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.
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…
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NCERT Solutions for Class 12 Mathematics — Determinants. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Determinants — All 7 Questions Solved
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)(12) - (6)(10) = 48 - 60 = -12
VERIFICATION:
det(AB) = -12 = det(A) × det(B) ✓
CONCLUSION: The property det(AB) = det(A) × det(B) is verified.
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
Q7. Find |1 2 1; 0 3 2; 2 1 3| using expansion along the first column.
Given matrix: |1 2 1; 0 3 2; 2 1 3|
Expanding along the first column:
det = 1 × |3 2; 1 3| - 0 × |2 1; 1 3| + 2 × |2 1; 3 2|
Calculating the 2×2 minors:
|3 2; 1 3| = (3)(3) - (2)(1) = 9 - 2 = 7
|2 1; 1 3| = (2)(3) - (1)(1) = 6 - 1 = 5 (not needed since coefficient is 0)
|2 1; 3 2| = (2)(2) - (1)(3) = 4 - 3 = 1
det = 1(7) - 0(5) + 2(1)
= 7 + 0 + 2
= 9
CONCLUSION: |1 2 1; 0 3 2; 2 1 3| = 9
Determinants — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the value of the determinant of the matrix A = [[2, 3], [4, 5]]?
- A. -2
- B. 2
- C. 22
- D. -22
Answer: A. -2 — The determinant of a 2x2 matrix [[a, b], [c, d]] is ad - bc. For the given matrix, it is (2*5) - (3*4) = 10 - 12 = -2.
MCQ 2. If two rows of a determinant are identical, what is its value?
- A. 1
- B. 0
- C. -1
- D. Depends on elements
Answer: B. 0 — A fundamental property of determinants states that if any two rows or columns of a determinant are identical, its value is zero.
MCQ 3. If A is a square matrix of order 3 such that |A| = 5, then what is the value of |3A|?
- A. 15
- B. 45
- C. 125
- D. 135
Answer: D. 135 — For a square matrix A of order n, |kA| = k^n |A|. Here, n=3 and k=3, so |3A| = 3^3 |A| = 27 * 5 = 135.
MCQ 4. For the matrix A = [[1, 2, 3], [4, 5, 6], [7, 8, 9]], what is the minor M_11?
- A. -3
- B. 5
- C. 9
- D. -1
Answer: A. -3 — M_11 is the determinant of the submatrix obtained by deleting the 1st row and 1st column. This submatrix is [[5, 6], [8, 9]], so M_11 = (5*9) - (6*8) = 45 - 48 = -3.
MCQ 5. If A = [[2, 3], [1, 4]], what is the adjoint of A?
- A. [[4, -3], [-1, 2]]
- B. [[2, 1], [3, 4]]
- C. [[4, 1], [3, 2]]
- D. [[2, -1], [-3, 4]]
Answer: A. [[4, -3], [-1, 2]] — For a 2x2 matrix [[a, b], [c, d]], the adjoint is [[d, -b], [-c, a]]. So, for A, it's [[4, -3], [-1, 2]].
MCQ 6. If A is a square matrix of order 3 and |A| = 2, then what is |adj(A)|?
- A. 2
- B. 4
- C. 8
- D. 1
Answer: B. 4 — For a square matrix A of order n, |adj(A)| = |A|^(n-1). Here, n=3 and |A|=2, so |adj(A)| = 2^(3-1) = 2^2 = 4.
MCQ 7. Find the area of the triangle with vertices (1, 0), (6, 0), and (4, 3) using determinants.
- A. 7.5 sq units
- B. 15 sq units
- C. 3 sq units
- D. 9 sq units
Answer: A. 7.5 sq units — The area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is 1/2 |det([[x1, y1, 1], [x2, y2, 1], [x3, y3, 1]])|. For the given points, area = 1/2 |1(0-3) - 0(6-4) + 1(18-0)| = 1/2 |-3 + 18| = 1/2 * 15 = 7.5 sq units.
MCQ 8. For what value of k are the points (k, 2-2k), (-k+1, 2k), and (-4-k, 6-2k) collinear?
- A. k=1 or k=2
- B. k=0 or k=1
- C. k=-1 or k=1/2
- D. k=0 or k=1/2
Answer: D. k=0 or k=1/2 — Points are collinear if the area of the triangle formed by them is zero. This means det([[k, 2-2k, 1], [-k+1, 2k, 1], [-4-k, 6-2k, 1]]) = 0. After applying row operations and expanding, we find k=0 or k=1/2.
MCQ 9. A square matrix A is said to be singular if:
- A. |A| = 1
- B. |A| = 0
- C. |A| is negative
- D. |A| is positive
Answer: B. |A| = 0 — A square matrix A is defined as singular if its determinant |A| is equal to zero. If |A| is non-zero, it is a non-singular matrix.
MCQ 10. For the matrix A = [[1, 2, 3], [0, 1, 4], [5, 6, 0]], what is the cofactor C_23?
- A. -4
- B. 4
- C. -6
- D. 6
Answer: B. 4 — C_ij = (-1)^(i+j) * M_ij. C_23 = (-1)^(2+3) * M_23 = (-1)^5 * det([[1, 2], [5, 6]]) = -1 * (1*6 - 2*5) = -1 * (6 - 10) = -1 * (-4) = 4.
MCQ 11. For a system of linear equations AX = B, if |A| ≠ 0, then the system has:
- A. No solution
- B. Infinitely many solutions
- C. A unique solution
- D. Exactly two solutions
Answer: C. A unique solution — According to Cramer's Rule, if the determinant of the coefficient matrix (|A|) is non-zero, then the system of linear equations has a unique solution.
MCQ 12. If A is a square matrix of order n and det(A) = 5, then det(adj(A^T)) is:
- A. 5^(n-1)
- B. 5^n
- C. 5^(n-2)
- D. 5
Answer: A. 5^(n-1) — We know that det(A^T) = det(A) and det(adj(B)) = det(B)^(n-1) for any square matrix B of order n. Therefore, det(adj(A^T)) = det(A^T)^(n-1) = det(A)^(n-1) = 5^(n-1).
MCQ 13. If A and B are two square matrices of order 3 such that |A| = -2 and |B| = 3, then |AB| is:
- A. -6
- B. 6
- C. 1
- D. -5
Answer: A. -6 — For any two square matrices A and B of the same order, |AB| = |A| * |B|. Therefore, |AB| = (-2) * (3) = -6.
MCQ 14. If we interchange any two rows or columns of a determinant, its value:
- A. Remains unchanged
- B. Changes its sign
- C. Becomes zero
- D. Is multiplied by -1
Answer: B. Changes its sign — A property of determinants states that if any two rows or two columns of a determinant are interchanged, the sign of the determinant changes.
MCQ 15. Find the value of x if |[[x, 2], [18, x]]| = |[[6, 2], [18, 6]]|.
- A. ±6
- B. 6
- C. -6
- D. ±3
Answer: A. ±6 — Evaluate both determinants. Left side: x*x - 2*18 = x^2 - 36. Right side: 6*6 - 2*18 = 36 - 36 = 0. So, x^2 - 36 = 0 => x^2 = 36 => x = ±6.
MCQ 16. For an invertible matrix A of order 2, which of the following is true?
- A. adj(A) = A
- B. adj(A) = |A|A^-1
- C. adj(A) = A^-1
- D. adj(A) = |A|^-1 A
Answer: B. adj(A) = |A|A^-1 — We know that A * adj(A) = |A| * I, where I is the identity matrix. If A is invertible, then A^-1 = (1/|A|) * adj(A). Multiplying by |A| gives adj(A) = |A| * A^-1.
MCQ 17. In a 3x3 determinant, how is a minor M_ij formed?
- A. By deleting the i-th column and j-th row
- B. By deleting the i-th row and j-th column
- C. By multiplying the i-th row and j-th column
- D. By adding elements of i-th row and j-th column
Answer: B. By deleting the i-th row and j-th column — The minor M_ij of an element a_ij in a determinant is the determinant of the submatrix obtained by deleting the i-th row and j-th column of the original determinant.
MCQ 18. What is the value of the determinant |[[1, x, y+z], [1, y, z+x], [1, z, x+y]]|?
- A. 0
- B. x+y+z
- C. xyz
- D. 1
Answer: A. 0 — Apply the column operation C3 -> C3 + C2. The determinant becomes |[[1, x, x+y+z], [1, y, x+y+z], [1, z, x+y+z]]|. Take out (x+y+z) as a common factor from C3. The resulting determinant has two identical columns (C1 and C3), making its value 0.
MCQ 19. If A is an invertible matrix, then det(A^-1) is equal to:
- A. det(A)
- B. 1/det(A)
- C. det(A)^-1
- D. det(A)^n
Answer: B. 1/det(A) — We know that A * A^-1 = I. Taking the determinant of both sides, |A * A^-1| = |I|. Since |A * A^-1| = |A| * |A^-1| and |I| = 1, we have |A| * |A^-1| = 1. Therefore, |A^-1| = 1/|A|.
MCQ 20. For what value of λ does the system of equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = 12 have no solution?
- A. λ = 1
- B. λ = 2
- C. λ = 3
- D. λ = 0
Answer: C. λ = 3 — For no solution, the determinant of the coefficient matrix |A| must be 0, and at least one of Dx, Dy, Dz must be non-zero. The determinant |A| = λ - 3. Setting |A|=0 gives λ=3. For λ=3, Dx = det([[6,1,1],[10,2,3],[12,2,3]]) = 2, which is non-zero. Hence, for λ=3, there is no solution.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.