Determinant of a 1 × 1 and 2 × 2 matrix
A determinant is a single number made from a square matrix. We write it as |A| or det A.
- 1 × 1: |[a]| = a.
- 2 × 2: |a b; c d| = ad − bc (down-right product minus up-right product).
Meaning: the columns (a, c) and (b, d) are arrows. The parallelogram they make has area |ad − bc|. See step 1 of the 3D. If the arrows lie on one line, the area is 0.
Determinant of a 3 × 3 matrix
Expand along any row or column. Along row 1:
|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃, where Aᵢⱼ are cofactors (next section).
Example: A = [2 1 3; 0 4 1; 1 2 0]. |A| = 2(4·0 − 1·2) − 1(0·0 − 1·1) + 3(0·2 − 4·1) = −4 + 1 − 12 = −15.
Tip: expand along the row or column with the most zeros – less work. Every choice gives the same answer. Useful facts: |Aᵀ| = |A|, |AB| = |A||B|, and for an n × n matrix |kA| = kⁿ|A|.
Minors and cofactors
The minor Mᵢⱼ of aᵢⱼ is the determinant left after deleting row i and column j.
The cofactor Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ. The sign follows a chessboard pattern: + − + / − + − / + − +.
In step 3 of the 3D, row 1 is expanded: M₁₁ = −2 → A₁₁ = −2; M₁₂ = −1 → A₁₂ = +1; M₁₃ = −4 → A₁₃ = −4. Then |A| = 2(−2) + 1(1) + 3(−4) = −15.
Useful fact: if you multiply a row by the cofactors of a different row and add, you always get 0.
Area of a triangle using determinants
For corners (x₁, y₁), (x₂, y₂), (x₃, y₃):
Area = ½ |det [x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1]|
Take the absolute value because area cannot be negative. If the determinant is 0, the three points are collinear (on one line) – no triangle. This also gives the equation of a line through two points: put (x, y) as the third point and set the determinant to 0.
Adjoint of a matrix
The adjoint adj A is the transpose of the matrix of cofactors.
For 2 × 2: adj [a b; c d] = [d −b; −c a] (swap a and d, change the signs of b and c).
Main result: A(adj A) = (adj A)A = |A| I. Also |adj A| = |A|ⁿ⁻¹ for an n × n matrix.
Inverse using the adjoint
If |A| ≠ 0 (A is non-singular): A⁻¹ = (1/|A|) adj A.
If |A| = 0 (A is singular), there is no inverse. Why? The columns squash the area (or volume) to 0, and nothing can "un-squash" it. Step 2 of the 3D shows the parallelogram going flat.
Always check: A · A⁻¹ = I.
Consistency of a system of linear equations
Write the equations as AX = B (A = coefficients, X = unknowns, B = right side).
- Consistent: has at least one solution. Inconsistent: no solution.
- |A| ≠ 0 → consistent with a unique solution.
- |A| = 0 and (adj A)B ≠ O → inconsistent (no solution).
- |A| = 0 and (adj A)B = O → infinitely many solutions or none; check by another method.
In 3D pictures: three planes meeting at one point (step 5) is the unique case.
Solving 2- and 3-variable systems by the inverse
- Write AX = B.
- Find |A|. If it is 0, stop and use the consistency test.
- Find the cofactors and adj A.
- A⁻¹ = adj A / |A|.
- X = A⁻¹B. Read x, y (and z).
- Check in every equation.
The free-play step in the 3D does steps 1 to 5 for any 2 × 2 system you make.
Try it: measure a triangle on the floor
Put three coins on floor tiles. Take one corner of the room as (0, 0) and count tiles to get each coin's (x, y). Use ½|det| to find the area of the coin triangle in tiles. Then check by counting whole and half tiles inside the triangle. In the 3D free play, set a, c and b, d to two sides of your triangle: the parallelogram area is twice your triangle's area.
Exam tip: expect a 5-mark question solving 3 equations by A⁻¹, plus 1–2 mark questions on |kA|, |adj A|, area and singular matrices.
Key formulas and definitions
- |a b; c d| = ad − bc
- |A| (3 × 3) = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ (expand along any row/column)
- Minor Mᵢⱼ = det after deleting row i, column j; cofactor Aᵢⱼ = (−1)ⁱ⁺ʲMᵢⱼ
- Area of triangle = ½ |det [x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1]|; = 0 ⇒ collinear
- adj A = (matrix of cofactors)ᵀ; A(adj A) = (adj A)A = |A| I
- A⁻¹ = (1/|A|) adj A, only if |A| ≠ 0
- |Aᵀ| = |A|, |AB| = |A||B|, |kA| = kⁿ|A|, |adj A| = |A|ⁿ⁻¹
- AX = B ⇒ X = A⁻¹B when |A| ≠ 0
- |A| = 0 and (adj A)B ≠ O ⇒ inconsistent
Worked examples
1. Find |4 3; 2 5|.
ad − bc = 4 × 5 − 3 × 2 = 20 − 6 = 14.
2. Write the minors and cofactors of A = [1 2; 3 4].
Delete row 1, col 1 → M₁₁ = 4; delete row 1, col 2 → M₁₂ = 3; M₂₁ = 2; M₂₂ = 1. Signs + − / − +: A₁₁ = 4, A₁₂ = −3, A₂₁ = −2, A₂₂ = 1.
3. Find |A| for A = [2 1 3; 0 4 1; 1 2 0].
Along row 1: 2·(4·0 − 1·2) − 1·(0·0 − 1·1) + 3·(0·2 − 4·1) = 2(−2) − 1(−1) + 3(−4) = −4 + 1 − 12 = −15.
4. Find the area of the triangle with corners (1, 1), (5, 1) and (1, 4).
det [1 1 1; 5 1 1; 1 4 1] = 1(1 − 4) − 1(5 − 1) + 1(20 − 1) = −3 − 4 + 19 = 12. Area = ½ × 12 = 6 square units. (Check: base 4, height 3, ½ × 4 × 3 = 6.)
5. The triangle with corners (k, 0), (4, 0), (0, 2) has area 4. Find k.
det [k 0 1; 4 0 1; 0 2 1] = k(0 − 2) − 0 + 1(8 − 0) = 8 − 2k. ½|8 − 2k| = 4 ⇒ |8 − 2k| = 8 ⇒ 8 − 2k = 8 or −8 ⇒ k = 0 or k = 8.
6. Find adj A and A⁻¹ for A = [2 1; 5 3].
|A| = 6 − 5 = 1. adj A = [3 −1; −5 2]. A⁻¹ = (1/1) adj A = [3 −1; −5 2]. Check: A·A⁻¹ = [6 − 5 −2 + 2; 15 − 15 −5 + 6] = I ✓.
7. Solve by the matrix method: 2x + y = 5, 5x + 3y = 13.
A = [2 1; 5 3], B = [5; 13]. From the last example A⁻¹ = [3 −1; −5 2]. X = A⁻¹B = [3·5 − 13; −5·5 + 2·13] = [2; 1]. So x = 2, y = 1. Check: 4 + 1 = 5 ✓, 10 + 3 = 13 ✓.
8. Solve x + y + z = 6, x − y + z = 2, 2x + y − z = 1 using A⁻¹.
A = [1 1 1; 1 −1 1; 2 1 −1], |A| = 1(1 − 1) − 1(−1 − 2) + 1(1 + 2) = 6. Cofactors: A₁₁ = 0, A₁₂ = 3, A₁₃ = 3, A₂₁ = 2, A₂₂ = −3, A₂₃ = 1, A₃₁ = 2, A₃₂ = 0, A₃₃ = −2. adj A = [0 2 2; 3 −3 0; 3 1 −2]. (adj A)B = [0 + 4 + 2; 18 − 6 + 0; 18 + 2 − 2] = [6; 12; 18]. X = (1/6)[6; 12; 18] = [1; 2; 3]. So x = 1, y = 2, z = 3.
9. Is the system x + 2y = 3, 2x + 4y = 7 consistent?
|A| = 1·4 − 2·2 = 0. adj A = [4 −2; −2 1]. (adj A)B = [4·3 − 2·7; −2·3 + 1·7] = [−2; 1] ≠ O. So the system is inconsistent: no solution (the lines are parallel).
Common mistakes
- Forgetting the minus sign on the middle term when expanding a 3 × 3 determinant (+ − +).
- Using the cofactor matrix itself as adj A; you must take its transpose.
- Writing |kA| = k|A|; for a 3 × 3 matrix it is k³|A|.
- Leaving the triangle area negative, or forgetting the ½.