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:**
For a 2×2 matrix A = [a b; c d], the determinant is calculated as |A| = ad - bc
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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.
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For a matrix A = [a₁₁ a₁₂; a₂₁ a₂₂] of order 2×2:
**|A| = a₁₁a₂₂ - a₂₁a₁₂**
**Worked Example 1:** Evaluate |2 4; -1 2|
**Worked Example 2:** Evaluate |x x+1; x-1 x|
**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).
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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.
**|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).
The determinant can be expanded along:
**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.
Since column 3 has two zeros, expand along C₃:
Expand along R₁:
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|
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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:**
Area = (1/2)|3 8 1; -4 2 1; 5 1 1|
Expanding along R₁:
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**
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The **minor Mᵢⱼ** of element aᵢⱼ is the determinant obtained by deleting the iᵗʰ row and jᵗʰ column from the original determinant.
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.
**Minors:**
**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.
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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:
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)
Using the shortcut formula:
**adj A = |4 -2; -3 1|**
Verification:
**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.
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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|
**Example of Non-Singular Matrix:** A = |1 2; 3 4|
**Significance:**
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**A square matrix A is invertible if and only if A is non-singular** (i.e., |A| ≠ 0)
If A is invertible:
**A⁻¹ = (1/|A|) × adj A**
**Verification Theorem:** If A is invertible, then:
**AA⁻¹ = A⁻¹A = I**
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|**
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:
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**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.
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Q1. Evaluate the determinant |2 4; −1 2|.
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|?
Answer: C — For 2×2 matrix: |3A| = 3² × |A| = 9 × 5 = 45.
Q3. Which determinant property states that swapping two rows changes the sign?
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|.
Answer: B — Expanding along R1 yields 0 − sinα(−sinβcosα) − cosα(sinαsinβ) = sinαsinβcosα − sinαsinβcosα = 0.
Q5. If |3 x; 2 4| = 3, find x.
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:
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?
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|
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.
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:
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.
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.
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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