If $A = left[ {begin{array}{*{20}{c}} { - 4}&{ - 1} 3&1 end{array}} right] , then

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3 and 4 .Determinants and Matrices

hard

If $A = \left[ {\begin{array}{*{20}{c}}
{ - 4}&{ - 1}\\
3&1
\end{array}} \right]$ , then the determinant of the matrix $\left( {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right)$ is

$-175$

$2014$

$2016$

$-25$

(JEE MAIN-2016)

Solution

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

$ \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$

Standard 12

Mathematics

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normal

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easy

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easy

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easy

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Solution

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If $A = \left[ {\begin{array}{*{20}{c}}
{ - 4}&{ - 1}\\
3&1
\end{array}} \right]$ , then the determinant of the matrix $\left( {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right)$ is

If $A = \left[ {\begin{array}{{20}{c}}{1/3}&2\\0&{2x – 3}\end{array}} \right],B = \left[ {\begin{array}{{20}{c}}3&6\\0&{ – 1}\end{array}} \right]$and $AB = I$, then $x =$