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Engineering Mathematics → Linear Algebra → Eigenvalues & Eigenvectors
Last updated 5 September 2026
Question The smallest eigenvalue and the corresponding eigenvector of the matrix
[ 2 − 2 − 1 6 ] \begin{bmatrix} 2 & -2 \\ -1 & 6 \end{bmatrix} [ 2 − 1 − 2 6 ] , respectively, are
Correct answer (A) 1.55 and Bmatrix 2.00 \ 0.45 Bmatrix
Solution Let
A = [ 2 − 2 − 1 6 ] A = \begin{bmatrix} 2 & -2 \\ -1 & 6 \end{bmatrix} A = [ 2 − 1 − 2 6 ] . The characteristic equation is
∣ A − λ I ∣ = 0 |A - \lambda I| = 0 ∣ A − λ I ∣ = 0 .
∣ 2 − λ − 2 − 1 6 − λ ∣ = 0 \begin{vmatrix} 2-\lambda & -2 \\ -1 & 6-\lambda \end{vmatrix} = 0 2 − λ − 1 − 2 6 − λ = 0 ( 2 − λ ) ( 6 − λ ) − ( − 2 ) ( − 1 ) = 0 (2-\lambda)(6-\lambda) - (-2)(-1) = 0 ( 2 − λ ) ( 6 − λ ) − ( − 2 ) ( − 1 ) = 0 12 − 8 λ + λ 2 − 2 = 0 12 - 8\lambda + \lambda^2 - 2 = 0 12 − 8 λ + λ 2 − 2 = 0 λ 2 − 8 λ + 10 = 0 \lambda^2 - 8\lambda + 10 = 0 λ 2 − 8 λ + 10 = 0 λ = 8 ± 64 − 40 2 = 8 ± 24 2 = 4 ± 6 ≈ 1.55 , 6.45 \lambda = \frac{8 \pm \sqrt{64 - 40}}{2} = \frac{8 \pm \sqrt{24}}{2} = 4 \pm \sqrt{6} \approx 1.55, 6.45 λ = 2 8 ± 64 − 40 = 2 8 ± 24 = 4 ± 6 ≈ 1.55 , 6.45 The smallest eigenvalue is
λ 1 ≈ 1.55 \lambda_1 \approx 1.55 λ 1 ≈ 1.55 .
To find the eigenvector
X = { x 1 x 2 } X = \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} X = { x 1 x 2 } for
λ 1 = 1.55 \lambda_1 = 1.55 λ 1 = 1.55 :
( A − 1.55 I ) X = 0 (A - 1.55I)X = 0 ( A − 1.55 I ) X = 0 [ 2 − 1.55 − 2 − 1 6 − 1.55 ] { x 1 x 2 } = { 0 0 } \begin{bmatrix} 2-1.55 & -2 \\ -1 & 6-1.55 \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} [ 2 − 1.55 − 1 − 2 6 − 1.55 ] { x 1 x 2 } = { 0 0 } [ 0.45 − 2 − 1 4.45 ] { x 1 x 2 } = { 0 0 } \begin{bmatrix} 0.45 & -2 \\ -1 & 4.45 \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} [ 0.45 − 1 − 2 4.45 ] { x 1 x 2 } = { 0 0 } From the first row:
0.45 x 1 − 2 x 2 = 0 ⇒ x 1 x 2 = 2 0.45 ≈ 4.44 0.45x_1 - 2x_2 = 0 \Rightarrow \frac{x_1}{x_2} = \frac{2}{0.45} \approx 4.44 0.45 x 1 − 2 x 2 = 0 ⇒ x 2 x 1 = 0.45 2 ≈ 4.44 .
Checking option A:
2.00 0.45 ≈ 4.44 \frac{2.00}{0.45} \approx 4.44 0.45 2.00 ≈ 4.44 . This matches.
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Correct answer (A) 1.55 and Bmatrix 2.00 \ 0.45 Bmatrix
Solution Let
A = [ 2 − 2 − 1 6 ] A = \begin{bmatrix} 2 & -2 \\ -1 & 6 \end{bmatrix} A = [ 2 − 1 − 2 6 ] . The characteristic equation is
∣ A − λ I ∣ = 0 |A - \lambda I| = 0 ∣ A − λ I ∣ = 0 .
∣ 2 − λ − 2 − 1 6 − λ ∣ = 0 \begin{vmatrix} 2-\lambda & -2 \\ -1 & 6-\lambda \end{vmatrix} = 0 2 − λ − 1 − 2 6 − λ = 0 ( 2 − λ ) ( 6 − λ ) − ( − 2 ) ( − 1 ) = 0 (2-\lambda)(6-\lambda) - (-2)(-1) = 0 ( 2 − λ ) ( 6 − λ ) − ( − 2 ) ( − 1 ) = 0 12 − 8 λ + λ 2 − 2 = 0 12 - 8\lambda + \lambda^2 - 2 = 0 12 − 8 λ + λ 2 − 2 = 0 λ 2 − 8 λ + 10 = 0 \lambda^2 - 8\lambda + 10 = 0 λ 2 − 8 λ + 10 = 0 λ = 8 ± 64 − 40 2 = 8 ± 24 2 = 4 ± 6 ≈ 1.55 , 6.45 \lambda = \frac{8 \pm \sqrt{64 - 40}}{2} = \frac{8 \pm \sqrt{24}}{2} = 4 \pm \sqrt{6} \approx 1.55, 6.45 λ = 2 8 ± 64 − 40 = 2 8 ± 24 = 4 ± 6 ≈ 1.55 , 6.45 The smallest eigenvalue is
λ 1 ≈ 1.55 \lambda_1 \approx 1.55 λ 1 ≈ 1.55 .
To find the eigenvector
X = { x 1 x 2 } X = \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} X = { x 1 x 2 } for
λ 1 = 1.55 \lambda_1 = 1.55 λ 1 = 1.55 :
( A − 1.55 I ) X = 0 (A - 1.55I)X = 0 ( A − 1.55 I ) X = 0 [ 2 − 1.55 − 2 − 1 6 − 1.55 ] { x 1 x 2 } = { 0 0 } \begin{bmatrix} 2-1.55 & -2 \\ -1 & 6-1.55 \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} [ 2 − 1.55 − 1 − 2 6 − 1.55 ] { x 1 x 2 } = { 0 0 } [ 0.45 − 2 − 1 4.45 ] { x 1 x 2 } = { 0 0 } \begin{bmatrix} 0.45 & -2 \\ -1 & 4.45 \end{bmatrix} \begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix} [ 0.45 − 1 − 2 4.45 ] { x 1 x 2 } = { 0 0 } From the first row:
0.45 x 1 − 2 x 2 = 0 ⇒ x 1 x 2 = 2 0.45 ≈ 4.44 0.45x_1 - 2x_2 = 0 \Rightarrow \frac{x_1}{x_2} = \frac{2}{0.45} \approx 4.44 0.45 x 1 − 2 x 2 = 0 ⇒ x 2 x 1 = 0.45 2 ≈ 4.44 .
Checking option A:
2.00 0.45 ≈ 4.44 \frac{2.00}{0.45} \approx 4.44 0.45 2.00 ≈ 4.44 . This matches.
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