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Question: \(A = \left( {\begin{array}{*{20}{c}} { - 4}&{ - 1} \\\ 3&1 \end{array}} \right)\), then...

A = \left( {\begin{array}{*{20}{c}} { - 4}&{ - 1} \\\ 3&1 \end{array}} \right), then the determinant of (A2016−2A2015−A2014)({A^{2016}} - 2{A^{2015}} - {A^{2014}}) is:
A) −175 B) 2014 C) 2016 D) −25  A) \, - 175 \\\ B) \,2014 \\\ C) \,2016 \\\ D) \, - 25 \\\

Explanation

Solution

Take A2014{A^{2014}} common in first step. Then you will get a matrix of power two, find the matrix. At last find its determinant and get your answer.

Complete step-by-step answer:
Let us assume that
B=(A2016−2A2015−A2014)B = ({A^{2016}} - 2{A^{2015}} - {A^{2014}})
Taking A2014{A^{2014}} in common
B=A2014(A2−2A−I)B = {A^{2014}}({A^2} - 2A - I)
Now we need to find (A2−2A−I)({A^2} - 2A - I)
(A2−2A−I)({A^2} - 2A - I) = \left( {\begin{array}{*{20}{c}} { - 4}&{ - 1} \\\ 3&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} { - 4}&{ - 1} \\\ 3&1 \end{array}} \right) - 2\left( {\begin{array}{*{20}{c}} { - 4}&{ - 1} \\\ 3&1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right)
({A^2} - 2A - I) = \left( {\begin{array}{*{20}{c}} {( - 4)( - 4) + ( - 1)(3)}&{( - 4)( - 1) + ( - 1)(1)} \\\ {( - 1)(3) + (3)(1)}&{(3)( - 1) + (1)(1)} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 8&2 \\\ { - 6}&{ - 2} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right) \\\ ({A^2} - 2A - I) = \left( {\begin{array}{*{20}{c}} {13}&3 \\\ { - 9}&{ - 2} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 8&2 \\\ { - 6}&{ - 2} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right) \\\ ({A^2} - 2A - I) = \left( {\begin{array}{*{20}{c}} {20}&5 \\\ { - 15}&{ - 5} \end{array}} \right) \\\
So B = {A^{2014}}\left( {\begin{array}{*{20}{c}} {20}&5 \\\ { - 15}&{ - 5} \end{array}} \right)
Taking determinant on both sides
\left| B \right| = \left| {{A^{2014}}\left( {\begin{array}{*{20}{c}} {20}&5 \\\ { - 15}&{ - 5} \end{array}} \right)} \right|
Since ∣A.B∣=∣A∣∣B∣\left| {A.B} \right| = \left| A \right|\left| B \right|
Therefore
\left| B \right| = \left| {{A^{2014}}} \right|\left| {\left( {\begin{array}{*{20}{c}} {20}&5 \\\ { - 15}&{ - 5} \end{array}} \right)} \right|
Using property ∣Bx∣=∣B∣x\left| {{B^x}} \right| = {\left| B \right|^x}
\left| B \right| = {\left| A \right|^{2014}}\left| {\left( {\begin{array}{*{20}{c}} {20}&5 \\\ { - 15}&{ - 5} \end{array}} \right)} \right| \\\ \left| B \right| = {\left| {\left( {\begin{array}{*{20}{c}} { - 4}&{ - 1} \\\ 3&1 \end{array}} \right)} \right|^{2014}}\left| {\left( {\begin{array}{*{20}{c}} {20}&5 \\\ { - 15}&{ - 5} \end{array}} \right)} \right| \\\
((−4)(1)−(−1)(3)2014) (25)({(-4)(1)-(-1)(3)}^{2014})\,(25)
∣B∣=(−1)2014(−25)=−25\left| B \right| = {( - 1)^{2014}}( - 25) = - 25

So, the correct answer is “Option D”.

Note: In order to solve such a question you must know all the properties of matrices and determinants. We used the properties ∣Ax∣=∣A∣x and ∣A.B∣=∣A∣∣B∣\left| {{A^x}} \right| = {\left| A \right|^x}{\text{ and }}\left| {A.B} \right| = \left| A \right|\left| B \right| in the question. In the next we may need AA−1=IA{A^{ - 1}} = I. Hence, remembering all these properties is quite important also, we need to be careful while multiplying matrices as it could lead to mistakes.