Question
If \[A = \left[ \begin{array}{l}
\,\,2\,\,\,\,\,\, - 3\\
- 4\,\,\,\,\,\,\,\,1
\end{array} \right],\] then $$adj\left( {3{A^2} + 12A} \right)$$ is equal to:
A.
\[\left[ \begin{array}{l}
\,\,72\,\,\,\,\,\, - 63\\
- 84\,\,\,\,\,\,\,\,51
\end{array} \right]\]
B.
\[\left[ \begin{array}{l}
\,\,72\,\,\,\,\,\, - 84\\
- 63\,\,\,\,\,\,\,\,51
\end{array} \right]\]
C.
\[\left[ \begin{array}{l}
51\,\,\,\,\,\,\,63\\
84\,\,\,\,\,\,\,72
\end{array} \right]\]
D.
\[\left[ \begin{array}{l}
51\,\,\,\,\,\,\,84\\
63\,\,\,\,\,\,\,72
\end{array} \right]\]
Answer :
\[\left[ \begin{array}{l}
51\,\,\,\,\,\,\,63\\
84\,\,\,\,\,\,\,72
\end{array} \right]\]
Solution :
\[{\rm{We \,\,have }}\,\,A = \left[ \begin{array}{l}
\,\,2\,\,\,\,\,\,\, - 3\\
- 4\,\,\,\,\,\,\,\,\,1
\end{array} \right]\]
\[ \Rightarrow \,\,{A^2} = \left[ \begin{array}{l}
\,\,16\,\,\,\,\,\,\, - 9\\
- 12\,\,\,\,\,\,\,13
\end{array} \right]\]
\[ \Rightarrow \,\,3{A^2} = \left[ \begin{array}{l}
\,\,48\,\,\,\,\,\,\, - 27\\
- 36\,\,\,\,\,\,\,\,\,39
\end{array} \right]\]
\[{\rm{Also\,\, 12}}A = \left[ \begin{array}{l}
\,\,24\,\,\,\,\,\, - 36\\
- 48\,\,\,\,\,\,\,\,\,12
\end{array} \right]\]
\[\therefore \,\,3{A^2} + 12A = \left[ \begin{array}{l}
\,\,48\,\,\,\,\,\,\, - 27\\
- 36\,\,\,\,\,\,\,\,\,39
\end{array} \right] + \left[ \begin{array}{l}
\,\,24\,\,\,\,\,\, - 36\\
- 48\,\,\,\,\,\,\,\,\,12
\end{array} \right]\]
\[\,\,\,\,\,\, = \left[ \begin{array}{l}
\,\,72\,\,\,\,\,\, - 63\\
- 84\,\,\,\,\,\,\,\,51
\end{array} \right]\]
\[adj\left( {3{A^2} + 12A} \right) = \left[ \begin{array}{l}
51\,\,\,\,\,\,\,63\\
84\,\,\,\,\,\,\,72
\end{array} \right]\]