Question:

If \(A=\begin{bmatrix}2&-3\\-4&1\end{bmatrix}\), then adj\((3A^2+12A)\) is equal to

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Compute \(3A^2+12A\) first, then use \(\mathrm{adj}\begin{bmatrix}p&q\\r&s\end{bmatrix}=\begin{bmatrix}s&-q\\-r&p\end{bmatrix}\).
Updated On: Nov 11, 2026
  • \(\begin{bmatrix}51&63\\84&72\end{bmatrix}\)
  • \(\begin{bmatrix}51&84\\63&72\end{bmatrix}\)
  • \(\begin{bmatrix}72&-63\\-84&51\end{bmatrix}\)
  • \(\begin{bmatrix}72&-84\\-63&51\end{bmatrix}\)
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The Correct Option is A

Solution and Explanation

\(A^2=\begin{bmatrix}2&-3\\-4&1\end{bmatrix}\begin{bmatrix}2&-3\\-4&1\end{bmatrix}=\begin{bmatrix}16&-9\\-12&13\end{bmatrix}\).
\(3A^2=\begin{bmatrix}48&-27\\-36&39\end{bmatrix}\), \(12A=\begin{bmatrix}24&-36\\-48&12\end{bmatrix}\).
\(3A^2+12A=\begin{bmatrix}72&-63\\-84&51\end{bmatrix}\).
For a 2x2 matrix, swap the diagonal entries and change the sign of the off-diagonal entries: adj \(=\begin{bmatrix}51&63\\84&72\end{bmatrix}\).
Answer: (1) \(\begin{bmatrix}51&63\\84&72\end{bmatrix}\)
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