JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q65

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If $A=\begin{bmatrix}-4&-1\\3&1\end{bmatrix}$ , then the determinant of the matrix $\left(A^{2016}-2A^{2015}-A^{2014}\right)$ is :

2014

−175

2016

−25

Answer and worked solution

Official answer: D

$A^{2016}-2A^{2015}-A^{2014}=A^{2014}(A^2-2A-I)$ , with $|A|=-1$ . $A^2=\begin{bmatrix}13&3\\-9&-2\end{bmatrix}$ , so $A^2-2A-I=\begin{bmatrix}20&5\\-15&-5\end{bmatrix}$ with determinant −25. The result is $(-1)^{2014}(-25)=-25$ .

Topics:Determinants and properties

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 65