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. $I-P$
• C. $2I+P$
• D. $2I-P$

Answer 1

The correct option is C $2I+P$

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

Then the matrix is $P^3+2P^2$

The characteristic equation of $P$ is $|P-\lambda |=0$

$\Rightarrow \left|\begin{array}{ccc}1-\lambda & 0& 0\\ 0& -1-\lambda & 0\\ 0& 0& -2-\lambda \end{array}\right|=0$

$\Rightarrow (1-\lambda )\left\{\right(1+\lambda \left)\right(2+\lambda \left)\right\}=0$

$\Rightarrow (1-{\lambda }^2)(2+\lambda )=0$

$\Rightarrow 2-2{\lambda }^2+\lambda -{\lambda }^3=0$

$\Rightarrow {\lambda }^3+2{\lambda }^2-\lambda -2=0$

We know that, Caylay Hamilton theorem states that 'Every square matrix satisfy its characteristic equation'

$\therefore$ $P^3+2P^2-P-2I=0$

$\Rightarrow$ $P^3+2P^2=P+2I$