10 min readConcept sheet1-mark MCQ and assertion-reason; often a 2-mark question

Determinant of a square matrix

Every square matrix A has a number linked to it, its determinant |A|. Expand along any row or column; the value is the same.

2 × 2 determinant|a, b; c, d| = ad − bc (we write rows separated by a semicolon)

3 × 3 along row 1Δ = a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃, where each M is the 2 × 2 determinant left after deleting that element's row and column

Scalar multiple|kA| = kⁿ|A| for a square matrix A of order n

Singular matrixA square matrix with |A| = 0; it is non-singular if |A| ≠ 0

What a determinant is

A determinant is a single number linked to a square matrix (same number of rows and columns). We write it |A|, det A or Δ. For a 1 × 1 matrix A = [a], |A| = a. Here the bars do not mean modulus: a determinant can be negative. In Haya we write a matrix row by row, with rows separated by a semicolon: [2, 3; 1, 4] has rows 2, 3 and 1, 4.

How to expand a 3 × 3 determinant along row 1

Take Δ = |a₁, a₂, a₃; b₁, b₂, b₃; c₁, c₂, c₃|.

1
Take a₁. Cover its row and column. The 2 × 2 determinant left is b₂c₃ − b₃c₂. Write + a₁(b₂c₃ − b₃c₂).
2
Take a₂. Cover its row and column. Write − a₂(b₁c₃ − b₃c₁).
The middle term of row 1 always gets a minus sign.
3
Take a₃. Cover its row and column. Write + a₃(b₁c₂ − b₂c₁).
4★
Add: Δ = a₁(b₂c₃ − b₃c₂) − a₂(b₁c₃ − b₃c₁) + a₃(b₁c₂ − b₂c₁).
5
Along any other row or column, take the signs from the pattern + − + ; − + − ; + − +.

★ key step

Rules to remember

Only a square matrix has a determinant.
Expanding along any row or column gives the same value. Pick the line with the most zeros to save work.
|kA| = kⁿ|A| for order n, not k|A|. For a 3 × 3 matrix, |2A| = 8|A|.
If |A| = 0, A is singular. If |A| ≠ 0, A is non-singular.

Example 1: a 3 × 3 determinant

Problem

Evaluate Δ = |2, −1, 3; 0, 4, 0; 1, 5, −2|.

Working

Row 2 is 0, 4, 0. It has two zeros, so expand along row 2.
The sign at position (2, 2) is +.
Delete row 2 and column 2. What is left is |2, 3; 1, −2|.
|2, 3; 1, −2| = 2 × (−2) − 3 × 1 = −4 − 3 = −7
Δ = 0 + 4 × (−7) + 0 = −28
Check along row 1: 2(−8 − 0) − (−1)(0 − 0) + 3(0 − 4) = −16 + 0 − 12 = −28.

Answer

Δ = −28

Example 2: find x

Problem

Find x if |2x, 5; 8, x| = |6, −2; 7, 3|.

Working

Left side: 2x × x − 5 × 8 = 2x² − 40
Right side: 6 × 3 − (−2) × 7 = 18 + 14 = 32
So 2x² − 40 = 32, which gives 2x² = 72.
x² = 36, so x = 6 or x = −6.

Answer

x = ±6

Example 3: determinant of kA

Problem

A is a square matrix of order 3 and |A| = −4. Find |3A|.

Working

Use |kA| = kⁿ|A| with n = 3 and k = 3.
|3A| = 3³ × (−4) = 27 × (−4) = −108

Answer

|3A| = −108

In real life

When a phone app stretches or rotates a photo, it uses a matrix. The determinant of that matrix tells how many times the picture's area grows; a determinant of 0 squashes the picture flat into a line.

Exam angle

1 marks: “If A is a square matrix of order 3 and |A| = −2, find |4A|.” Use |kA| = kⁿ|A| with n = 3: |4A| = 4³ × (−2) = 64 × (−2) = −128.

2 marks: “For what values of x is the matrix [3 − x, 2; 1, 4 − x] singular?” Set |A| = 0: (3 − x)(4 − x) − 2 = 0, so x² − 7x + 10 = 0, that is (x − 2)(x − 5) = 0. x = 2 or x = 5.

1 marks: “Evaluate |cos θ, −sin θ; sin θ, cos θ|.” cos θ × cos θ − (−sin θ)(sin θ) = cos²θ + sin²θ = 1.

Remember it

Plus, minus, plus along the top row: cover the row and column, then multiply by what is left.