Class 12 Mathematics Notes Chapter 4 (Determinants) – Mathematics Part-I Book
Alright class, let's begin our focused revision of Chapter 4: Determinants, keeping in mind the requirements for your government exam preparation. This chapter is crucial as it links directly to matrices and solving systems of linear equations.
Chapter 4: Determinants - Detailed Notes
1. Introduction:
- Every square matrix (a matrix with an equal number of rows and columns) can be associated with a unique number (real or complex) called its determinant.
- It is denoted by det(A) or |A| or Δ.
- Important: Only square matrices have determinants. For a matrix A, |A| is read as the determinant of A, not the modulus of A.
2. Determinant of a Matrix:
- Order 1: If A = [a], then det(A) = |a| = a.
- Order 2: If A =
[ a b ]
[ c d ]
then det(A) = |A| = ad - bc. - Order 3: If A =
[ a11 a12 a13 ]
[ a21 a22 a23 ]
[ a31 a32 a33 ]
The determinant can be calculated by expanding along any row or column.- Expansion along Row 1 (R1):
|A| = a11 * C11 + a12 * C12 + a13 * C13
where Cij is the cofactor of the element aij. - Expansion along Column 1 (C1):
|A| = a11 * C11 + a21 * C21 + a31 * C31
- Expansion along Row 1 (R1):
3. Minors and Cofactors:
-
Minor (Mij): The minor of an element aij in a determinant is the determinant obtained by deleting the i-th row and j-th column in which aij lies.
- Example: For the 3x3 matrix A above, M11 = determinant of
[ a22 a23 ]
[ a32 a33 ]= a22a33 - a23a32.
- Example: For the 3x3 matrix A above, M11 = determinant of
-
Cofactor (Cij or Aij): The cofactor of an element aij is defined as:
Cij = (-1)^(i+j) * Mij- The sign (-1)^(i+j) follows a checkerboard pattern:
[ + - + ]
[ - + - ]
[ + - + ] - So, C11 = +M11, C12 = -M12, C21 = -M21, etc.
- The sign (-1)^(i+j) follows a checkerboard pattern:
-
Value of Determinant using Cofactors:
- The sum of the product of elements of any row (or column) with their corresponding cofactors gives the value of the determinant |A|.
- Important Property: The sum of the product of elements of any row (or column) with the cofactors of the corresponding elements of another row (or column) is zero.
Example: a11C21 + a12C22 + a13*C23 = 0.
4. Properties of Determinants (Extremely Important for Simplification & MCQs):
- (P1) Reflection Property: The value of the determinant remains unchanged if its rows and columns are interchanged. |A| = |Aᵀ|.
- (P2) All-Zero Property: If all elements of a row (or column) are zero, then the value of the determinant is zero.
- (P3) Proportionality/Identical Property: If any two rows (or columns) of a determinant are identical or proportional (one is a scalar multiple of the other), then its value is zero.
- (P4) Switching Property: If any two rows (or columns) of a determinant are interchanged, then the sign of the determinant changes.
- (P5) Scalar Multiple Property: If each element of a row (or column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.
- Corollary: If A is a square matrix of order n, then |kA| = kⁿ |A|.
- (P6) Sum Property: If some or all elements of a row (or column) are expressed as the sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants.
Example:| a1+λ1 a2+λ2 a3+λ3 || a1 a2 a3 || λ1 λ2 λ3 |
| b1 b2 b3 |=| b1 b2 b3 |+| b1 b2 b3 |
| c1 c2 c3 || c1 c2 c3 || c1 c2 c3 | - (P7) Invariance Property: The value of the determinant remains the same if we apply the operation Rᵢ → Rᵢ + kRⱼ or Cᵢ → Cᵢ + kCⱼ. (Adding a multiple of one row/column to another row/column). This is very useful for introducing zeros to simplify calculations.
- (P8) Triangle Property: If all elements above or below the main diagonal are zero (triangular matrix), the determinant is the product of the diagonal elements.
5. Area of a Triangle:
- The area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by:
Area (Δ) = ½ | x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂) | - Using Determinants:
Area (Δ) = ½ | det(M) | where M =
[ x1 y1 1 ]
[ x2 y2 1 ]
[ x3 y3 1 ] - Note: Since area is always positive, we take the absolute value of the determinant calculation.
- Condition for Collinearity: Three points (x₁, y₁), (x₂, y₂), and (x₃, y₃) are collinear if and only if the area of the triangle formed by them is zero.
i.e.,[ x1 y1 1 ]
[ x2 y2 1 ]= 0
[ x3 y3 1 ]
6. Adjoint and Inverse of a Matrix:
- Adjoint (adj A): The adjoint of a square matrix A = [aij]n×n is the transpose of the matrix [Cij]n×n, where Cij is the cofactor of the element aij.
- Steps:
- Find the minor of each element.
- Find the cofactor of each element.
- Construct the matrix of cofactors.
- Transpose this cofactor matrix to get adj A.
- Steps:
- Important Theorems for Adjoint: For any square matrix A of order n:
- A (adj A) = (adj A) A = |A| Iₙ (where Iₙ is the identity matrix of order n).
- |adj A| = |A|ⁿ⁻¹
- adj(AB) = (adj B)(adj A)
- adj(Aᵀ) = (adj A)ᵀ
- Singular Matrix: A square matrix A is singular if |A| = 0.
- Non-Singular Matrix: A square matrix A is non-singular if |A| ≠ 0.
- Inverse of a Matrix (A⁻¹): A square matrix A is invertible (i.e., its inverse exists) if and only if A is non-singular (|A| ≠ 0).
The inverse is given by:
A⁻¹ = (1 / |A|) * adj A - Properties of Inverse:
- (A⁻¹)⁻¹ = A
- (AB)⁻¹ = B⁻¹ A⁻¹ (Reversal Law)
- (Aᵀ)⁻¹ = (A⁻¹)ᵀ
- |A⁻¹| = 1 / |A| = |A|⁻¹
7. Applications to Systems of Linear Equations (Matrix Method):
Consider a system of n linear equations in n variables:
a₁₁x₁ + a₁₂x₂ + ... + a₁nxn = d₁
a₂₁x₁ + a₂₂x₂ + ... + a₂nxn = d₂
...
an₁x₁ + an₂x₂ + ... + annxn = dn
This system can be written in matrix form as AX = B, where:
A = [ aij ] (coefficient matrix)
X = [ xj ] (variable matrix)
B = [ di ] (constant matrix)
-
Case 1: If |A| ≠ 0 (A is non-singular):
The system has a unique solution given by X = A⁻¹B.
The system is consistent. -
Case 2: If |A| = 0 (A is singular):
Calculate (adj A)B.- (a) If (adj A)B ≠ O (Null matrix):
The system has no solution.
The system is inconsistent. - (b) If (adj A)B = O (Null matrix):
The system may be either consistent with infinitely many solutions or inconsistent (no solution). Further investigation might be needed, but often points towards infinite solutions in typical problems.
- (a) If (adj A)B ≠ O (Null matrix):
Summary of Consistency:
- |A| ≠ 0 => Consistent, Unique Solution
- |A| = 0 and (adj A)B ≠ O => Inconsistent, No Solution
- |A| = 0 and (adj A)B = O => Consistent, Infinitely Many Solutions (Usually) OR Inconsistent
Multiple Choice Questions (MCQs):
-
If A =
[ 2 -1 ]
[ 4 3 ], then |A| is:
(A) 10
(B) 2
(C) -10
(D) 6 -
For what value of x is the matrix
[ 6-x 4 ]singular?
[ 3-x 1 ]
(A) 1
(B) 2
(C) -1
(D) -2 -
If A is a square matrix of order 3 and |A| = 5, then |adj A| is:
(A) 5
(B) 125
(C) 25
(D) 1/5 -
The area of the triangle with vertices (-2, -3), (3, 2), and (-1, -8) is:
(A) 15 sq. units
(B) 30 sq. units
(C) 15/2 sq. units
(D) 0 sq. units -
If A =
[ cosθ -sinθ ]
[ sinθ cosθ ], then A⁻¹ is equal to:
(A) A
(B) adj A
(C) -A
(D) Aᵀ -
If A is an invertible matrix of order 3 and |A| = 7, then |A⁻¹| is:
(A) 7
(B) 1/7
(C) 49
(D) 1/49 -
Consider the system of equations: x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30. This system has:
(A) A unique solution
(B) No solution
(C) Infinitely many solutions
(D) Exactly two solutions -
If A is a 3x3 matrix such that |A| = 4, then |2A| is equal to:
(A) 8
(B) 12
(C) 32
(D) 24 -
If
| x 2 |=| 3 2 |
| 18 x || 18 3 |, then x is equal to:
(A) 3
(B) 6
(C) ±6
(D) ±3 -
If A =
[ 1 2 ]
[ 3 4 ], then A(adj A) is equal to:
(A)[ -2 0 ]
[ 0 -2 ]
(B)[ 2 0 ]
[ 0 2 ]
(C)[ -2 1 ]
[ 1 -2 ]
(D)[ 1 0 ]
[ 0 1 ]
Answers to MCQs:
- (A) |A| = (2)(3) - (-1)(4) = 6 + 4 = 10
- (B) Singular means det = 0. (6-x)(1) - 4(3-x) = 0 => 6 - x - 12 + 4x = 0 => 3x - 6 = 0 => x = 2.
- (C) |adj A| = |A|ⁿ⁻¹ = 5³⁻¹ = 5² = 25.
- (D) Area = ½ | det
[ -2 -3 1 ]| = ½ | -2(-16 - (-2)) - (-3)(24 - (-1)) + 1(-24 - (-2)) | = ½ | -2(-14) + 3(25) + 1(-22) | = ½ | 28 + 75 - 22 | = ½ | 81 | = 40.5. Let's recheck calculation.
Area = ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |
= ½ | -2(2 - (-8)) + 3(-8 - (-3)) + (-1)(-3 - 2) |
= ½ | -2(10) + 3(-5) + (-1)(-5) |
= ½ | -20 - 15 + 5 | = ½ | -30 | = 15 sq. units. Correction: Previous determinant calculation was wrong. Let's re-calculate the determinant way:
Area = ½ | det[ -2 -3 1 ]|
[ 3 2 1 ]
[ -1 -8 1 ]
= ½ | -2(2 - (-8)) - (-3)(3 - (-1)) + 1(-24 - (-2)) |
= ½ | -2(10) + 3(4) + 1(-22) |
= ½ | -20 + 12 - 22 | = ½ | -30 | = 15 sq. units. My first answer C was correct, the check calculation was wrong initially. Let's re-evaluate the options. The answer is 15. So (A) is 15. Wait, the first check calculation result was 15, matching A. The second check calculation was also 15. Why did I write D earlier? Let's check for collinearity:det = -30. So area is 15. Answer is (A). - (B) or (D). |A| = cos²θ - (-sin²θ) = cos²θ + sin²θ = 1.
adj A =[ cosθ sinθ ]ᵀ =[ cosθ sinθ ]
[ -sinθ cosθ ][ -sinθ cosθ ]
A⁻¹ = (1/|A|) adj A = adj A.
Also, Aᵀ =[ cosθ sinθ ]
[ -sinθ cosθ ]. So A⁻¹ = adj A = Aᵀ. Both (B) and (D) seem correct based on calculation. However, the standard formula involves adj A. Let's stick with (B) as it directly relates to the inverse definition steps. But Aᵀ is also numerically correct. Often in MCQs, if multiple options are mathematically equivalent, check the context or standard definition. A⁻¹ = adj A / |A|. Here |A|=1, so A⁻¹ = adj A. Also, for orthogonal matrices like this rotation matrix, A⁻¹ = Aᵀ. Let's choose (B) as it's derived from the adjoint definition. Reconsidering: Aᵀ is also a valid representation. Which is 'more' correct? Both are true. Let's check NCERT convention. NCERT defines A⁻¹ using adjoint. Let's re-examine Aᵀ vs adj A. They are indeed identical here. Let's select D as Aᵀ is a property of orthogonal matrices. Final decision: (D) Aᵀ. - (B) |A⁻¹| = 1 / |A| = 1/7.
- (C) A =
[ 1 1 1 ], |A| = 1(14-12) - 1(7-3) + 1(4-2) = 1(2) - 1(4) + 1(2) = 2 - 4 + 2 = 0.
[ 1 2 3 ]
[ 1 4 7 ]
Calculate adj A. C11=2, C12=-4, C13=2, C21=-3, C22=6, C23=-3, C31=1, C32=-2, C33=1.
adj A =[ 2 -3 1 ]
[ -4 6 -2 ]
[ 2 -3 1 ]
B =[ 6 ]
[ 14 ]
[ 30 ]
(adj A)B =[ 2(6) - 3(14) + 1(30) ]=[ 12 - 42 + 30 ]=[ 0 ]
[ -4(6) + 6(14) - 2(30) ][ -24 + 84 - 60 ][ 0 ]
[ 2(6) - 3(14) + 1(30) ][ 12 - 42 + 30 ][ 0 ]
Since |A| = 0 and (adj A)B = O, the system has infinitely many solutions. - (C) |kA| = kⁿ |A|. Here n=3, k=2. |2A| = 2³ |A| = 8 * 4 = 32.
- (C) x(x) - 2(18) = 3(3) - 2(18) => x² - 36 = 9 - 36 => x² = 9 => x = ±3. Wait, calculation error.
x² - 36 = 9 - 36
x² = 9
x = ±3. Let me re-read the question.
| x 2 |=| 3 2 |
| 18 x || 18 3 |
x² - (18)(2) = (3)(3) - (2)(18)
x² - 36 = 9 - 36
x² = 9
x = ±3. The calculation seems correct. Let me check the options again. Options are 3, 6, ±6, ±3. My result is ±3. So option (D). Hold on, let me re-evaluate the second determinant.
| 3 2 |= 33 - 218 = 9 - 36 = -27.
| 18 3 |
So, x² - 36 = -27
x² = 36 - 27
x² = 9
x = ±3. The answer is (D). - (A) A(adj A) = |A| I.
|A| = (1)(4) - (2)(3) = 4 - 6 = -2.
|A| I = -2 *[ 1 0 ]=[ -2 0 ]
[ 0 1 ][ 0 -2 ].
Corrected Answers:
- A
- B
- C
- A (Area = 15 sq units)
- D (Aᵀ is equivalent to adj A here, and a property of orthogonal matrices)
- B
- C
- C
- D (x = ±3)
- A
Make sure you understand the properties thoroughly, as they often provide shortcuts for solving problems quickly, which is essential in competitive exams. Practice calculating determinants, adjoints, inverses, and solving systems of equations efficiently. Good luck!