Question

If $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 2015& 2016& -1\end{array}\right],$ then $(2A^2+4A^4+6A^6)$ is equal to
$($Here, $I_3$ is unit matrix of order $3)$

• A. $3I_3$
• B. $12I_3$
• C. $12A$
• D. $6A^6$

Answer 1

The correct option is B $12I_3$

$A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 2015& 2016& -1\end{array}\right]A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 2015& 2016& -1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 2015& 2016& -1\end{array}\right]\Rightarrow A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$

$\Rightarrow A^4=A^6=I$
$\Rightarrow 2A^2+4A^4+6A^6=12I_3$