A 4 × 4 matrix [P] is given below [P] = {bmatrix} 0 & 1 & 3 & 0 \\ -2 & 3 & 0 & 4 \\ 0 & 0 & 6 & 1 \\ 0 & 0 & 1 & 6 {bmatrix} The eigenvalues of [P] are

Correct Answer is : 1,2,5,71, 2, 5, 71,2,5,7

Correct Answer is : 1,2,5,71, 2, 5, 71,2,5,7

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GATE Civil Engineering (CE) 2020 Shift 2 Solved Paper

Question : 39 of 65

Marks: +1, -0

4 \times 4

[P] = \begin{bmatrix} 0 & 1 & 3 & 0 \\ -2 & 3 & 0 & 4 \\ 0 & 0 & 6 & 1 \\ 0 & 0 & 1 & 6 \end{bmatrix}

[P]

are

Solution:

|P| = -(-2) \begin{vmatrix} 1 & 3 & 0 \\ 0 & 6 & 1 \\ 0 & 1 & 6 \end{vmatrix}

= 2 \begin{vmatrix} 1 & 3 & 0 \\ 0 & 6 & 1 \\ 0 & 1 & 6 \end{vmatrix}

|P| = 2 \left\{ 1(36-1) - 3(0-0) - 0(0-0) \right\}

|P| = 70

As, the determinant of the given matrix is non-zero

Hence, the matrix is singular.

The product of eigen values

= |A|

\lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4} = 70

sum of eigen values

=

trace of matrix

P

trace of matrix

P = 0 + 3 + 6 + 6

=15

\lambda_{1} + \lambda_{2} + \lambda_{3} + \lambda_{4} = 50

Checking the above options, only (B) option is correct.

\lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4}, = 1, 2, 5, 7

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