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Determinants
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Class 12MathsChapter 4NCERTCBSE
Determinants – Class 12 Maths Chapter 4
📌 Quick Overview: Chapter 4 builds directly on Chapter 3 (Matrices). You will learn how to calculate the determinant of a matrix, use its properties to simplify calculations, find the adjoint and inverse of a matrix using determinants, calculate the area of a triangle, and solve a system of linear equations using matrices. This is one of the highest-scoring chapters in CBSE board exams.
Topics Covered:
- > Determinant of a Matrix
- > Determinant of 2x2 Matrix
- > Determinant of 3x3 Matrix
- > Properties of Determinants
- > Minors and Cofactors
- > Adjoint of a Matrix
- > Inverse using Determinants
- > Area of Triangle
- > Consistency of Equations
- > Solving Equations (Matrix Method)
- > Important Board Questions
- > MCQ Practice
1. What is a Determinant?
Definition: To every square matrix A of order n, we can associate a number (real or complex) called the determinant of A. It is denoted by det(A) or |A|.
Note: Only square matrices have determinants. A non-square matrix does not have a determinant.
Note: Only square matrices have determinants. A non-square matrix does not have a determinant.
1.1 Determinant of a 1x1 Matrix
If A = [a], then |A| = a.
1.2 Determinant of a 2x2 Matrix
If A = [a, b / c, d], then:
|A| = ad - bc
|A| = ad - bc
Example: A = [3, 4 / 2, 5]
|A| = (3)(5) - (4)(2) = 15 - 8 = 7
|A| = (3)(5) - (4)(2) = 15 - 8 = 7
1.3 Determinant of a 3x3 Matrix (Expansion along Row 1) ⭐
If A = [a1, b1, c1 / a2, b2, c2 / a3, b3, c3], then:
|A| = a1(b2c3 - b3c2) - b1(a2c3 - a3c2) + c1(a2b3 - a3b2)
|A| = a1(b2c3 - b3c2) - b1(a2c3 - a3c2) + c1(a2b3 - a3b2)
Example: Find the determinant of A = [2, -3, 1 / 4, 0, -2 / 1, 5, 3]
|A| = 2(0x3 - (-2)x5) - (-3)(4x3 - (-2)x1) + 1(4x5 - 0x1)
= 2(0 + 10) + 3(12 + 2) + 1(20 - 0)
= 2(10) + 3(14) + 20
= 20 + 42 + 20
= 82
|A| = 2(0x3 - (-2)x5) - (-3)(4x3 - (-2)x1) + 1(4x5 - 0x1)
= 2(0 + 10) + 3(12 + 2) + 1(20 - 0)
= 2(10) + 3(14) + 20
= 20 + 42 + 20
= 82
⭐ Sign Pattern for 3x3 Expansion:
+ - +
- + -
+ - +
The signs alternate starting with + at position (1,1). You can expand along ANY row or column.
+ - +
- + -
+ - +
The signs alternate starting with + at position (1,1). You can expand along ANY row or column.
2. Properties of Determinants ⭐ (Very Important)
Property 1: The value of the determinant remains unchanged if its rows and columns are interchanged.
|A| = |A'|
|A| = |A'|
Property 2: If any two rows (or columns) of a determinant are interchanged, the sign of the determinant changes.
Ri ↔ Rj => |A| becomes -|A|
Ri ↔ Rj => |A| becomes -|A|
Property 3: If any two rows (or columns) of a determinant are identical, the value of the determinant is zero.
Property 4: If each element of a row (or column) is multiplied by a constant k, the value of the determinant is multiplied by k.
|kA| = kn|A| for an n x n matrix.
|kA| = kn|A| for an n x n matrix.
Property 5: If some or all elements of a row (or column) of a determinant are expressed as the sum of two terms, then the determinant can be expressed as the sum of two determinants.
Property 6: If each element of a row (or column) is added to a constant multiple of the corresponding elements of another row (or column), the value of the determinant remains unchanged.
Ri → Ri + kRj does NOT change |A|
Ri → Ri + kRj does NOT change |A|
Property 7: If any row (or column) of a determinant is zero, the value of the determinant is zero.
⭐ Key Results:
> |AB| = |A| . |B|
> |kA| = kn|A| where A is n x n
> If A is a triangular matrix (upper or lower), |A| = product of diagonal elements
> |A-1| = 1/|A| = |A|-1
> |AB| = |A| . |B|
> |kA| = kn|A| where A is n x n
> If A is a triangular matrix (upper or lower), |A| = product of diagonal elements
> |A-1| = 1/|A| = |A|-1
3. Minors and Cofactors ⭐
3.1 Minor
Definition: The minor of element aij (denoted Mij) is the determinant of the submatrix obtained by deleting the i-th row and j-th column of A.
3.2 Cofactor
Definition: The cofactor of element aij (denoted Aij) is:
Aij = (-1)i+j Mij
Aij = (-1)i+j Mij
Example: For A = [1, 2, 3 / 4, 5, 6 / 7, 8, 9], find M11 and A11.
M11 = |5, 6 / 8, 9| = (5)(9) - (6)(8) = 45 - 48 = -3
A11 = (-1)1+1 M11 = (+1)(-3) = -3
M12 = |4, 6 / 7, 9| = 36 - 42 = -6
A12 = (-1)1+2(-6) = (-1)(-6) = 6
M11 = |5, 6 / 8, 9| = (5)(9) - (6)(8) = 45 - 48 = -3
A11 = (-1)1+1 M11 = (+1)(-3) = -3
M12 = |4, 6 / 7, 9| = 36 - 42 = -6
A12 = (-1)1+2(-6) = (-1)(-6) = 6
Expansion using cofactors (along row 1):
|A| = a11.A11 + a12.A12 + a13.A13
Sum of products of any row with cofactors of ANOTHER row = 0
(This is used in proving adjoint properties)
4. Adjoint of a Matrix
Definition: The adjoint of a square matrix A (written as adj A) is the transpose of the matrix of cofactors of A.
adj A = [Aij]T
adj A = [Aij]T
Example: Find adj A for A = [1, 2 / 3, 4].
Cofactors:
A11 = 4, A12 = -3, A21 = -2, A22 = 1
Matrix of cofactors = [4, -3 / -2, 1]
adj A = Transpose = [4, -2 / -3, 1]
Cofactors:
A11 = 4, A12 = -3, A21 = -2, A22 = 1
Matrix of cofactors = [4, -3 / -2, 1]
adj A = Transpose = [4, -2 / -3, 1]
⭐ Key Property: A(adj A) = (adj A)A = |A| . I
For a 2x2 matrix A = [a, b / c, d]:
adj A = [d, -b / -c, a] (swap diagonal, negate off-diagonal)
For a 2x2 matrix A = [a, b / c, d]:
adj A = [d, -b / -c, a] (swap diagonal, negate off-diagonal)
5. Inverse of a Matrix using Determinants ⭐
Formula: If |A| ≠ 0, the matrix A is invertible and:
A-1 = (1/|A|) . adj A
A-1 = (1/|A|) . adj A
⭐ Singular vs Non-Singular Matrix:
> If |A| = 0 → A is called a singular matrix → inverse does NOT exist.
> If |A| ≠ 0 → A is called a non-singular matrix → inverse EXISTS.
> If |A| = 0 → A is called a singular matrix → inverse does NOT exist.
> If |A| ≠ 0 → A is called a non-singular matrix → inverse EXISTS.
Example: Find A-1 for A = [2, 1 / 5, 3].
|A| = (2)(3) - (1)(5) = 6 - 5 = 1
adj A = [3, -1 / -5, 2]
A-1 = (1/1)[3, -1 / -5, 2] = [3, -1 / -5, 2]
Verify: AA-1 = [2,1/5,3][3,-1/-5,2] = [6-5,-2+2/15-15,-5+6] = [1,0/0,1] = I ✓
|A| = (2)(3) - (1)(5) = 6 - 5 = 1
adj A = [3, -1 / -5, 2]
A-1 = (1/1)[3, -1 / -5, 2] = [3, -1 / -5, 2]
Verify: AA-1 = [2,1/5,3][3,-1/-5,2] = [6-5,-2+2/15-15,-5+6] = [1,0/0,1] = I ✓
6. Area of a Triangle ⭐
Formula: Area of triangle with vertices (x1,y1), (x2,y2), (x3,y3) is:
Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Using determinant form:
Area = (1/2) |det[x1, y1, 1 / x2, y2, 1 / x3, y3, 1]|
Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Using determinant form:
Area = (1/2) |det[x1, y1, 1 / x2, y2, 1 / x3, y3, 1]|
Example: Find the area of triangle with vertices A(1,0), B(6,0), C(4,3).
Area = (1/2)|1(0-3) + 6(3-0) + 4(0-0)|
= (1/2)|(-3) + 18 + 0|
= (1/2)(15) = 7.5 sq. units
Area = (1/2)|1(0-3) + 6(3-0) + 4(0-0)|
= (1/2)|(-3) + 18 + 0|
= (1/2)(15) = 7.5 sq. units
⭐ Collinearity Condition: Three points (x1,y1), (x2,y2), (x3,y3) are collinear if and only if:
det[x1, y1, 1 / x2, y2, 1 / x3, y3, 1] = 0 (area = 0)
det[x1, y1, 1 / x2, y2, 1 / x3, y3, 1] = 0 (area = 0)
7. Solution of System of Linear Equations ⭐
A system of linear equations AX = B can be solved using matrices:
Given: AX = B
Solution: X = A^-1 . B (provided |A| is not equal to 0)
7.1 Consistency of System
| Condition | Type of System | Solutions |
|---|---|---|
| |A| ≠ 0 | Consistent | Unique solution: X = A-1B |
| |A| = 0 and (adj A)B = O | Consistent | Infinitely many solutions |
| |A| = 0 and (adj A)B ≠ O | Inconsistent | No solution |
Example: Solve: x + 2y = 5, 3x + 4y = 11 using matrix method.
Matrix form: [1,2/3,4][x/y] = [5/11]
A = [1,2/3,4], X = [x/y], B = [5/11]
|A| = (1)(4) - (2)(3) = 4 - 6 = -2 ≠ 0, so unique solution exists.
adj A = [4,-2/-3,1]
A-1 = (1/-2)[4,-2/-3,1] = [-2, 1 / 3/2, -1/2]
X = A-1B = [-2,1/3/2,-1/2][5/11]
x = (-2)(5) + (1)(11) = -10 + 11 = 1
y = (3/2)(5) + (-1/2)(11) = 15/2 - 11/2 = 4/2 = 2
∴ x = 1, y = 2
Matrix form: [1,2/3,4][x/y] = [5/11]
A = [1,2/3,4], X = [x/y], B = [5/11]
|A| = (1)(4) - (2)(3) = 4 - 6 = -2 ≠ 0, so unique solution exists.
adj A = [4,-2/-3,1]
A-1 = (1/-2)[4,-2/-3,1] = [-2, 1 / 3/2, -1/2]
X = A-1B = [-2,1/3/2,-1/2][5/11]
x = (-2)(5) + (1)(11) = -10 + 11 = 1
y = (3/2)(5) + (-1/2)(11) = 15/2 - 11/2 = 4/2 = 2
∴ x = 1, y = 2
8. Important Board Exam Questions
Q1. If A = [1, 2, -3 / 2, 3, 2 / 3, -3, -4], find A-1 and use it to solve x + 2y - 3z = -4, 2x + 3y + 2z = 2, 3x - 3y - 4z = 11.
Step 1: Find |A|
|A| = 1(3x(-4) - 2x(-3)) - 2(2x(-4) - 2x3) + (-3)(2x(-3) - 3x3)
= 1(-12+6) - 2(-8-6) + (-3)(-6-9)
= 1(-6) - 2(-14) + (-3)(-15)
= -6 + 28 + 45 = 67 ≠ 0, so A-1 exists.
Step 2: Find cofactors and adj A (compute all 9 cofactors, then transpose)
Step 3: A-1 = (1/67)(adj A), then X = A-1B gives the solution.
|A| = 1(3x(-4) - 2x(-3)) - 2(2x(-4) - 2x3) + (-3)(2x(-3) - 3x3)
= 1(-12+6) - 2(-8-6) + (-3)(-6-9)
= 1(-6) - 2(-14) + (-3)(-15)
= -6 + 28 + 45 = 67 ≠ 0, so A-1 exists.
Step 2: Find cofactors and adj A (compute all 9 cofactors, then transpose)
Step 3: A-1 = (1/67)(adj A), then X = A-1B gives the solution.
Q2. Using properties of determinants, prove that:
|a, a+b, a+b+c / 2a, 3a+2b, 4a+3b+2c / 3a, 6a+3b, 10a+6b+3c| = a³
|a, a+b, a+b+c / 2a, 3a+2b, 4a+3b+2c / 3a, 6a+3b, 10a+6b+3c| = a³
Apply R2 → R2 - 2R1 and R3 → R3 - 3R1:
= |a, a+b, a+b+c / 0, a, 2a+b / 0, 3a, 7a+3b+3c... | → continue applying operations → eventually = a³
= |a, a+b, a+b+c / 0, a, 2a+b / 0, 3a, 7a+3b+3c... | → continue applying operations → eventually = a³
Q3. Find the area of triangle whose vertices are (2,7), (1,1), (10,8).
Area = (1/2)|2(1-8) + 1(8-7) + 10(7-1)|
= (1/2)|2(-7) + 1(1) + 10(6)|
= (1/2)|-14 + 1 + 60|
= (1/2)|47| = 47/2 = 23.5 sq. units
= (1/2)|2(-7) + 1(1) + 10(6)|
= (1/2)|-14 + 1 + 60|
= (1/2)|47| = 47/2 = 23.5 sq. units
Q4. If A = [2, -3 / 3, 4], show that A² - 6A + 17I = O. Hence find A-1.
A² = A.A = [2,-3/3,4][2,-3/3,4] = [4-9,-6-12/6+12,-9+16] = [-5,-18/18,7]
6A = [12,-18/18,24]
17I = [17,0/0,17]
A² - 6A + 17I = [-5-12+17, -18+18+0 / 18-18+0, 7-24+17] = [0,0/0,0] = O ✓
From A² - 6A + 17I = O:
Multiply both sides by A-1: A - 6I + 17A-1 = O
17A-1 = 6I - A = [6,0/0,6] - [2,-3/3,4] = [4,3/-3,2]
∴ A-1 = (1/17)[4,3/-3,2]
6A = [12,-18/18,24]
17I = [17,0/0,17]
A² - 6A + 17I = [-5-12+17, -18+18+0 / 18-18+0, 7-24+17] = [0,0/0,0] = O ✓
From A² - 6A + 17I = O:
Multiply both sides by A-1: A - 6I + 17A-1 = O
17A-1 = 6I - A = [6,0/0,6] - [2,-3/3,4] = [4,3/-3,2]
∴ A-1 = (1/17)[4,3/-3,2]
Q5. If the points (k,2-2k), (1-k,2k), (-k-4, 6-2k) are collinear, find k.
For collinearity, area of triangle = 0:
(1/2)|k(2k-(6-2k)) + (1-k)((6-2k)-(2-2k)) + (-k-4)((2-2k)-2k)| = 0
|k(4k-6) + (1-k)(4) + (-k-4)(2-4k)| = 0
|4k²-6k + 4-4k + (-2k+4k²-8+16k)| = 0
|4k²-6k + 4-4k + 4k²+14k-8| = 0
|8k²+4k-4| = 0
8k²+4k-4 = 0 → 2k²+k-1 = 0 → (2k-1)(k+1) = 0
∴ k = 1/2 or k = -1
(1/2)|k(2k-(6-2k)) + (1-k)((6-2k)-(2-2k)) + (-k-4)((2-2k)-2k)| = 0
|k(4k-6) + (1-k)(4) + (-k-4)(2-4k)| = 0
|4k²-6k + 4-4k + (-2k+4k²-8+16k)| = 0
|4k²-6k + 4-4k + 4k²+14k-8| = 0
|8k²+4k-4| = 0
8k²+4k-4 = 0 → 2k²+k-1 = 0 → (2k-1)(k+1) = 0
∴ k = 1/2 or k = -1
9. Key Formulas at a Glance
DETERMINANT:
2x2: |a,b/c,d| = ad - bc
3x3: Expand along row/column using cofactors
COFACTOR: A_ij = (-1)^(i+j) . M_ij
ADJOINT: adj A = Transpose of cofactor matrix
For 2x2: [a,b/c,d] → adj = [d,-b/-c,a]
KEY PROPERTY: A(adj A) = (adj A)A = |A|.I
INVERSE:
A^-1 = (1/|A|) . adj A (only if |A| is not 0)
|A^-1| = 1/|A|
|AB| = |A|.|B|
|kA| = k^n |A| (n x n matrix)
AREA OF TRIANGLE:
= (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
Collinear if area = 0
SYSTEM AX = B:
Unique solution: X = A^-1 B (when |A| is not 0)
10. MCQ Practice (1 Mark)
1. If A is a square matrix of order 3 and |A| = 5, then |2A| equals:
(a) 10 (b) 25 (c) 40 (d) 20
Answer: (c) 40 — because |2A| = 2³|A| = 8 x 5 = 40
2. If A is a singular matrix, then adj A is:
(a) Also singular (b) Non-singular (c) Symmetric (d) Zero matrix
Answer: (a) Also singular
3. For a 3x3 matrix A, |adj A| equals:
(a) |A| (b) |A|² (c) |A|³ (d) 1/|A|
Answer: (b) |A|² — because |adj A| = |A|^(n-1) = |A|²
4. If A = [1,2/4,2] and f(A) = A² - 3A + I, then f(A) equals:
(a) O (b) I (c) 5I (d) 3A
Answer: (a) O — compute A² and substitute to verify
5. Area of a triangle with vertices (1,0), (0,1), (0,0) is:
(a) 1 sq. unit (b) 1/2 sq. unit (c) 2 sq. units (d) 0
Answer: (b) 1/2 sq. unit
11. Exam Tips
- For 3x3 determinants in board exams, always use row/column operations to create zeros before expanding — it saves time and reduces errors.
- For "prove" determinant questions, use Property 6 (Ri → Ri + kRj) to create a triangular form or factor out common terms.
- For 2x2 matrix inverse: use the shortcut adj formula — swap diagonal, negate off-diagonal.
- Always check |A| ≠ 0 FIRST before attempting to find inverse or solve a system — if |A| = 0, state the system is singular.
- In collinearity problems, set up the 3x3 determinant and equate to zero — do not use the area formula approach, the determinant form is faster.
- Remember |adj A| = |A|^(n-1) for an n x n matrix — this appears in MCQs regularly.
- For equation-solving questions (5 marks), write the matrix form AX = B clearly first before finding A-1.
✅ Summary: Determinants is one of the most important chapters for CBSE board exams. The 5-mark "solve using matrix method" and "prove using properties" questions are almost guaranteed every year. Master Property 6 (adding multiples of rows), the adjoint-inverse formula, and the area of triangle formula. Practice at least 5 full problems of each type before the exam.
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