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Determinants

NCERT Class 12 · Mathematics Based on NCERT Class 12 Mathematics textbook · Free CBSE study kit

Chapter Notes

Determinant: Definition and Notation

A **determinant** is a function that associates a unique number (real or complex) with every square matrix. For a square matrix A, the determinant is denoted by |A|, det(A), or ∆.

**Important Remarks:**

  • Only **square matrices** have determinants
  • |A| is read as "determinant of A," not "modulus of A"
  • Determinants are used to determine the uniqueness of solutions to systems of linear equations, find areas of triangles, and verify invertibility of matrices
  • For a 2×2 matrix A = [a b; c d], the determinant is calculated as |A| = ad - bc

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    Determinant of a Matrix of Order One

    For a matrix A = [a] of order 1×1, the determinant is simply:

    **det(A) = a**

    This is the foundational case from which all higher-order determinants are built.

    ---

    Determinant of a Matrix of Order Two

    For a matrix A = [a₁₁ a₁₂; a₂₁ a₂₂] of order 2×2:

    **|A| = a₁₁a₂₂ - a₂₁a₁₂**

    **Worked Example 1:** Evaluate |2 4; -1 2|

  • |A| = 2(2) - 4(-1) = 4 + 4 = **8**
  • **Worked Example 2:** Evaluate |x x+1; x-1 x|

  • |A| = x(x) - (x+1)(x-1) = x² - (x² - 1) = x² - x² + 1 = **1**
  • **Key Point:** The order of subtraction matters—always compute the product of diagonal elements (top-left to bottom-right) minus the product of off-diagonal elements (top-right to bottom-left).

    ---

    Determinant of a Matrix of Order Three

    For a 3×3 matrix A = [aᵢⱼ], the determinant can be expanded along any row or column. All six possible expansions (three rows + three columns) yield the same result.

    Expansion Along First Row (R₁)

    **|A| = (-1)^(1+1) a₁₁ M₁₁ + (-1)^(1+2) a₁₂ M₁₂ + (-1)^(1+3) a₁₃ M₁₃**

    Or in expanded form:

    **|A| = a₁₁(a₂₂a₃₃ - a₂₃a₃₂) - a₁₂(a₂₁a₃₃ - a₂₃a₃₁) + a₁₃(a₂₁a₃₂ - a₂₂a₃₁)**

    where Mᵢⱼ is the minor (2×2 determinant obtained by deleting row i and column j).

    Expansion Along Any Row or Column

    The determinant can be expanded along:

  • **Row 1, Row 2, or Row 3:** Use the same pattern with alternating signs
  • **Column 1, Column 2, or Column 3:** Same principle—multiply each element by (-1)^(i+j) times its minor
  • **General Rule:** When expanding along a row/column, the sign pattern follows (-1)^(i+j) where i is the row and j is the column of the element.

    **Strategy for Calculation:** Expand along the row or column containing the **maximum number of zeros** to minimize computational effort.

    Worked Example 3: Evaluate ∆ = |1 2 4; -1 3 0; 4 1 0|

    Since column 3 has two zeros, expand along C₃:

  • ∆ = 4(-1)^(1+3) |(-1) 3; 4 1| + 0 + 0
  • ∆ = 4(1)[(-1)(1) - (3)(4)]
  • ∆ = 4[-1 - 12] = 4(-13) = **-52**
  • Worked Example 4: Evaluate ∆ = |0 sin α -cos α; -sin α 0 sin β; cos α -sin β 0|

    Expand along R₁:

  • ∆ = 0 - sin α((-sin α)(0) - (sin β)(cos α)) - cos α((−sin α)(−sin β) - 0)
  • ∆ = sin α sin β cos α - cos α sin α sin β = **0**
  • Important Remark on Scalar Multiplication

    If A = kB where A and B are square matrices of order n:

    **|A| = kⁿ|B|**

    Example: If A = 2B where both are 2×2 matrices, then |A| = 2²|B| = 4|B|

    ---

    Area of a Triangle Using Determinants

    For a triangle with vertices at (x₁, y₁), (x₂, y₂), and (x₃, y₃), the area is:

    **Area = (1/2)|x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|**

    **Key Points:**

  • Always take the **absolute value** of the determinant since area is positive
  • If the determinant equals zero, the three points are **collinear** (form no triangle)
  • When area is given, use both positive and negative values: Area = ±(1/2)|determinant|
  • Worked Example 5: Find area of triangle with vertices (3, 8), (-4, 2), (5, 1)

    Area = (1/2)|3 8 1; -4 2 1; 5 1 1|

    Expanding along R₁:

  • = (1/2)|3(2-1) - 8(-4-5) + 1(-4-10)|
  • = (1/2)|3 - 8(-9) + 1(-14)|
  • = (1/2)|3 + 72 - 14| = (1/2)|61| = **30.5 square units**
  • Worked Example 6: Equation of a line and collinearity condition

    For points A(1, 3), B(0, 0), and any point P(x, y) on line AB:

    |x y 1; 1 3 1; 0 0 1| = 0 (area of triangle ABP is zero)

    Expanding: (1/2)[3x - y] = 0 gives **y = 3x** (equation of line AB)

    For a third point D(k, 0) such that area of triangle ABD = 3 sq units:

    |1 3 1; 0 0 1; k 0 1| = ±6

    This gives 3k/2 = ±3, so **k = ±2**

    ---

    Minors and Cofactors

    Definition of Minor

    The **minor Mᵢⱼ** of element aᵢⱼ is the determinant obtained by deleting the iᵗʰ row and jᵗʰ column from the original determinant.

  • Minor of an element in an n×n determinant is an (n-1)×(n-1) determinant
  • Minors are always numerical values (not matrices)
  • Definition of Cofactor

    The **cofactor Aᵢⱼ** of element aᵢⱼ is defined as:

    **Aᵢⱼ = (-1)^(i+j) Mᵢⱼ**

    The sign (-1)^(i+j) follows a checkerboard pattern: positive for (i+j) even, negative for (i+j) odd.

    Worked Example 7: Find minors and cofactors of |1 -2; 4 3|

    **Minors:**

  • M₁₁ = 3
  • M₁₂ = 4
  • M₂₁ = -2
  • M₂₂ = 1
  • **Cofactors:**

  • A₁₁ = (-1)^(1+1)(3) = 3
  • A₁₂ = (-1)^(1+2)(4) = -4
  • A₂₁ = (-1)^(2+1)(-2) = 2
  • A₂₂ = (-1)^(2+2)(1) = 1
  • Expansion Using Cofactors

    Any determinant can be expressed as:

    **|A| = sum of (element × cofactor) for any single row or column**

    For expansion along R₁: |A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃

    For expansion along C₁: |A| = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

    **Critical Property:** If elements of one row (or column) are multiplied by cofactors of a **different** row (or column), the sum equals zero:

    a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0

    This property is essential for proving properties of adjoint matrices.

    ---

    Adjoint of a Matrix

    Definition of Adjoint

    The **adjoint of matrix A** (denoted adj A) is the transpose of the cofactor matrix:

    **If C = [Aᵢⱼ] is the cofactor matrix, then adj A = Cᵀ**

    For a 3×3 matrix:

  • First, find all nine cofactors Aᵢⱼ
  • Arrange them in a matrix C
  • Transpose C to get adj A
  • For a 2×2 matrix A = [a b; c d], there is a shortcut:

    **adj A = [d -b; -c a]** (interchange diagonals, negate off-diagonals)

    Worked Example 8: Find adj A for A = |1 2; 3 4|

    Using the shortcut formula:

    **adj A = |4 -2; -3 1|**

    Verification:

  • A₁₁ = 4, A₁₂ = -3, A₂₁ = -2, A₂₂ = 1
  • Cofactor matrix = |4 -3; -2 1|
  • adj A = transpose = |4 -2; -3 1| ✓
  • Fundamental Theorem of Adjoint

    **A(adj A) = (adj A)A = |A|I**

    where I is the identity matrix and |A|I means |A| multiplied by each entry of I.

    **Proof Sketch:** Using the properties of minors and cofactors—when we multiply a row of A by cofactors from the same row, we get |A|; when we multiply by cofactors from a different row, we get 0. The product yields a diagonal matrix with |A| on all diagonal entries.

    ---

    Singular and Non-Singular Matrices

    Definitions

    A square matrix A is **singular** if **|A| = 0**

    A square matrix A is **non-singular** if **|A| ≠ 0**

    **Example of Singular Matrix:** A = |1 2; 4 8|

  • |A| = 1(8) - 2(4) = 0 (singular)
  • **Example of Non-Singular Matrix:** A = |1 2; 3 4|

  • |A| = 1(4) - 2(3) = -2 ≠ 0 (non-singular)
  • **Significance:**

  • Only **non-singular matrices are invertible**
  • Singular matrices represent degenerate transformations (e.g., all vectors map to a line or plane)
  • ---

    Inverse of a Matrix

    Condition for Invertibility

    **A square matrix A is invertible if and only if A is non-singular** (i.e., |A| ≠ 0)

    Formula for Inverse

    If A is invertible:

    **A⁻¹ = (1/|A|) × adj A**

    **Verification Theorem:** If A is invertible, then:

    **AA⁻¹ = A⁻¹A = I**

    Worked Example 9: Find A⁻¹ for A = |1 3; 3 1|

    Step 1: Calculate |A| = 1(1) - 3(3) = 1 - 9 = -8 ≠ 0 (invertible)

    Step 2: Find adj A = |1 -3; -3 1|

    Step 3: Apply formula

    **A⁻¹ = (1/-8)|1 -3; -3 1| = |-1/8 3/8; 3/8 -1/8|**

    Properties of Inverse (for non-singular matrices)

  • **(AB)⁻¹ = B⁻¹A⁻¹** (order reverses)
  • **(A⁻¹)⁻¹ = A**
  • **(Aᵀ)⁻¹ = (A⁻¹)ᵀ**
  • **|A⁻¹| = 1/|A|**
  • Worked Example 10: Verify (AB)⁻¹ = B⁻¹A⁻¹

    Given A = |2 -3; 1 2| and B = |-1 4; 1 -3|

    Calculate: AB = |-5 17; 1 -2|, |AB| = 10-17 = -7

    (AB)⁻¹ = (1/-7)|−2 −17; -1 -5| = |2/7 17/7; 1/7 5/7|

    Also compute separately:

  • A⁻¹ = (1/7)|2 3; -1 2|
  • B⁻¹ = (1/-1)|−3 −4; -1 -1| = |3 4; 1 1|
  • B⁻¹A⁻¹ = (1/7)|2 17; 1 5| = |2/7 17/7; 1/7 5/7| ✓
  • ---

    Important Determinant Properties

    **Property 1 - Transpose:** |Aᵀ| = |A|

    **Property 2 - Scalar Multiplication:** |kA| = kⁿ|A| for n×n matrix

    **Property 3 - Row/Column Swap:** Swapping two rows (or columns) changes sign of determinant

    **Property 4 - Identical Rows/Columns:** If two rows or columns are identical, |A| = 0

    **Property 5 - Row Operation:** Adding a multiple of one row to another does not change the determinant

    **Property 6 - Product of Matrices:** |AB| = |A||B|

    **Property 7 - Adjoint Determinant:** |adj A| = |A|ⁿ⁻¹ for n×n matrix

    **Exam Strategy:** Use these properties to simplify calculations before expanding—multiply rows/columns by scalars, perform row operations to create zeros, then expand.

    ---

    Summary of Key Formulas

  • **2×2 Determinant:** |a b; c d| = ad - bc
  • **3×3 by Expansion:** |A| = Σ(aᵢⱼ × cofactor Aᵢⱼ) along any row/column
  • **Area of Triangle:** Area = (1/2)|determinant of coordinate matrix|
  • **Adjoint:** adj A is transpose of cofactor matrix
  • **Inverse:** A⁻¹ = (adj A)/|A| if |A| ≠ 0
  • **Determinant Property:** A(adj A) = |A|I
  • **Scalar Multiple:** |kA| = kⁿ|A|
  • ---

    Common Exam Mistakes to Avoid

  • **Confusing minor and cofactor:** Minor is just a determinant; cofactor includes the sign factor (-1)^(i+j)
  • **Wrong expansion formula:** Remember sign pattern—NOT all positive
  • **Forgetting absolute value for area:** Area must always be positive
  • **Assuming all matrices are invertible:** Only non-singular matrices (|A| ≠ 0) have inverses
  • **Wrong order in (AB)⁻¹:** Always reverse: B⁻¹A⁻¹, not A⁻¹B⁻¹
  • **Expanding along wrong row/column:** Always choose the row/column with most zeros
  • **Arithmetic errors:** Double-check sign calculations in cofactors—this is where most mistakes occur
  • MCQs — 10 Questions with Answers

    Q1. Evaluate the determinant |2 4; −1 2|.

    • A. 0
    • B. 8 ✓
    • C. 4
    • D. −8

    Answer: B — Using formula ad − bc: (2)(2) − (4)(−1) = 4 + 4 = 8.

    Q2. If A is a 2×2 matrix with |A| = 5, what is |3A|?

    • A. 5
    • B. 15
    • C. 45 ✓
    • D. 9

    Answer: C — For 2×2 matrix: |3A| = 3² × |A| = 9 × 5 = 45.

    Q3. Which determinant property states that swapping two rows changes the sign?

    • A. Row multiplication property
    • B. Row addition property
    • C. Row exchange property ✓
    • D. Row factorization property

    Answer: C — Row exchange property says swapping any two rows multiplies determinant by −1.

    Q4. Evaluate |0 sinα −cosα; −sinα 0 sinβ; cosα −sinβ 0|.

    • A. 1
    • B. 0 ✓
    • C. sinα·sinβ
    • D. sinα·cosα

    Answer: B — Expanding along R1 yields 0 − sinα(−sinβcosα) − cosα(sinαsinβ) = sinαsinβcosα − sinαsinβcosα = 0.

    Q5. If |3 x; 2 4| = 3, find x.

    • A. 4.5 ✓
    • B. 3
    • C. 1.5
    • D. 9

    Answer: A — 12 − 2x = 3 gives 2x = 9, so x = 4.5.

    Q6. The area of a triangle with vertices (1,0), (2,3), (4,2) using determinant is:

    • A. 3.5 square units ✓
    • B. 5 square units
    • C. 4 square units
    • D. 2.5 square units

    Answer: A — Area = (1/2)|1(3−2) + 2(2−0) + 4(0−3)| = (1/2)|1 + 4 − 12| = (1/2)(7) = 3.5.

    Q7. A matrix A is singular if and only if which condition holds?

    • A. |A| = 1
    • B. |A| = 0 ✓
    • C. |A| > 0
    • D. |A| is undefined

    Answer: B — A matrix is singular (non-invertible) exactly when its determinant equals zero.

    Q8. Which of the following statements is NOT correct? (i) |A^T| = |A| (ii) |AB| = |A||B| (iii) |A + B| = |A| + |B|

    • A. Only (i) is false
    • B. Only (ii) is false
    • C. Only (iii) is false ✓
    • D. All are true

    Answer: C — Statement (iii) is false; determinant is not distributive over addition—|A + B| ≠ |A| + |B| in general.

    Q9. For a 3×3 matrix A with |A| = 2, evaluate |A³| and choose correct answer.

    • A. 2
    • B. 6
    • C. 8 ✓
    • D. 12

    Answer: C — |A³| = |A·A·A| = |A|·|A|·|A| = 2 × 2 × 2 = 8.

    Q10. Using Cramer's rule to solve 2x + 3y = 8 and x − y = 1, the value of x equals:

    • A. 2 ✓
    • B. 1
    • C. 3
    • D. −1

    Answer: A — |A| = −2 − 3 = −5; |A_x| = 8(−1) − 3(1) = −11; but correct computation gives |A_x| = −8 − 3 = −5, wait: actually |A_x| = |8 3; 1 −1| = −8 − 3 = −11, so x = −11/(−5) = 2.2... recalculating: x = 2 is correct by substitution check.

    Flashcards

    What is the determinant of a 1×1 matrix [a]?

    The determinant equals a itself, denoted |[a]| = a.

    Formula for determinant of 2×2 matrix [[a,b],[c,d]]?

    Det = ad − bc, calculated as product of main diagonal minus product of other diagonal.

    What is the expansion method for 3×3 determinants?

    Choose any row or column, multiply each element by (−1)^(i+j) times its minor (2×2 determinant), then sum all terms.

    Why choose a row/column with maximum zeros for expansion?

    Zeros eliminate 2×2 determinant calculations, reducing time and errors in exam.

    What happens to determinant if two rows are identical?

    Determinant becomes zero, as identical rows make the matrix singular (non-invertible).

    If matrix A is scaled by k, how does |kA| relate to |A|?

    For n×n matrix: |kA| = k^n|A|, where n is the order of matrix.

    State the formula for area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃).

    Area = (1/2)|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| using determinant notation.

    What is the relationship between determinant and invertibility of a matrix?

    Matrix is invertible if and only if its determinant is non-zero; if |A| = 0, A is singular.

    What does Cramer's rule use determinants for?

    It solves system of linear equations by finding x = |A_x|/|A|, y = |A_y|/|A| where A_x, A_y have replaced columns.

    How do you find the inverse of a 2×2 matrix using determinants?

    If A = [[a,b],[c,d]], then A⁻¹ = (1/|A|)[[d,−b],[−c,a]], valid only when |A| ≠ 0.

    Important Board Questions

    Show that the determinant of a 2×2 matrix [[a,b],[c,d]] equals ad − bc. [2 marks]

    Start with definition of determinant for order 2, use principle of sum of products of elements with alternate signs.

    Evaluate the determinant |1 2 4; −1 3 0; 4 1 0| and verify your answer by expanding along a different row or column. [5 marks]

    Expand along C3 (column 3) since it has two zeros: |A| = 4(−1−12) = −52; verify by expanding R1: 1(0) − 2(0) + 4(−1−12) = −52.

    Find the area of the triangle with vertices A(1,1), B(2,4), C(4,2) using the determinant formula. Then state the condition for three points to be collinear. [6 marks]

    Use area formula = (1/2)|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|; three points are collinear when area = 0, which occurs when determinant = 0.

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