MHT CET · Maths · Matrices
If \(A=\left[\begin{array}{cc}2 & 3 \\ -4 & 1\end{array}\right]\), then adj \(\left(3 A^2+12 A\right)\) is equal to
- A \(\left[\begin{array}{cc}-21 & 63 \\ 84 & 0\end{array}\right]\)
- B \(\left[\begin{array}{cc}21 & 63 \\ 84 & 0\end{array}\right]\)
- C \(\left[\begin{array}{cc}21 & -63 \\ 84 & 0\end{array}\right]\)
- D \(\left[\begin{array}{cc}-21 & -63 \\ 84 & 0\end{array}\right]\)
Answer & Solution
Correct Answer
(D) \(\left[\begin{array}{cc}-21 & -63 \\ 84 & 0\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & 3 A^2+12 A=3\left[\begin{array}{cc}2 & 3 \\ -4 & 1\end{array}\right]^2+12 \\ & =3\left[\begin{array}{cc}-8 & 9 \\ -12 & -11\end{array}\right]+12\left[\begin{array}{cc}2 & 3 \\ -4 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}-24+24 & 27+36 \\ -36-48 & -33+12\end{array}\right] \\ & =\left[\begin{array}{cc}0 & 63 \\ -84 & -21\end{array}\right] \\ & \Rightarrow \operatorname{adj}\left(3 A^2+12 A\right)=\left[\begin{array}{cc}-21 & -63 \\ 84 & 0\end{array}\right]\end{aligned}\)
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