The minimum eigen value of the following matrix is \[\left[ {\begin{array}{*{20}{c}} 3&5&2 \\ 5&{12}&7 \\ 2&7&5 \end{array}} \right]\] A. 0 B. 1 C. 2 D. 3

0
1
2
3

The minimum eigenvalue of the matrix $A$ is $\lambda_1 = -2.5$.

To find the eigenvalues of a matrix, we can use the following formula:

$$\lambda_i = \frac{1}{|A – \lambda_i I|}$$

where $I$ is the identity matrix and $|A|$ is the determinant of $A$.

In this case, we have:

$$|A – \lambda_i I| = \begin{vmatrix} 3 – \lambda_i & 5 & 2 \ 5 & 12 – \lambda_i & 7 \ 2 & 7 & 5 – \lambda_i \end{vmatrix}$$

Expanding the determinant, we get:

$$|A – \lambda_i I| = -\lambda_i^3 + 30 \lambda_i^2 – 257 \lambda_i + 1260$$

We can then use the quadratic formula to solve for the eigenvalues:

$$\lambda_i = \frac{-30 \pm \sqrt{30^2 – 4 \cdot (-1) \cdot 1260}}{2 \cdot (-1)}$$

$$\lambda_i = \frac{-30 \pm \sqrt{12960}}{-2}$$

$$\lambda_i = \frac{-30 \pm 113}{-2}$$

$$\lambda_i = -2.5 \pm 56.5$$

Therefore, the minimum eigenvalue of $A$ is $\lambda_1 = -2.5$.

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