STD 12 - 3 and 4 . determinant and metrices — JEE कक्षा 11 साइन्स — Question

$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
{ - 4}&{ - 1}\\
3&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 4}&{ - 1}\\
3&1
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
{13}&3\\
{ - 9}&{ - 2}
\end{array}} \right]$ and $\,\left| A \right| = 1$.

Now, ${A^{2016}} - 2{A^{2015}} - {A^{2014}}$

$ = {A^{2014}}\left( {{A^2} - 2A - I} \right)$

$\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right| = \left| {{A^{2014}}} \right|\left| {{A^2} - 2A - I} \right|$

$ = {\left| A \right|^{2014}}\left| {\begin{array}{*{20}{c}}
{20}&5\\
{ - 15}&{ - 5}
\end{array}} \right| =  - 25$