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Let $A=\left(\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3\end{array}\right) .$ Then, the roots of the equation $\operatorname{det}\left(A-\lambda I_{3}\right)=0$ (where $I_{3}$ is the identity matrix of order 3 ) are
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Verified Answer
The correct answers are:
0,3,6
We have,
$$
A=\left[\begin{array}{lll}
3 & 0 & 3 \\
0 & 3 & 0 \\
3 & 0 & 3
\end{array}\right]
$$
Again, we have
$$
\left|\begin{array}{ccc}
3-\lambda & 0 & 3 \\
0 & 3-\lambda & 0 \\
3 & 0 & 3-\lambda
\end{array}\right|=0
$$
$\Rightarrow(3-\lambda)\left[(3-\lambda)^{2}-0\right]-0+3[0-3(3-\lambda)]=0$
$\Rightarrow (3-\lambda)^{3}-9(3-\lambda)=0$
$\Rightarrow (3-\lambda)\left[(3-\lambda)^{2}-9\right]=0$
$\Rightarrow (3-\lambda)\left(9+\lambda^{2}-6 \lambda-9\right)=0$
$\Rightarrow (3-\lambda)\left(\lambda^{2}-6 \lambda\right)=0$
$\Rightarrow \quad(3-\lambda) \lambda(\lambda-6)=0$
${\lambda}=3,0,6$
$$
A=\left[\begin{array}{lll}
3 & 0 & 3 \\
0 & 3 & 0 \\
3 & 0 & 3
\end{array}\right]
$$
Again, we have
$$
\left|\begin{array}{ccc}
3-\lambda & 0 & 3 \\
0 & 3-\lambda & 0 \\
3 & 0 & 3-\lambda
\end{array}\right|=0
$$
$\Rightarrow(3-\lambda)\left[(3-\lambda)^{2}-0\right]-0+3[0-3(3-\lambda)]=0$
$\Rightarrow (3-\lambda)^{3}-9(3-\lambda)=0$
$\Rightarrow (3-\lambda)\left[(3-\lambda)^{2}-9\right]=0$
$\Rightarrow (3-\lambda)\left(9+\lambda^{2}-6 \lambda-9\right)=0$
$\Rightarrow (3-\lambda)\left(\lambda^{2}-6 \lambda\right)=0$
$\Rightarrow \quad(3-\lambda) \lambda(\lambda-6)=0$
${\lambda}=3,0,6$
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