PYQs / GATE CE / 2015 / Set 2 / Q38 GATE CE 2015 Set 2 — Question 38 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Eigenvalues & Eigenvectors Linear Algebra Engineering Mathematics
Engineering Mathematics → Linear Algebra → Eigenvalues & Eigenvectors
Last updated 5 September 2026
Question The two Eigen values of the matrix
[ 2 1 1 p ] \begin{bmatrix} 2 & 1 \\ 1 & p \end{bmatrix} [ 2 1 1 p ] have a ratio of 3:1 for
p = 2 p = 2 p = 2 . What is another value of
p p p for which the Eigen values have the same ratio of 3:1?
Solution Let the eigenvalues be
λ 1 \lambda_1 λ 1 and
λ 2 \lambda_2 λ 2 . For a
2 × 2 2 \times 2 2 × 2 matrix, the sum of eigenvalues equals the trace and the product equals the determinant.
Trace:
λ 1 + λ 2 = 2 + p \lambda_1 + \lambda_2 = 2 + p λ 1 + λ 2 = 2 + p Determinant:
λ 1 λ 2 = 2 p − 1 \lambda_1 \lambda_2 = 2p - 1 λ 1 λ 2 = 2 p − 1 Given the ratio
λ 1 : λ 2 = 3 : 1 \lambda_1 : \lambda_2 = 3 : 1 λ 1 : λ 2 = 3 : 1 , let
λ 1 = 3 λ 2 \lambda_1 = 3\lambda_2 λ 1 = 3 λ 2 .
From the sum:
3 λ 2 + λ 2 = 2 + p ⟹ 4 λ 2 = 2 + p ⟹ λ 2 = 2 + p 4 3\lambda_2 + \lambda_2 = 2 + p \implies 4\lambda_2 = 2 + p \implies \lambda_2 = \frac{2+p}{4} 3 λ 2 + λ 2 = 2 + p ⟹ 4 λ 2 = 2 + p ⟹ λ 2 = 4 2 + p From the product:
( 3 λ 2 ) ( λ 2 ) = 2 p − 1 ⟹ 3 λ 2 2 = 2 p − 1 (3\lambda_2)(\lambda_2) = 2p - 1 \implies 3\lambda_2^2 = 2p - 1 ( 3 λ 2 ) ( λ 2 ) = 2 p − 1 ⟹ 3 λ 2 2 = 2 p − 1 Substitute
λ 2 \lambda_2 λ 2 :
3 ( 2 + p 4 ) 2 = 2 p − 1 3 \left( \frac{2+p}{4} \right)^2 = 2p - 1 3 ( 4 2 + p ) 2 = 2 p − 1 3 ( 4 + 4 p + p 2 16 ) = 2 p − 1 3 \left( \frac{4 + 4p + p^2}{16} \right) = 2p - 1 3 ( 16 4 + 4 p + p 2 ) = 2 p − 1 3 p 2 + 12 p + 12 = 32 p − 16 3p^2 + 12p + 12 = 32p - 16 3 p 2 + 12 p + 12 = 32 p − 16 3 p 2 − 20 p + 28 = 0 3p^2 - 20p + 28 = 0 3 p 2 − 20 p + 28 = 0 Solving the quadratic equation:
p = 20 ± ( − 20 ) 2 − 4 ( 3 ) ( 28 ) 2 ( 3 ) = 20 ± 400 − 336 6 = 20 ± 64 6 = 20 ± 8 6 p = \frac{20 \pm \sqrt{(-20)^2 - 4(3)(28)}}{2(3)} = \frac{20 \pm \sqrt{400 - 336}}{6} = \frac{20 \pm \sqrt{64}}{6} = \frac{20 \pm 8}{6} p = 2 ( 3 ) 20 ± ( − 20 ) 2 − 4 ( 3 ) ( 28 ) = 6 20 ± 400 − 336 = 6 20 ± 64 = 6 20 ± 8 p 1 = 28 6 = 14 3 p_1 = \frac{28}{6} = \frac{14}{3} p 1 = 6 28 = 3 14 and
p 2 = 12 6 = 2 p_2 = \frac{12}{6} = 2 p 2 = 6 12 = 2 .
The other value of
p p p is
14 / 3 14/3 14/3 .
Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Solution Let the eigenvalues be
λ 1 \lambda_1 λ 1 and
λ 2 \lambda_2 λ 2 . For a
2 × 2 2 \times 2 2 × 2 matrix, the sum of eigenvalues equals the trace and the product equals the determinant.
Trace:
λ 1 + λ 2 = 2 + p \lambda_1 + \lambda_2 = 2 + p λ 1 + λ 2 = 2 + p Determinant:
λ 1 λ 2 = 2 p − 1 \lambda_1 \lambda_2 = 2p - 1 λ 1 λ 2 = 2 p − 1 Given the ratio
λ 1 : λ 2 = 3 : 1 \lambda_1 : \lambda_2 = 3 : 1 λ 1 : λ 2 = 3 : 1 , let
λ 1 = 3 λ 2 \lambda_1 = 3\lambda_2 λ 1 = 3 λ 2 .
From the sum:
3 λ 2 + λ 2 = 2 + p ⟹ 4 λ 2 = 2 + p ⟹ λ 2 = 2 + p 4 3\lambda_2 + \lambda_2 = 2 + p \implies 4\lambda_2 = 2 + p \implies \lambda_2 = \frac{2+p}{4} 3 λ 2 + λ 2 = 2 + p ⟹ 4 λ 2 = 2 + p ⟹ λ 2 = 4 2 + p From the product:
( 3 λ 2 ) ( λ 2 ) = 2 p − 1 ⟹ 3 λ 2 2 = 2 p − 1 (3\lambda_2)(\lambda_2) = 2p - 1 \implies 3\lambda_2^2 = 2p - 1 ( 3 λ 2 ) ( λ 2 ) = 2 p − 1 ⟹ 3 λ 2 2 = 2 p − 1 Substitute
λ 2 \lambda_2 λ 2 :
3 ( 2 + p 4 ) 2 = 2 p − 1 3 \left( \frac{2+p}{4} \right)^2 = 2p - 1 3 ( 4 2 + p ) 2 = 2 p − 1 3 ( 4 + 4 p + p 2 16 ) = 2 p − 1 3 \left( \frac{4 + 4p + p^2}{16} \right) = 2p - 1 3 ( 16 4 + 4 p + p 2 ) = 2 p − 1 3 p 2 + 12 p + 12 = 32 p − 16 3p^2 + 12p + 12 = 32p - 16 3 p 2 + 12 p + 12 = 32 p − 16 3 p 2 − 20 p + 28 = 0 3p^2 - 20p + 28 = 0 3 p 2 − 20 p + 28 = 0 Solving the quadratic equation:
p = 20 ± ( − 20 ) 2 − 4 ( 3 ) ( 28 ) 2 ( 3 ) = 20 ± 400 − 336 6 = 20 ± 64 6 = 20 ± 8 6 p = \frac{20 \pm \sqrt{(-20)^2 - 4(3)(28)}}{2(3)} = \frac{20 \pm \sqrt{400 - 336}}{6} = \frac{20 \pm \sqrt{64}}{6} = \frac{20 \pm 8}{6} p = 2 ( 3 ) 20 ± ( − 20 ) 2 − 4 ( 3 ) ( 28 ) = 6 20 ± 400 − 336 = 6 20 ± 64 = 6 20 ± 8 p 1 = 28 6 = 14 3 p_1 = \frac{28}{6} = \frac{14}{3} p 1 = 6 28 = 3 14 and
p 2 = 12 6 = 2 p_2 = \frac{12}{6} = 2 p 2 = 6 12 = 2 .
The other value of
p p p is
14 / 3 14/3 14/3 .
Understand the concept, then try another question Revisit Engineering Mathematics with concept notes, common mistakes and an original worked example before your next attempt.
More questions on Linear Algebra ← Q37 Full paper Q39 → 