Minors and cofactors of a determinant
Crossing out one row and one column of a seating chart leaves a smaller chart. The value of that smaller chart is the minor of the seat you crossed out.
The minor Mᵢⱼ of an entry aᵢⱼ is the determinant left after deleting row i and column j. The cofactor is Aᵢⱼ = (−1)^(i + j) Mᵢⱼ, so it carries the sign from the checkerboard + − + / − + − / + − +.
Formula: Aᵢⱼ = (−1)^(i + j) Mᵢⱼ
Rule 1: Find a minor and a cofactor — Delete, evaluate, then apply the sign.
- Easy: For |3 5 ; 2 7|: M₁₂ = 2, A₁₂ = −2; A₁₁ = 7 and A₂₂ = 3.
- Medium: For [2 0 1 ; 3 1 2 ; 0 4 5]: M₁₁ = 5 − 8 = −3, so A₁₁ = −3.
- Exam-type: For the same matrix M₁₂ = |3 2 ; 0 5| = 15 and A₁₂ = −15.
Rule 2: Use cofactors to get the determinant — Multiply each entry of a row by its own cofactor.
- Easy: For [2 0 1 ; 3 1 2 ; 0 4 5]: A₁₃ = +|3 1 ; 0 4| = 12.
- Medium: |A| = 2(−3) + 0(−15) + 1(12) = 6.
- Exam-type: Direct check: 2(5 − 8) − 0 + 1(12 − 0) = 6.
The sign of position (i, j) is plus when i + j is even and minus when it is odd.
Forgetting the sign when changing a minor into a cofactor.
Minor = just delete. Cofactor = delete and add the checkerboard sign.
Practice questions
- 1Find M₁₂ for |3 5 ; 2 7|.
- 2Find A₁₂ for |3 5 ; 2 7|.
- 3Find A₁₁ and A₂₂ for |3 5 ; 2 7|.
- 4For [1 2 3 ; 0 1 4 ; 5 6 0], find M₂₃ and A₂₃.
- 5For [2 0 1 ; 3 1 2 ; 0 4 5], find M₁₁.
- 6For the same matrix find A₁₂.
- 7For the same matrix find A₁₃.
- 8Use the cofactors of row 1 to find the determinant of [2 0 1 ; 3 1 2 ; 0 4 5].
Show answers (try them yourself first)
- 12
- 2−2
- 3A₁₁ = 7 and A₂₂ = 3
- 4M₂₃ = |1 2 ; 5 6| = −4 and A₂₃ = +4
- 5M₁₁ = 5 − 8 = −3
- 6A₁₂ = −|3 2 ; 0 5| = −15
- 7A₁₃ = +|3 1 ; 0 4| = 12
- 82(−3) + 0(−15) + 1(12) = 6
Expanding a determinant with cofactors
To count the votes of a school, you may count row by row or column by column and get the same total. A determinant can also be expanded along any row or column.
Multiplying each entry of any row (or column) by its cofactor and adding gives |A|. Multiplying the entries of one row by the cofactors of a different row gives 0.
Formula: |A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃; a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0
Rule 1: Expand along a row — Add entry × cofactor over the row.
- Easy: |2 5 ; 1 3| = 2 × 3 + 5 × (−1) = 1.
- Medium: |1 0 2 ; 3 1 0 ; 0 4 5| = 1(5) + 0 + 2(12) = 29.
- Exam-type: For that matrix, a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 1(8) + 0(5) + 2(−4) = 0.
Rule 2: Pick the best row or column — Choose the one with the most zeros.
- Easy: |5 0 0 ; 2 3 1 ; 1 4 2| = 5(6 − 4) = 10.
- Medium: |2 0 3 ; 1 0 4 ; 5 7 6|: column 2 has one non-zero entry; |A| = 7 × (−1) × |2 3 ; 1 4| = −35.
- Exam-type: |1 2 3 ; 0 0 4 ; 5 6 7| = 4 × (−1) × |1 2 ; 5 6| = 16.
Expanding along any row or column gives the same value; choose the cheapest.
Using the cofactors of another row and expecting the determinant (you get 0 instead).
Same row: entry × cofactor adds to the determinant; different row: adds to zero.
Practice questions
- 1For A = [2 1 0 ; 1 3 2 ; 0 1 4], find the cofactor A₁₁.
- 2For the same A find the cofactor A₁₂.
- 3For the same A find the cofactor A₁₃.
- 4Find |A| using row 1 of the same A.
- 5Find the cofactors A₂₁, A₂₂, A₂₃ of the same A.
- 6Find |A| using row 2 and compare with your answer from row 1.
- 7Find a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ for the same A.
- 8Evaluate |0 3 0 ; 2 1 5 ; 4 6 7| using row 1.
Show answers (try them yourself first)
- 1A₁₁ = 12 − 2 = 10
- 2A₁₂ = −(4 − 0) = −4
- 3A₁₃ = +(1 − 0) = 1
- 42(10) + 1(−4) + 0(1) = 16
- 5A₂₁ = −4, A₂₂ = 8, A₂₃ = −2
- 61(−4) + 3(8) + 2(−2) = 16, the same value
- 72(−4) + 1(8) + 0(−2) = 0
- 8Only a₁₂ = 3 counts: 3 × (−1) × |2 5 ; 4 7| = −3 × (−6) = 18