Minors and cofactors
The minor Mᵢⱼ is the determinant left after deleting row i and column j. The cofactor Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ adds a sign.
Minor and cofactor
Pick any element aᵢⱼ of a square matrix (i is its row, j its column). Delete row i and column j. The determinant of what is left is the minor Mᵢⱼ. Attach the sign (−1)ⁱ⁺ʲ to get the cofactor Aᵢⱼ. In a 3 × 3 matrix each minor is a 2 × 2 determinant. In a 2 × 2 matrix each minor is a single number. We write matrices with rows separated by a semicolon.
Sign of the cofactor in a 3 × 3 matrix
| Column 1 | Column 2 | Column 3 | |
|---|---|---|---|
| Row 1 | + | − | + |
| Row 2 | − | + | − |
| Row 3 | + | − | + |
Example 1: a 2 × 2 matrix
Find all the minors and cofactors of A = [2, −3; 4, 5], and check that a₁₁A₁₁ + a₁₂A₁₂ = |A|.
- M₁₁: delete row 1 and column 1; 5 is left. M₁₁ = 5, so A₁₁ = +5.
- M₁₂: delete row 1 and column 2; 4 is left. M₁₂ = 4, so A₁₂ = −4.
- M₂₁ = −3, so A₂₁ = −(−3) = 3.
- M₂₂ = 2, so A₂₂ = +2.
- a₁₁A₁₁ + a₁₂A₁₂ = 2 × 5 + (−3) × (−4) = 10 + 12 = 22
- |A| = 2 × 5 − (−3) × 4 = 10 + 12 = 22. The two agree.
M₁₁ = 5, M₁₂ = 4, M₂₁ = −3, M₂₂ = 2; A₁₁ = 5, A₁₂ = −4, A₂₁ = 3, A₂₂ = 2; |A| = 22
Example 2: Δ from the cofactors of one row
For Δ = |1, 2, 3; 0, 4, −1; 2, 1, 5|, find the cofactors of the elements of the second row and use them to find Δ.
- A₂₁ = (−1)³ |2, 3; 1, 5| = −(10 − 3) = −7
- A₂₂ = (−1)⁴ |1, 3; 2, 5| = 5 − 6 = −1
- A₂₃ = (−1)⁵ |1, 2; 2, 1| = −(1 − 4) = 3
- Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ = 0 × (−7) + 4 × (−1) + (−1) × 3
- Δ = 0 − 4 − 3 = −7
- Check the zero-sum rule: row 1 with the row 2 cofactors gives 1(−7) + 2(−1) + 3(3) = −7 − 2 + 9 = 0.
A₂₁ = −7, A₂₂ = −1, A₂₃ = 3 and Δ = −7
Minor vs cofactor
Δ is found with cofactors, never with bare minors.
Graphics software for mobile games often inverts small 3 × 3 and 4 × 4 matrices using cofactors. For such small sizes, the cofactor formula is quick and exact.
1 marks: “Write the cofactor of a₁₂ in [3, 1, 2; −4, 5, 0; 6, −2, 7].” M₁₂ = |−4, 0; 6, 7| = −28 − 0 = −28. A₁₂ = (−1)³ × (−28) = 28.
2 marks: “If Aᵢⱼ is the cofactor of aᵢⱼ in Δ = |2, 1, 3; 4, −1, 0; 5, 2, 6|, find a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃.” These are the elements of row 1 with the cofactors of row 2, a different row, so the sum is 0. (Check: A₂₁ = 0, A₂₂ = −3, A₂₃ = 1, and 2(0) + 1(−3) + 3(1) = 0.)
3 marks: “Using the cofactors of the elements of the second column, evaluate Δ = |2, −1, 3; 1, 0, 4; −2, 5, 1|.” A₁₂ = −9, A₂₂ = 8, A₃₂ = −5. Δ = (−1)(−9) + 0(8) + 5(−5) = 9 − 25 = −16.
Minor = no sign; cofactor = minor with its sign. Even i + j keeps the sign, odd i + j flips it.