Determinant of order 1 and 2
The area of a parallelogram with sides along (a, c) and (b, d) is a number. That number is the determinant of the 2 × 2 matrix [a b ; c d].
The determinant of a square matrix is a single number. For order 1 it is the entry itself. For order 2 it is the product of the diagonal entries minus the product of the other two. A determinant is written with straight bars, here as |a b ; c d|.
Formula: |a b ; c d| = ad − bc
Rule 1: Evaluate a 2 × 2 determinant — Multiply down the main diagonal, subtract the other diagonal.
- Easy: |3 5 ; 2 4| = 12 − 10 = 2.
- Medium: |−2 3 ; 4 1| = −2 − 12 = −14.
- Exam-type: |sin θ −cos θ ; cos θ sin θ| = sin²θ + cos²θ = 1.
Rule 2: Use it to find x — Form the equation ad − bc = given number.
- Easy: |x 3 ; 2 x − 1| = 0 gives x² − x − 6 = 0, so x = 3 or x = −2.
- Medium: |2x 3 ; 4 x| = 20 gives 2x² − 12 = 20, so x = 4 or −4.
- Exam-type: |A| = −2 for A = [1 2 ; 3 4] and |2A| = |2 4 ; 6 8| = −8, which is 2² × |A| (order 2).
A 2 × 2 determinant is zero exactly when the rows are proportional.
Subtracting in the wrong order (bc − ad instead of ad − bc).
"Down minus up": main diagonal minus the other diagonal.
Practice questions
- 1Evaluate |4 1 ; 2 3|.
- 2Evaluate |5 −2 ; 3 0|.
- 3Evaluate |2 3 ; 4 6|.
- 4Find x if |x 2 ; 5 3| = 8.
- 5Find x if |x 3 ; 2 x − 1| = 0.
- 6Find x if |x + 3 2 ; 4 x + 1| = 0.
- 7For A = [2 −1 ; 3 4] find |A| and |3A|.
- 8Evaluate |sin 30° cos 30° ; −cos 30° sin 30°|.
Show answers (try them yourself first)
- 112 − 2 = 10
- 20 + 6 = 6
- 312 − 12 = 0
- 43x − 10 = 8, so x = 6
- 5x² − x − 6 = 0, so x = 3 or −2
- 6(x + 3)(x + 1) − 8 = x² + 4x − 5 = 0, so x = 1 or −5
- 7|A| = 8 + 3 = 11 and |3A| = 9 × 11 = 99
- 8sin²30° + cos²30° = 1
Determinant of order 3
A 3 × 3 determinant gives the volume of a box built from three vectors. It is computed by breaking the problem into three 2 × 2 determinants.
Expand along any row or column. Each entry is multiplied by the determinant left after deleting its row and column (its minor), with signs following the pattern + − + / − + − / + − +. Expanding along a row or column with zeros saves work.
Formula: |A| = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁)
Rule 1: Expand along the first row — Use the signs + − +.
- Easy: |1 2 3 ; 0 4 5 ; 0 0 6| = 24, the product of the diagonal (zeros below).
- Medium: |2 0 1 ; 1 3 2 ; 0 5 4| = 2(12 − 10) − 0 + 1(5 − 0) = 9.
- Exam-type: |1 2 3 ; 4 5 6 ; 7 8 9| = 1(45 − 48) − 2(36 − 42) + 3(32 − 35) = 0.
Rule 2: Choose the row or column with most zeros — Fewer terms, fewer mistakes.
- Easy: |3 0 0 ; 2 5 1 ; 4 1 2| = 3(10 − 1) = 27.
- Medium: |1 2 0 ; 3 4 0 ; 5 6 7| expanded along column 3 gives 7 (4 − 6) = −14.
- Exam-type: |0 2 1 ; 3 0 4 ; 5 1 0| = −2(0 − 20) + 1(3 − 0) = 43.
For a triangular matrix (all zeros above or below the diagonal) the determinant is the product of the diagonal entries.
Forgetting the middle minus sign in + − +.
Checkerboard of signs: plus, minus, plus.
Practice questions
- 1Evaluate |1 0 0 ; 0 2 0 ; 0 0 3|.
- 2Evaluate |2 1 3 ; 0 4 5 ; 0 0 6|.
- 3Evaluate |1 2 3 ; 4 5 6 ; 7 8 9|.
- 4Evaluate |1 2 0 ; 0 3 1 ; 4 0 2|.
- 5Evaluate |0 1 2 ; 1 0 3 ; 2 3 0|.
- 6Evaluate |2 −1 3 ; 1 0 4 ; 3 2 1|.
- 7Find x if |x 1 0 ; 0 x 1 ; 1 0 x| = 0 (real x).
- 8Evaluate |3 1 2 ; 0 2 1 ; 0 0 4|.
Show answers (try them yourself first)
- 16
- 22 × 4 × 6 = 48
- 30
- 41(6 − 0) − 2(0 − 4) + 0 = 14
- 50 − 1(0 − 6) + 2(3 − 0) = 12
- 62(0 − 8) + 1(1 − 12) + 3(2 − 0) = −21
- 7x³ + 1 = 0, so x = −1
- 83 × 2 × 4 = 24
Properties of determinants
Doing the same tidy-up on a table of numbers (swapping rows, scaling a row, subtracting rows) changes the determinant in predictable ways. This lets us simplify before expanding.
(1) |A′| = |A|. (2) Swapping two rows (or columns) changes the sign. (3) If two rows are equal or proportional, |A| = 0. (4) Multiplying one row by k multiplies |A| by k. (5) Adding a multiple of one row to another changes nothing. (6) |AB| = |A||B|. (7) For order n, |kA| = kⁿ|A|. (8) |A⁻¹| = 1/|A|.
Formula: |AB| = |A||B|, |kA| = kⁿ |A|, |A⁻¹| = 1 / |A|
Rule 1: Use row and column operations — Create zeros without changing the value.
- Easy: |2 4 ; 1 3| = 2, and swapping rows gives |1 3 ; 2 4| = −2.
- Medium: Multiplying row 1 of |2 4 ; 1 3| by 3 gives |6 12 ; 1 3| = 6 = 3 × 2.
- Exam-type: R₂ → R₂ − 2R₁ on |1 2 ; 3 4| gives |1 2 ; 1 0| = −2, the same value.
Rule 2: Use the product and power rules — No expansion needed.
- Easy: A = [1 2 ; 0 1], B = [2 0 ; 1 3]: |A| = 1, |B| = 6, and AB = [4 6 ; 1 3] has |AB| = 6.
- Medium: If |A| = 5 for a 3 × 3 matrix, |2A| = 2³ × 5 = 40.
- Exam-type: If |A| = 4, then |A⁻¹| = ¼.
Scaling the whole 3 × 3 matrix by k scales the determinant by k³, not k.
Writing |kA| = k|A| for a matrix of order 2 or 3.
Same row twice means zero: the box has no volume.
Practice questions
- 1If |A| = 7, find |A′|.
- 2Evaluate |1 2 ; 2 4| without expanding.
- 3For a 2 × 2 matrix with |A| = −3, find |4A|.
- 4For 3 × 3 matrices with |A| = 2 and |B| = 5, find |AB|.
- 5For a 3 × 3 matrix with |A| = 6, find |A⁻¹|.
- 6Two rows of a determinant with value 9 are swapped. What is the new value?
- 7Evaluate |3 6 9 ; 1 2 3 ; 4 5 6| without expanding.
- 8Show |1 2 3 ; 1 3 5 ; 1 4 7| = 0 without expanding.
Show answers (try them yourself first)
- 17
- 20, since row 2 is twice row 1
- 316 × (−3) = −48
- 410
- 51/6
- 6−9
- 70, since row 1 is three times row 2
- 8Column 3 equals 2 × column 2 − column 1 (3 = 4 − 1, 5 = 6 − 1, 7 = 8 − 1), so the determinant is 0