GATE CS 2017 Set 2 Q.52 (CS Q.52): Linear Algebra

If the characteristic polynomial of a $3 \times 3$ matrix $M$ over $\mathbb{R}$ (the set of real numbers) is $\lambda^3 - 4\lambda^2 + a\lambda + 30$ , $a \in \mathbb{R}$ , and one eigenvalue of $M$ is 2, then the largest among the absolute values of the eigenvalues of $M$ is

Your answer

Official answer and solution

Official answer: 5
From: Official GATE 2017 answer key

Solution

Substitute the known eigenvalue $\lambda = 2$ into the characteristic polynomial to determine $a$ , factor the polynomial to find all three eigenvalues, and identify the maximum absolute value.

Given characteristic polynomial $P(\lambda) = \lambda^3 - 4\lambda^2 + a\lambda + 30 = 0$ .
Since $\lambda = 2$ is an eigenvalue, it must satisfy $P(2) = 0$ :

2^3 - 4(2^2) + a(2) + 30 = 0

8 - 16 + 2a + 30 = 0

22 + 2a = 0 \implies a = -11

Substitute $a = -11$ into the polynomial:

P(\lambda) = \lambda^3 - 4\lambda^2 - 11\lambda + 30 = 0

Divide by $(\lambda - 2)$ :

\lambda^3 - 4\lambda^2 - 11\lambda + 30 = (\lambda - 2)(\lambda^2 - 2\lambda - 15) = 0

Factor the quadratic factor:

\lambda^2 - 2\lambda - 15 = (\lambda - 5)(\lambda + 3) = 0

Thus, the three eigenvalues of $M$ are:

\lambda_1 = 2, \quad \lambda_2 = 5, \quad \lambda_3 = -3

The absolute values of the eigenvalues are:

|\lambda_1| = 2, \quad |\lambda_2| = 5, \quad |\lambda_3| = |-3| = 3

The largest among the absolute values is 5.

Final answer

Common trap

Forgetting that absolute value makes $|-3| = 3$ , though 5 is still the maximum.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 52 of 65, printed as CS Q.52
Section: Computer Science and Information Technology

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