Determinants — Class 12 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Determinants" — 7 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Determinants" — 7 important questions with detailed answers for CBSE board exam pr…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Find the determinant of the matrix A = [2 3; 1 4].
Find the determinant of B = [1 2 3; 0 4 5; 1 0 2].
If A = [3 2; 1 4], find |2A| (the determinant of 2A).
Verify that det(AB) = det(A) × det(B) for A = [1 2; 3 4] and B = [2 0; 1 3].
Find the value of x such that |x 2; 3 4| = 0.
If A = [1 0 0; 0 2 0; 0 0 3], find det(A).
+ 1 more questions in the full chapter

Solutions Summary:

Question	Status
Find the determinant of the matrix A = [2 3; 1 4].	✓ Solved
Find the determinant of B = [1 2 3; 0 4 5; 1 0 2].	✓ Solved
If A = [3 2; 1 4], find \|2A\| (the determinant of 2A).	✓ Solved
Verify that det(AB) = det(A) × det(B) for A = [1 2; 3 4] …	✓ Solved
Find the value of x such that \|x 2; 3 4\| = 0.	✓ Solved
If A = [1 0 0; 0 2 0; 0 0 3], find det(A).	✓ Solved

Showing 6 of 7 questions

Q1: Find the determinant of the matrix A = [2 3; 1 4].

Given: A = [2 3; 1 4] For a 2×2 matrix [a b; c d], the determinant is ad - bc. Here: a = 2, b = 3, c = 1, d = 4 det(A) = (2)(4) - (3)(1) = 8 - 3 = 5 CONCLUSION: det(A) = 5

Q2: Find the determinant of B = [1 2 3; 0 4 5; 1 0 2].

Given: B = [1 2 3; 0 4 5; 1 0 2] We use expansion along the first column (since it has a zero): det(B) = 1 × |4 5; 0 2| - 0 × |2 3; 0 2| + 1 × |2 3; 4 5| Calculating the 2×2 determinants: |4 5; 0 2| = (4)(2) - (5)(0) = 8 - 0 = 8 |2 3; 4 5| = (2)(5) - (3)(4) = 10 - 12 = -2 det(B) = 1(8) - 0 + 1(-2) = 8 + 0 - 2 = 6 CONCLUSION: det(B) = 6

Q3: If A = [3 2; 1 4], find |2A| (the determinant of 2A).

Given: A = [3 2; 1 4] First, find 2A: 2A = [6 4; 2 8] Now find det(2A): det(2A) = (6)(8) - (4)(2) = 48 - 8 = 40 ALTERNATIVE METHOD (using property): For an n×n matrix A and scalar k: det(kA) = k^n × det(A) Here n = 2 and k = 2. First find det(A): det(A) = (3)(4) - (2)(1) = 12 - 2 = 10 Then: det(2A) = 2² × det(A) = 4 × 10 = 40 CONCLUSION: |2A| = 40

Q4: Verify that det(AB) = det(A) × det(B) for A = [1 2; 3 4] and B = [2 0; 1 3].

Given: A = [1 2; 3 4] and B = [2 0; 1 3] Step 1: Find det(A) det(A) = (1)(4) - (2)(3) = 4 - 6 = -2 Step 2: Find det(B) det(B) = (2)(3) - (0)(1) = 6 - 0 = 6 Step 3: Calculate det(A) × det(B) det(A) × det(B) = (-2)(6) = -12 Step 4: Find AB AB = [1 2; 3 4] × [2 0; 1 3] Element (1,1): (1)(2) + (2)(1) = 2 + 2 = 4 Element (1,2): (1)(0) + (2)(3) = 0 + 6 = 6 Element (2,1): (3)(2) + (4)(1) = 6 + 4 = 10 Element (2,2): (3)(0) + (4)(3) = 0 + 12 = 12 AB = [4 6; 10 12] Step 5: Find det(AB) det(AB) = (4...

Q5: Find the value of x such that |x 2; 3 4| = 0.

We need to find x such that the determinant equals zero. |x 2; 3 4| = 0 Using the 2×2 determinant formula: (x)(4) - (2)(3) = 0 4x - 6 = 0 4x = 6 x = 6/4 x = 3/2 VERIFICATION: When x = 3/2: |3/2 2; 3 4| = (3/2)(4) - (2)(3) = 6 - 6 = 0 ✓ CONCLUSION: x = 3/2

Q6: If A = [1 0 0; 0 2 0; 0 0 3], find det(A).

Given: A = [1 0 0; 0 2 0; 0 0 3] (diagonal matrix) For a diagonal matrix, the determinant is the product of diagonal elements. det(A) = 1 × 2 × 3 = 6 ALTERNATIVE METHOD (using expansion): Expanding along the first row: det(A) = 1 × |2 0; 0 3| - 0 × |0 0; 0 3| + 0 × |0 2; 0 0| = 1 × (2×3 - 0×0) - 0 + 0 = 1 × 6 = 6 CONCLUSION: det(A) = 6

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