Question:

The matrix \(A^2+4A-5I\), where \(I\) is identity matrix and \(A=\begin{bmatrix}1&2\\4&-3\end{bmatrix}\), equals :

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Compute A^2 directly, or use the Cayley-Hamilton theorem for A.
Updated On: Oct 31, 2026
  • \(4\begin{bmatrix}2&1\\2&0\end{bmatrix}\)
  • \(4\begin{bmatrix}0&-1\\2&2\end{bmatrix}\)
  • \(32\begin{bmatrix}2&1\\2&0\end{bmatrix}\)
  • \(32\begin{bmatrix}1&1\\1&0\end{bmatrix}\)
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The Correct Option is A

Solution and Explanation

\(A^2=\begin{bmatrix}1&2\\4&-3\end{bmatrix}\begin{bmatrix}1&2\\4&-3\end{bmatrix}=\begin{bmatrix}1+8&2-6\\4-12&8+9\end{bmatrix}=\begin{bmatrix}9&-4\\-8&17\end{bmatrix}\).
\(4A=\begin{bmatrix}4&8\\16&-12\end{bmatrix}\), \(5I=\begin{bmatrix}5&0\\0&5\end{bmatrix}\).
\(A^2+4A-5I=\begin{bmatrix}9+4-5&-4+8\\-8+16&17-12-5\end{bmatrix}=\begin{bmatrix}8&4\\8&0\end{bmatrix}=4\begin{bmatrix}2&1\\2&0\end{bmatrix}\).
Answer: (1) \(4\begin{bmatrix}2&1\\2&0\end{bmatrix}\)
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