GATE 2022 IN Q.49: solution and answer

GATE 2022 · INQ.49 · 2 marks · MSQ

The matrix $A = [49 3 - 2]$ has eigenvalues $- 5$ and $7$ . The eigenvector(s) is/are ________

(A)
$[11]$
(B)
$[34]$
(C)
$[2 - 6]$
(D)
$[28]$

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Answer and solution

ANSWER

(A) and (C). $[11]$ is an eigenvector for $λ = 7$ , and $[2 - 6]$ is an eigenvector for $λ = - 5$ .

Official answer: A, CSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Check the given eigenvalues against the trace and the determinant: $tr (A) = 4 + (- 2) = 2 = 7 + (- 5)$ and $det A = (4) (- 2) - (3) (9) = - 35 = (7) (- 5)$ . Both agree.
For $λ = 7$ , solve $(A - 7 I) x = 0$ : $[- 3 9 3 - 9] [x_{1} x_{2}] = 0 \Rightarrow x_{1} = x_{2}$ . Every eigenvector for $λ = 7$ is a nonzero multiple of $[11]$ , which is option (A).
For $λ = - 5$ , solve $(A + 5 I) x = 0$ : $[9933] [x_{1} x_{2}] = 0 \Rightarrow 9 x_{1} + 3 x_{2} = 0 \Rightarrow x_{2} = - 3 x_{1}$ . Every eigenvector for $λ = - 5$ is a nonzero multiple of $[1 - 3]$ . Since $[2 - 6] = 2 [1 - 3]$ , option (C) is one of them.
Check (A) and (C) directly: $A [11] = [4 + 3 9 - 2] = [77] = 7 [11]$ and $A [2 - 6] = [8 - 18 18 + 12] = [- 10 30] = - 5 [2 - 6]$ .
Rule out (B) and (D): $A [34] = [12 + 12 27 - 8] = [2419]$ and $A [28] = [8 + 24 18 - 16] = [322]$ . Neither result is a scalar multiple of the vector it came from, so neither is an eigenvector. (Equivalently, neither satisfies $x_{2} = x_{1}$ or $x_{2} = - 3 x_{1}$ .)
The correct options are (A), for $λ = 7$ , and (C), for $λ = - 5$ .

KEY IDEA

A nonzero vector $x$ is an eigenvector of $A$ exactly when $A x$ is a scalar multiple of $x$ . For a $2 \times 2$ matrix with known eigenvalues, one row of $(A - λ I) x = 0$ gives the direction for each eigenvalue: here $x_{2} = x_{1}$ for $λ = 7$ and $x_{2} = - 3 x_{1}$ for $λ = - 5$ . Any nonzero scalar multiple of an eigenvector is also an eigenvector, so $[2 - 6]$ counts even though it is not the simplest form $[1 - 3]$ .

WATCH OUT FOR

Rejecting (C) because it is not written in its simplest form $[1 - 3]$ . Scaling an eigenvector by a nonzero number gives another eigenvector. The other common slip is a sign error in $(A + 5 I)$ , which gives $x_{2} = 3 x_{1}$ instead of $x_{2} = - 3 x_{1}$ and can steer you towards a vector like (B).

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GATE 2022 IN Q.49: solution and answer

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