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

Answer Verified Verified
Hint:Take ${A^{2014}}$ 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
$B = ({A^{2016}} - 2{A^{2015}} - {A^{2014}})$
Taking ${A^{2014}}$ in common
$B = {A^{2014}}({A^2} - 2A - I)$
Now we need to find $({A^2} - 2A - I)$
$({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 $\left| {A.B} \right| = \left| A \right|\left| B \right|$
Therefore
$\left| B \right| = \left| {{A^{2014}}} \right|\left| {\left( {\begin{array}{*{20}{c}}
  {20}&5 \\
  { - 15}&{ - 5}
\end{array}} \right)} \right|$
Using property $\left| {{B^x}} \right| = {\left| B \right|^x}$
$
  \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| \\
$
$({(-4)(1)-(-1)(3)}^{2014})\,(25)$
 $\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 $\left| {{A^x}} \right| = {\left| A \right|^x}{\text{ and }}\left| {A.B} \right| = \left| A \right|\left| B \right|$ in the question. In the next we may need $A{A^{ - 1}} = I$. Hence, remembering all these properties is quite important also, we need to be careful while multiplying matrices as it could lead to mistakes.