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Linear Algebra

GATE CSE 2015 Set 3 · Q15

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In the given matrix [1−12010121]\begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}, one of the eigenvalues is 1. The eigenvectors corresponding to the eigenvalue 1 are

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Official answer: B) { (-4,2,1) | ≠ 0, ∈ R}

Linear Algebra

Given the matrix:

A=[1−12010121]A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}

For eigenvalue λ=1\lambda = 1, the corresponding eigenvectors satisfy (A−I)X=0(A - I)X = 0:

[1−1−1201−10121−1][xyz]=[000]\begin{bmatrix} 1-1 & -1 & 2 \\ 0 & 1-1 & 0 \\ 1 & 2 & 1-1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
[0−12000120][xyz]=[000]\begin{bmatrix} 0 & -1 & 2 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

From row 1:

−y+2z=0  ⟹  y=2z-y + 2z = 0 \implies y = 2z

From row 3:

x+2y=0  ⟹  x=−2y=−2(2z)=−4zx + 2y = 0 \implies x = -2y = -2(2z) = -4z

Letting z=αz = \alpha (where α≠0,α∈R\alpha \neq 0, \alpha \in \mathbb{R}):

X=[xyz]=[−4α2αα]=α[−421]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -4\alpha \\ 2\alpha \\ \alpha \end{bmatrix} = \alpha \begin{bmatrix} -4 \\ 2 \\ 1 \end{bmatrix}

Thus, the set of eigenvectors is {α(−4,2,1)∣α≠0,α∈R}\{\alpha(-4, 2, 1) \mid \alpha \neq 0, \alpha \in \mathbb{R}\}.

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This Linear Algebra previous year question is Q15 of the GATE CSE 2015 Set 3 paper. Find more GATE CSE PYQs on Linear Algebra in Engineering Mathematics, or browse all GATE CSE PYQs.

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Linear Algebra in GATE CSE

Linear Algebra has 46 GATE CSE PYQs across 26 papers from 2011 to 2026. 13 of them are 1-mark MCQs like this one, and GATE CSE 2015 Set 3 has 2.

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GateAI solution cardGATE CSE 2015 Set 3 · Q15 · MCQ

In the given matrix 1 & -1 & 2 0 & 1 & 0 1 & 2 & 1 , one of the eigenvalues is 1. The eigenvectors corresponding to the eigenvalue 1 are

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