For $A_{3\times 3}$ matrix we, can write the characteristics equation
$$\mid A - \lambda I\mid = 0$$
$$\textbf{(OR)}$$
$$\lambda^{3}-\bigg(\Sigma \big(L\big)\bigg)\lambda^{2} + \bigg(\Sigma \big(PC\big)\bigg) \lambda - \mid A \mid = 0$$
Where $\Sigma \big(L\big) =\text{Sum of leading diagonal elements (or) trace}$
and $\Sigma \big(PC\big) = \text{Sum of the leading diagonal cofactors}$
$\implies \lambda^{3} - 6\lambda^{2} + 9\lambda - 4 = 0\rightarrow(1)$
To find the roots$:$
Put the value $\lambda = 1,$ it satisfy the equation.So, $\lambda = 1$ is the one of the root of the equation.
$\lambda^{2}(\lambda - 1) - 5\lambda (\lambda - 1) + 4 (\lambda - 1) = 0$
$\implies (\lambda-1) (\lambda^{2} - 5\lambda + 4) = 0$
$\implies (\lambda-1) (\lambda^{2} - 4\lambda - \lambda + 4) = 0$
$\implies (\lambda-1) \big[\lambda(\lambda - 4) - 1 (\lambda - 4)\big] = 0$
$\implies (\lambda-1) \big[(\lambda - 4) (\lambda - 1)\big] = 0$
$\implies (\lambda-1)(\lambda-1)(\lambda-4) = 0$
$\implies \lambda = 1,1,4$
So, the correct answer is $(A).$
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Important properties of Eigen values:
- Sum of all eigen values$=$Sum of leading diagonal(principle diagonal) elements$=$Trace of the matrix.
- Product of all Eigen values$=Det(A)= \mid A \mid$
- Any square diagonal(lower triangular or upper triangular) matrix eigen values are leading diagonal (principle diagonal)elements itself.
Example:$A=\begin{bmatrix} 1& 0& 0\\ 0&1 &0 \\ 0& 0& 1\end{bmatrix}$
Diagonal matrix
Eigenvalues are $1,1,1$
$B=\begin{bmatrix} 1& 9& 6\\ 0&1 &12 \\ 0& 0& 1\end{bmatrix}$
Upper triangular matrix
Eigenvalues are $1,1,1$
$C=\begin{bmatrix} 1& 0& 0\\ 8&1 &0 \\ 2& 3& 1\end{bmatrix}$
Lower triangular matrix
Eigenvalues are $1,1,1$