Let A=[[|gathered 2 2+p 2+p+q 4 6+2p 8+3p+2q 6 12+3p 20+6p+3q gathered]]|.

Question

Let $A=[[|\begin{gathered} 2 \quad  2+p \quad  2+p+q \ 4 \quad  6+2p \quad  8+3p+2q \ 6 \quad  12+3p \quad  20+6p+3q \end{gathered}]]|$.

If $det(((3A)))={2}^{m}\cdot {3}^{n}$, $m,n\in N$, then $m+n$ is equal to

Accepted Answer

$|A|=\begin{vmatrix} 2 \quad  2+p \quad  2+p+q \ 4 \quad  6+2p \quad  8+3p+2q \ 6 \quad  12+3p \quad  20+6p+3q \end{vmatrix}$

$\begin{gathered} {C}_{3}	o {C}_{3}-{C}_{2}-{C}_{1}	imes \frac{q}{2} \ {C}_{3}	o {C}_{2}-{C}_{1}X(1+\frac{p}{2}) \ \Rightarrow |A|=|\begin{array}{ccc}2 \quad  0 \quad  0 \ 4 \quad  2 \quad  2+p \ 6 \quad  6 \quad  8+3p\end{array}| \end{gathered}$

$\begin{gathered} \Rightarrow |A|=2(16+6p-12-6p)=8={2}^{3} \ |adj(adj(3A))|=|3A{|}^{(3-1{)}^{2}}=|3A{|}^{4} \ ={({3}^{3}|A|)}^{4}={({3}^{3}	imes {2}^{3})}^{4}={2}^{12}	imes {3}^{12} \ \Rightarrow m+n=24 \end{gathered}$

Solutions · 1