GATE 2020 IN Q.26: solution and answer

GATE 2020 · INQ.26 · 2 marks · MCQ

Consider the matrix $M = 110 - 1 - 2 - 1 011$ . One of the eigenvectors of $M$ is

(A)
$1 - 1 1$
(B)
$11 - 1$
(C)
$- 1 1 - 1$
(D)
$111$

See the question in the paper

Answer and solution

ANSWER

The answer is (D), $111$ . Every row of $M$ sums to zero, so $M 111 = 0$ . That makes it an eigenvector with eigenvalue $λ = 0$ .

Official answer: DSource: Official answer key, GATE 2020 IN (IIT Delhi)

Definition: a nonzero vector $v$ is an eigenvector of $M$ if $M v = λ v$ for some scalar $λ$ . Multiply $M$ into each option and check whether the result is a scalar multiple of that option.
Quick observation: every row of $M$ sums to zero ( $1 - 1 + 0 = 0$ , $1 - 2 + 1 = 0$ , $0 - 1 + 1 = 0$ ). Multiplying $M$ by $111$ adds up each row, so $M 111 = 000 = 0 \cdot 111$ . Option (D) is an eigenvector with $λ = 0$ .
Cross-check with the determinant: $det M = 1 ((- 2) (1) - (1) (- 1)) - (- 1) ((1) (1) - (1) (0)) + 0 = - 1 + 1 = 0$ . So $λ = 0$ is an eigenvalue, as expected.
Rule out (A): $M 1 - 1 1 = 1 + 1 + 0 1 + 2 + 1 0 + 1 + 1 = 242$ . The first entry needs $λ (1) = 2$ , so $λ = 2$ . The second needs $λ (- 1) = 4$ , so $λ = - 4$ . No single $λ$ works, so (A) is not an eigenvector.
Rule out (B): $M 11 - 1 = 1 - 1 + 0 1 - 2 - 1 0 - 1 - 1 = 0 - 2 - 2$ . The first entry forces $λ (1) = 0$ , so $λ = 0$ . But then the second entry would have to be $0$ , not $- 2$ . So (B) is not an eigenvector.
Rule out (C): $M - 1 1 - 1 = - 1 - 1 + 0 - 1 - 2 - 1 0 - 1 - 1 = - 2 - 4 - 2$ . The first entry needs $λ (- 1) = - 2$ , so $λ = 2$ . The second needs $λ (1) = - 4$ , so $λ = - 4$ . That is a contradiction. (This fits, because (C) is just $- 1$ times (A).)
Only $111$ satisfies $M v = λ v$ (with $λ = 0$ ). The answer is (D).

KEY IDEA

Test each option directly against $M v = λ v$ : one value of $λ$ has to work in every entry. A useful shortcut: if every row of a matrix has the same sum $s$ , then $[11 \dots 1]^{T}$ is an eigenvector with eigenvalue $s$ . Here $s = 0$ .

WATCH OUT FOR

Stopping once one entry of $M v$ matches a multiple of $v$ . For option (A) the first and third entries of $M v = (2, 4, 2)^{T}$ fit $λ = 2$ , but the middle entry does not. Every entry has to give the same $λ$ . A second slip is to reject (D) because $M v = 0$ . The zero vector is not allowed as an eigenvector, but $λ = 0$ is a perfectly valid eigenvalue, and it occurs exactly when $M$ is singular.

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GATE 2020 IN Q.26: solution and answer

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