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

2014

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- 175

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2016

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- 25

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Solution

The correct option is D $- 25$
$| A^{2016} - 2 A^{2015} - A^{2014} |$
$= | A |^{2014} | A^{2} - 2 A - I |$
$Now, | A | = - 4 + 3 = - 1$
$and A^{2} - 2 A - I = [\begin{matrix} 20 & 5 - 15 & - 5 \end{matrix}]$
$\Rightarrow | A^{2} - 2 A - I | = - 25$
$∴ | A^{2016} - 2 A^{2015} - A^{2014} |$
$= (- 1)^{2014} \times (- 25) = - 25$

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If A=[−4−131], then the determinant of the matrix (A2016−2A2015−A2014) is

The correct option is D −25 |A2016−2A2015−A2014| =|A|2014|A2−2A−I| Now, |A|=−4+3=−1 and A2−2A−I=[205−15−5] ⇒|A2−2A−I|=−25 ∴|A2016−2A2015−A2014| =(−1)2014×(−25)=−25

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