GATE 2025 IN Q.32: solution and answer

GATE 2025 · INQ.32 · 1 mark · NAT

If one of the eigenvectors of the matrix $A = [- 1 x - 1 - 4]$ is along the direction of $[α 2 α]$ , where $α$ is any non-zero real number, then the value of $x$ is ______ (in integer).

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Answer and solution

ANSWER

$x = 2$ . The eigenvector $[12]$ belongs to the eigenvalue $λ = - 3$ .

Official answer: 2Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Any non-zero multiple of an eigenvector is also an eigenvector, so take $α = 1$ : $v = [12]$ .
Apply the definition of an eigenvector, $A v = λ v$ : $[- 1 x - 1 - 4] [12] = [- 1 - 2 x - 8] = [- 3 x - 8] = λ [12]$
The first row has no unknown in it, so it fixes the eigenvalue: $- 3 = λ \cdot 1$ , so $λ = - 3$ .
Second row: $x - 8 = 2 λ = - 6$ , so $x = 2$ .
Check with the characteristic equation $det (A - λ I) = 0$ for $A = [- 1 2 - 1 - 4]$ : $(- 1 - λ) (- 4 - λ) + 2 = λ^{2} + 5 λ + 6 = 0$ , so $λ = - 2$ or $λ = - 3$ . For $λ = - 3$ , $A + 3 I = [22 - 1 - 1]$ , which gives $2 v_{1} - v_{2} = 0$ , i.e. $v = [12]$ . The answer checks out.

KEY IDEA

If you are given an eigenvector but not its eigenvalue, write out $A v = λ v$ one row at a time. A row with no unknown in it gives you $λ$ straight away, and the other row then gives the unknown entry. The scale $α$ cancels, so set $α = 1$ .

WATCH OUT FOR

Assuming a value for $λ$ , or trying to solve the characteristic equation with $x$ still unknown, when $A v = λ v$ gives it directly. Another slip is multiplying $v$ into the columns rather than the rows, which gives $x + 2 (- 4)$ in the wrong place or a sign error in $- 1 - 2 = - 3$ . Also do not keep $α$ as a variable: it cancels from both sides, and it cannot fix $x$ .

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GATE 2025 IN Q.32: solution and answer

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