Higher Mathematics 1st Paper · Classes 11-12 · Chapter 1 · 1.5

Minors and Cofactors of a Determinant.

Pick one entry of a determinant and strike out its row and its column. What is left is a smaller determinant. That small determinant plus a sign is the cofactor, and cofactors let you open a determinant along any row or column. At the end comes a surprise: use the cofactors of the wrong row and the sum is exactly zero.

1. Pick an entry, strike out its row and column

Take any entry of a determinant. Delete the row it sits in and the column it sits in. The determinant built from the rows and columns that remain is the minor of that entry.

Put the proper sign in front of the minor and you have the cofactor of the entry. The sign of the (p, q)-th entry is (−1)^p+q, so cofactor = (−1)^p+q × minor. The sign is + when p + q is even and − when p + q is odd.

In a third-order determinant the signs sit like a chessboard: row 1 is + − +, row 2 is − + −, row 3 is + − +.

When the entries are small letters, their cofactors are written with capitals. Taking a for row 1, b for row 2 and c for row 3, the (1, 1)-th cofactor is A₁₁ = b₂₂c₃₃ − b₂₃c₃₂, and the (1, 2)-th is A₁₂ = −(b₂₁c₃₃ − b₂₃c₃₁).

2. Minors and cofactors by hand

In second order the minor is the entry diagonally opposite: for det[[1, 0], [−2, 5]] the (2, 2)-th minor is 1, the (1, 2)-th cofactor is −(−2) = 2, and the value is 5. In third order the (2, 3)-th minor of det[[1, 2, 3], [−1, 2, −3], [4, −5, 1]] is det[[1, 2], [4, −5]] = −5 − 8 = −13. The a₁₂ minor of det[[2, 1, 1], [0, 5, x], [1, −3, 4]] is det[[0, x], [1, 4]] = −x; if it equals −2 then x = 2. And the (3, 2)-th cofactor of det[[1, 2, 3], [2, 3, 4], [1, 5, 7]] is (−1)³⁺² det[[1, 3], [2, 4]] = −(4 − 6) = 2. Try it yourself in the grid below: pick an entry, the minor appears, and you choose the sign.

The same rule works on bigger grids. In det[[7, −7, 0], [−1, 2, −1], [5, p, 3]] the cofactor of p is (−1)³⁺² det[[7, 0], [−1, −1]] = −(−7) = 7, and the value of the determinant is 56 + 7p. In the fourth-order det[[2, −3, 7, 5], [3, 205, 1, 4], [3, −1, 97, 4], [0, −7, k, 7]] the cofactor of the entry 1 (2nd row, 3rd column) is (−1)²⁺³ det[[2, −3, 5], [3, −1, 4], [0, −7, 7]] = −(42 + 63 − 105) = 0.

3. Expanding with cofactors

With the cofactors known, opening a determinant is easy: D = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃. That is the expansion along the first row; the same works along any row or any column.

Two conclusions sit behind it. One: multiply the entries of a row (or column) by their own cofactors and add, and you get the value of the determinant. Two: multiply them by the cofactors of the corresponding entries of another row and add, and the sum is 0. Exercise 1(B) No. 8 is exactly this: a₂A₁ + b₂B₁ + c₂C₁ = 0. Below, write out the terms three ways on one determinant and compare.

4. The cofactor matrix

Replace every entry of a matrix by its cofactor and you get the cofactor matrix. For E = [[0, 1, 1], [1, 2, 0], [3, −1, 4]] the answer is [[8, −4, −7], [−5, −3, 3], [−2, 1, −1]]. Fill the nine cells yourself; the sign pattern is all you need to keep in mind. Note that this is the cofactor matrix itself; its transpose is the adjoint, which comes in a later topic.

Creative questions use the same arithmetic. In B = [[3, 5, 2], [4, −3, −4], [6, 1, −1]] the minor of the (3, 1) entry is det[[5, 2], [−3, −4]] = −14. The cofactors are B₁₃ = 22, B₂₁ = 7, B₃₂ = 20, B₂₂ = −15, so 2B₁₃ + 8B₂₁ − 2B₃₂ + 4B₂₂ = 44 + 56 − 40 − 60 = 0. In another, the (1, 2)-th minor of det[[4, 3, 2], [−1, 0, −3], [7, 5, −6]] is det[[−1, −3], [7, −6]] = 6 + 21 = 27, and its (3, 1)-th cofactor is det[[3, 2], [0, −3]] = −9.

One last example: for B = [[1, −2, 3], [2, 1, 2], [3, 2, 1]], B₁₁ = 1 − 4 = −3 and B₃₃ = 1 + 4 = 5, so B₁₁ + B₃₃ = 2.

5. Check yourself

The first question is in the exam's style.

In the second, watch the sign.

The third is the last turn.

The whole note on seven cards.

Source: Higher Mathematics 1st Paper, Classes 11-12 (NCTB), Chapter 1 “Matrices and Determinants”, section 1.5.