Question

If $P=\left(\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& -3\end{array}\right)$ then $P^5$ equals to

• A. $P$
• B. $2P$
• C. $-P$
• D. $-2P$

Answer 1

Answer: (A)

Given, $P=\left(\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& -3\end{array}\right)$
$P^2=P\cdot P=\left(\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& -3\end{array}\right)\left(\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& -3\end{array}\right)$
$=\left(\begin{array}{ccc}4+2-4& -4-6+8& -8-8+12\\ -2-3+4& 2+9-8& 4+12-12\\ 2+2-3& -2-6+6& -4-8+9\end{array}\right)$
$=\left(\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& -3\end{array}\right)=P$
$\therefore P^4=P^2=P$
$\Rightarrow P^5=P^2=P$