GATE CS 2016 Set 1 Q.15 (CS Q.5): Linear Algebra

Two eigenvalues of a $3 \times 3$ real matrix $P$ are $(2 + \sqrt{-1})$ and $3$ . The determinant of $P$ is _____.

Your answer

Official answer and solution

Official answer: 15
From: Official GATE 2016 answer key

Solution

Use the property that complex eigenvalues of real matrices occur in conjugate pairs and that the determinant equals the product of eigenvalues.

Let $P$ be a $3 \times 3$ real matrix. The characteristic polynomial of $P$ has real coefficients. Consequently, any non-real complex eigenvalues must occur in conjugate pairs.

We are given that $\lambda_1 = 2 + \sqrt{-1} = 2 + i$ is an eigenvalue. Its complex conjugate $\lambda_2 = 2 - i$ must also be an eigenvalue.

The third given eigenvalue is real: $\lambda_3 = 3$ .

The determinant of an $n \times n$ matrix is the product of all its eigenvalues:

\det(P) = \lambda_1 \times \lambda_2 \times \lambda_3 = (2 + i)(2 - i)(3)

Using $(2 + i)(2 - i) = 2^2 - i^2 = 4 - (-1) = 5$ :

\det(P) = 5 \times 3 = 15

Final answer

Common trap

Assuming the third eigenvalue is zero or overlooking the complex conjugate eigenvalue.

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

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 1
Question: 15 of 65, printed as CS Q.5
Section: Computer Science and Information Technology

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