Question:
If \(A=\begin{bmatrix}-4&-1\\3&1\end{bmatrix}\), then the determinant of the matrix \((A^{2016}-2A^{2015}-A^{2014})\) is :
If \(A=\begin{bmatrix}-4&-1\\3&1\end{bmatrix}\), then the determinant of the matrix \((A^{2016}-2A^{2015}-A^{2014})\) is :
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Use Cayley-Hamilton to get \(A^2=-3A+I\), factor out \(A^{2014}\) and use multiplicativity of the determinant.
Updated On: Nov 11, 2026
- \(2014\)
- \(-175\)
- \(2016\)
- \(-25\)
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The Correct Option is D
Solution and Explanation
Trace \(=-3\), \(\det A=-4+3=-1\). Cayley-Hamilton: \(A^2-(\mathrm{tr}A)A+(\det A)I=O\Rightarrow A^2+3A-I=O\), i.e. \(A^2=-3A+I\).
\(A^{2016}-2A^{2015}-A^{2014}=A^{2014}(A^2-2A-I)=A^{2014}(-3A+I-2A-I)=A^{2014}(-5A)\).
\(\det=(\det A)^{2014}\cdot\det(-5A)=(-1)^{2014}\cdot(-5)^2\det A=25\times(-1)=-25\).
Answer: (4) \(-25\)
\(A^{2016}-2A^{2015}-A^{2014}=A^{2014}(A^2-2A-I)=A^{2014}(-3A+I-2A-I)=A^{2014}(-5A)\).
\(\det=(\det A)^{2014}\cdot\det(-5A)=(-1)^{2014}\cdot(-5)^2\det A=25\times(-1)=-25\).
Answer: (4) \(-25\)
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