The minimum eigenvalue of the following matrix is

$$\begin{bmatrix} 3& 5& 2\\5 &12 &7 \\2 &7 & 5\end{bmatrix}$$

1. $0$
2. $1$
3. $2$
4. $3$

edited

+1 vote

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 $\Sigma \big(PC\big) = \text{Sum of the leading diagonal cofactors}$

Now $,\lambda^{3} – 20\lambda^{2} + 33\lambda – 0 = 0$

$\implies \lambda^{3} – 20\lambda^{2} + 33\lambda = 0$

$\implies \lambda(\lambda^{2} – 20\lambda + 33) = 0$

$\therefore \lambda_{min} = 0$

So, the correct answer is $(A).$

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Important properties of Eigen values:

1. Sum of all eigen values$=$Sum of leading diagonal(principle diagonal) elements$=$Trace of the matrix.
2. Product of all Eigen values$=Det(A)= \mid A \mid$
3. 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 $1,1,1$

by (190 points)