PYQs / GATE CE / 2014 / Set 2 / Q12 GATE CE 2014 Set 2 — Question 12 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +1 / -0 Medium Determinants Linear Algebra Engineering Mathematics
Engineering Mathematics → Linear Algebra → Determinants
Last updated 5 September 2026
Question The determinant of matrix
[ 0 1 2 3 1 0 3 0 2 3 0 1 3 0 1 2 ] \begin{bmatrix} 0 & 1 & 2 & 3 \\ 1 & 0 & 3 & 0 \\ 2 & 3 & 0 & 1 \\ 3 & 0 & 1 & 2 \end{bmatrix} 0 1 2 3 1 0 3 0 2 3 0 1 3 0 1 2 is
_____________ Solution Let
A = [ 0 1 2 3 1 0 3 0 2 3 0 1 3 0 1 2 ] A = \begin{bmatrix} 0 & 1 & 2 & 3 \\ 1 & 0 & 3 & 0 \\ 2 & 3 & 0 & 1 \\ 3 & 0 & 1 & 2 \end{bmatrix} A = 0 1 2 3 1 0 3 0 2 3 0 1 3 0 1 2 . We can find the determinant by expanding along the first row:
∣ A ∣ = 0 ⋅ M 11 − 1 ⋅ ∣ 1 3 0 2 0 1 3 1 2 ∣ + 2 ⋅ ∣ 1 0 0 2 3 1 3 0 2 ∣ − 3 ⋅ ∣ 1 0 3 2 3 0 3 0 1 ∣ |A| = 0 \cdot M_{11} - 1 \cdot \begin{vmatrix} 1 & 3 & 0 \\ 2 & 0 & 1 \\ 3 & 1 & 2 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & 0 & 0 \\ 2 & 3 & 1 \\ 3 & 0 & 2 \end{vmatrix} - 3 \cdot \begin{vmatrix} 1 & 0 & 3 \\ 2 & 3 & 0 \\ 3 & 0 & 1 \end{vmatrix} ∣ A ∣ = 0 ⋅ M 11 − 1 ⋅ 1 2 3 3 0 1 0 1 2 + 2 ⋅ 1 2 3 0 3 0 0 1 2 − 3 ⋅ 1 2 3 0 3 0 3 0 1 1. ∣ 1 3 0 2 0 1 3 1 2 ∣ = 1 ( 0 − 1 ) − 3 ( 4 − 3 ) + 0 = − 1 − 3 = − 4 \begin{vmatrix} 1 & 3 & 0 \\ 2 & 0 & 1 \\ 3 & 1 & 2 \end{vmatrix} = 1(0-1) - 3(4-3) + 0 = -1 - 3 = -4 1 2 3 3 0 1 0 1 2 = 1 ( 0 − 1 ) − 3 ( 4 − 3 ) + 0 = − 1 − 3 = − 4 2. ∣ 1 0 0 2 3 1 3 0 2 ∣ = 1 ( 6 − 0 ) − 0 + 0 = 6 \begin{vmatrix} 1 & 0 & 0 \\ 2 & 3 & 1 \\ 3 & 0 & 2 \end{vmatrix} = 1(6-0) - 0 + 0 = 6 1 2 3 0 3 0 0 1 2 = 1 ( 6 − 0 ) − 0 + 0 = 6 3. ∣ 1 0 3 2 3 0 3 0 1 ∣ = 1 ( 3 − 0 ) − 0 + 3 ( 0 − 9 ) = 3 − 27 = − 24 \begin{vmatrix} 1 & 0 & 3 \\ 2 & 3 & 0 \\ 3 & 0 & 1 \end{vmatrix} = 1(3-0) - 0 + 3(0-9) = 3 - 27 = -24 1 2 3 0 3 0 3 0 1 = 1 ( 3 − 0 ) − 0 + 3 ( 0 − 9 ) = 3 − 27 = − 24
Substituting these back:
∣ A ∣ = − 1 ( − 4 ) + 2 ( 6 ) − 3 ( − 24 ) = 4 + 12 + 72 = 88 |A| = -1(-4) + 2(6) - 3(-24) = 4 + 12 + 72 = 88 ∣ A ∣ = − 1 ( − 4 ) + 2 ( 6 ) − 3 ( − 24 ) = 4 + 12 + 72 = 88 .
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
A = [ 0 1 2 3 1 0 3 0 2 3 0 1 3 0 1 2 ] A = \begin{bmatrix} 0 & 1 & 2 & 3 \\ 1 & 0 & 3 & 0 \\ 2 & 3 & 0 & 1 \\ 3 & 0 & 1 & 2 \end{bmatrix} A = 0 1 2 3 1 0 3 0 2 3 0 1 3 0 1 2 . We can find the determinant by expanding along the first row:
∣ A ∣ = 0 ⋅ M 11 − 1 ⋅ ∣ 1 3 0 2 0 1 3 1 2 ∣ + 2 ⋅ ∣ 1 0 0 2 3 1 3 0 2 ∣ − 3 ⋅ ∣ 1 0 3 2 3 0 3 0 1 ∣ |A| = 0 \cdot M_{11} - 1 \cdot \begin{vmatrix} 1 & 3 & 0 \\ 2 & 0 & 1 \\ 3 & 1 & 2 \end{vmatrix} + 2 \cdot \begin{vmatrix} 1 & 0 & 0 \\ 2 & 3 & 1 \\ 3 & 0 & 2 \end{vmatrix} - 3 \cdot \begin{vmatrix} 1 & 0 & 3 \\ 2 & 3 & 0 \\ 3 & 0 & 1 \end{vmatrix} ∣ A ∣ = 0 ⋅ M 11 − 1 ⋅ 1 2 3 3 0 1 0 1 2 + 2 ⋅ 1 2 3 0 3 0 0 1 2 − 3 ⋅ 1 2 3 0 3 0 3 0 1 1. ∣ 1 3 0 2 0 1 3 1 2 ∣ = 1 ( 0 − 1 ) − 3 ( 4 − 3 ) + 0 = − 1 − 3 = − 4 \begin{vmatrix} 1 & 3 & 0 \\ 2 & 0 & 1 \\ 3 & 1 & 2 \end{vmatrix} = 1(0-1) - 3(4-3) + 0 = -1 - 3 = -4 1 2 3 3 0 1 0 1 2 = 1 ( 0 − 1 ) − 3 ( 4 − 3 ) + 0 = − 1 − 3 = − 4 2. ∣ 1 0 0 2 3 1 3 0 2 ∣ = 1 ( 6 − 0 ) − 0 + 0 = 6 \begin{vmatrix} 1 & 0 & 0 \\ 2 & 3 & 1 \\ 3 & 0 & 2 \end{vmatrix} = 1(6-0) - 0 + 0 = 6 1 2 3 0 3 0 0 1 2 = 1 ( 6 − 0 ) − 0 + 0 = 6 3. ∣ 1 0 3 2 3 0 3 0 1 ∣ = 1 ( 3 − 0 ) − 0 + 3 ( 0 − 9 ) = 3 − 27 = − 24 \begin{vmatrix} 1 & 0 & 3 \\ 2 & 3 & 0 \\ 3 & 0 & 1 \end{vmatrix} = 1(3-0) - 0 + 3(0-9) = 3 - 27 = -24 1 2 3 0 3 0 3 0 1 = 1 ( 3 − 0 ) − 0 + 3 ( 0 − 9 ) = 3 − 27 = − 24
Substituting these back:
∣ A ∣ = − 1 ( − 4 ) + 2 ( 6 ) − 3 ( − 24 ) = 4 + 12 + 72 = 88 |A| = -1(-4) + 2(6) - 3(-24) = 4 + 12 + 72 = 88 ∣ A ∣ = − 1 ( − 4 ) + 2 ( 6 ) − 3 ( − 24 ) = 4 + 12 + 72 = 88 .
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 ← Q11 Full paper Q13 → 