If A = [[ pmatrix - 4 & - 1 3 & 1 pmatrix ]], then the determinant of the matrix (A2016 - …

Question

If    A = $$\left[ {\begin{pmatrix} { - 4} & { - 1}  \
   3 & 1  \ \end{pmatrix} } \right]$$,

then the determinant of the matrix (A2016 - 2A2015 - A2014) is :

Accepted Answer

Given,

$$A = \left[ {\begin{pmatrix} { - 4} & { - 1}  \
   3 & 1  \

\end{pmatrix} } \right]$$

$${A^2} = \left[ {\begin{pmatrix} { - 4} & { - 1}  \
   3 & 1  \

\end{pmatrix} } \right]\left[ {\begin{pmatrix} { - 4} & { - 1}  \
   3 & 1  \

\end{pmatrix} } \right]$$

$$ = \left[ {\begin{pmatrix} {13} & 3  \
   { - 9} & { - 2}  \

\end{pmatrix} } \right]$$

A2 - 2A - I

$$ = \left[ {\begin{pmatrix} {13} & 3  \
   { - 9} & { - 2}  \

\end{pmatrix} } \right] - \left[ {\begin{pmatrix} { - 8} & { - 2}  \
   6 & 2  \

\end{pmatrix} } \right] - \left[ {\begin{pmatrix} 1 & 0  \
   0 & 1  \

\end{pmatrix} } \right]$$

$$ = \left[ {\begin{pmatrix} {20} & 5  \
   { - 15} & { - 5}  \

\end{pmatrix} } \right]$$

$$\left| A \right| = \left| {\begin{pmatrix} { - 4} & { - 1}  \
   3 & 1  \

\end{pmatrix} } \right|$$ $=$ - 4 + 3 $=$ - 1

Now,

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

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

$$ = {\left( { - 1} \right)^{2014}}\left| {\begin{pmatrix} {20} & 5  \
   { - 15} & { - 5}  \

\end{pmatrix} } \right|$$

$=$ 1 $$ \times $$ (- 100 + 75)

$=$ - 25

Solutions · 1