Let A=[[|gathered 1 2 2 3 0 0 4 5 gathered]]| and B=[[|gathered 1 0 0 0 -5 0 0 4 -2 gather…

Question

Let $A=[[|\begin{gathered} \alpha  \quad  1 \quad  2 \ 2 \quad  3 \quad  0 \ 0 \quad  4 \quad  5 \end{gathered}]]|$ and $B=[[|\begin{gathered} 1 \quad  0 \quad  0 \ 0 \quad  -5\alpha  \quad  0 \ 0 \quad  4\alpha  \quad  -2\alpha \end{gathered}]]|+adj(A)$. If $\det (B)=66$, then $\det (adj(A))$ equals :

Accepted Answer

$A=[[|\begin{gathered} \alpha  \quad  1 \quad  2 \ 2 \quad  3 \quad  0 \ 0 \quad  4 \quad  5 \end{gathered}]]|$

$\Rightarrow |A|=15\alpha -10+16=15\alpha +6$

Now adj $A=[[|\begin{gathered} 15 \quad  3 \quad  -6 \ -10 \quad  5\alpha  \quad  4 \ 8 \quad  -4\alpha  \quad  3\alpha -2 \end{gathered}]]|$

$	herefore B=[[|\begin{gathered} 1 \quad  0 \quad  0 \ 0 \quad  -5\alpha  \quad  0 \ 0 \quad  4\alpha  \quad  -2\alpha \end{gathered}]]|+adj(A)$

$B=[[|\begin{gathered} 16 \quad  3 \quad  -6 \ -10 \quad  0 \quad  4 \ 8 \quad  0 \quad  \alpha -2 \end{gathered}]]|$

$\begin{gathered} |B|=6(5\alpha +6)=66 \ \Rightarrow \alpha =1 \ \Rightarrow |A|=21 \end{gathered}$

$A\cdot Adj(A)=|A|I\Rightarrow |A\cdot Adj(A)|=\parallel A|I|=(|A{|}^{3})|I|=|A{|}^{3}$

$\Rightarrow$ $|A||Adj(A)|=|A{|}^{3}\Rightarrow |Adj(A)|=|A{|}^{2}=(21{)}^{2}=441$

Solutions · 1