Let A=pmatrix2&2+p&2+p+q4&6+2p&8+3p+2q6&12+3p&20+6p+3qpmatrix.

Question

Let $A=\begin{pmatrix}2&2+p&2+p+q\4&6+2p&8+3p+2q\6&12+3p&20+6p+3q\end{pmatrix}$. If $\det(\operatorname{adj}(\operatorname{adj}(3A)))=2^m\cdot3^n$, $m,n\in\mathbb{N}$, then $m+n$ is equal to:

Accepted Answer

$|A|=\left|\begin{array}{ccc} 2 & 2+p & 2+p+q \ 4 & 6+2 p & 8+3 p+2 q \ 6 & 12+3 p & 20+6 p+3 q \end{array}\right|$

$$\begin{aligned}
&\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2-\mathrm{C}_1 \times \frac{\mathrm{q}}{2}\
&\text { Then } \mathrm{C}_3 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right)\
&\Rightarrow|\mathrm{A}|=\left|\[\begin{array}{ccc} 2 & 0 & 0 \ 4 & 2 & 2+\mathrm{p} \ 6 & 6 & 8+3 \mathrm{p} \end{array}\end{aligned}$$\right|
\]

$$\begin{aligned}
& \Rightarrow|\mathrm{A}|=2(16+6 \mathrm{p}-12-6 \mathrm{p})=8=2^3 \
& |\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A}))|=|3 \mathrm{~A}|^{(3-1)^2}=|3 \mathrm{~A}|^4 \
& =\left(3^3|\mathrm{~A}|\right)^4=\left(3^3 \times 2^3\right)^4=2^{12} \times 3^{12} \
& \Rightarrow \mathrm{~m}+\mathrm{n}=24
\end{aligned}$$

Solutions · 1