adj}(\operatorname{AdjA})\right| = {32^n} \\
\Rightarrow & 2^3\left(2^6\right)^3|\operatorname{Adj}(\operatorname{Adj})| = {32^n} \\
\Rightarrow & 2^3 \cdot 2^{18}|\mathrm{~A}|^4 = {32^n} \\
\Rightarrow& 2^{21} \cdot 2^4 = {32^n}\\
\Rightarrow& 2^{25}  = {32^n}\\
\Rightarrow& \left(2^5\right)^5  = {32^n}\\
\Rightarrow& (32)^5  = {32^n}
\end{aligned}
$$

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

Solutions · 1