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If $I=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$ and $P=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2\end{array}\right)$ Then, the matrix $P^{3}+2 P^{2}$ is equal to
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Verified Answer
The correct answer is:
$2I+P$
Given, $I=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
and $P=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2\end{array}\right)$
The characteristic equation of $P$ is
$$
|P-\lambda|=0
$$
$\Rightarrow\left|\begin{array}{ccc}1-\lambda & 0 & 0 \\ 0 & -1-\lambda & 0 \\ 0 & 0 & -2-\lambda\end{array}\right|=0$
$\Rightarrow \quad(1-\lambda)\{(1+\lambda)(2+\lambda)\}=0$
$\Rightarrow \quad\left(1-\lambda^{2}\right)(2+\lambda)=0$
$\Rightarrow \quad 2-2 \lambda^{2}+\lambda-\lambda^{3}=0$
$\Rightarrow \quad \lambda^{3}+2 \lambda^{2}-\lambda-2=0$
We know that, Caylay Hamilton theorem states that Every square matrix satisfy its characteristic equation'.
$\therefore$
$$
P^{3}+2 P^{2}-P-2 I=0
$$
$\Rightarrow$\(p^{3}+2 p^{2}=2 I+p\)
and $P=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2\end{array}\right)$
The characteristic equation of $P$ is
$$
|P-\lambda|=0
$$
$\Rightarrow\left|\begin{array}{ccc}1-\lambda & 0 & 0 \\ 0 & -1-\lambda & 0 \\ 0 & 0 & -2-\lambda\end{array}\right|=0$
$\Rightarrow \quad(1-\lambda)\{(1+\lambda)(2+\lambda)\}=0$
$\Rightarrow \quad\left(1-\lambda^{2}\right)(2+\lambda)=0$
$\Rightarrow \quad 2-2 \lambda^{2}+\lambda-\lambda^{3}=0$
$\Rightarrow \quad \lambda^{3}+2 \lambda^{2}-\lambda-2=0$
We know that, Caylay Hamilton theorem states that Every square matrix satisfy its characteristic equation'.
$\therefore$
$$
P^{3}+2 P^{2}-P-2 I=0
$$
$\Rightarrow$\(p^{3}+2 p^{2}=2 I+p\)
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