Question

If $I=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),$ $P=\left(\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -2\end{array}\right),$ then the matrix $P^3+2P^2$ is equal to

• A. $P$
• B. $1-P$
• C. $2I+P$
• D. $2I-P$

Answer 1

Answer: (C)

$2I+P$

Given the matrix $P=\left(\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -2\end{array}\right)$.

Now,

$P^2=\left(\begin{array}{ccc}{1}^2& 0& 0\\ 0& (-1{)}^2& 0\\ 0& 0& (-2{)}^2\end{array}\right)$ [ Since $P$ is a diagonal matrix]

$\Rightarrow P^2=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 4\end{array}\right)$......(1).

Again,

$P^3=\left(\begin{array}{ccc}{1}^3& 0& 0\\ 0& (-1{)}^3& 0\\ 0& 0& (-2{)}^3\end{array}\right)$

$\Rightarrow P^3=\left(\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -8\end{array}\right)$.....(2).

So,

$P^3+2P^2$ $=\left(\begin{array}{ccc}1+2& 0& 0\\ 0& -1+2& 0\\ 0& 0& -8+8\end{array}\right)$ $=\left(\begin{array}{ccc}3& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)$ $=2I+P$.