Question
If \[P = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,3\\
1\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,3\\
2\,\,\,\,\,\,4\,\,\,\,\,\,\,4
\end{array} \right]\] is the adjoint of a $$3 \times 3$$ matrix $$A$$ and $$\left| A \right| = 4,$$ then $$\alpha $$ is equal to:
A.
4
B.
11
C.
5
D.
0
Answer :
11
Solution :
$$\eqalign{
& \left| P \right| = 1\left( {12 - 12} \right) - \alpha \left( {4 - 6} \right) + 3\left( {4 - 6} \right) = 2\alpha - 6 \cr
& {\text{Now, }}adj\,A = P \cr
& \Rightarrow \,\,\left| {adj\,A} \right| = \left| P \right| \cr
& \Rightarrow \,\,{\left| A \right|^2} = \left| P \right| \cr
& \Rightarrow \,\,\left| P \right| = 16 \cr
& \Rightarrow \,\,2\alpha - 6 = 16 \cr
& \Rightarrow \,\,\alpha = 11 \cr} $$