Minors and Cofactors

Section 1's expansion had two ingredients per element: a small determinant and a sign. This section names them — and the names carry the whole rest of the chapter, because the adjoint and the inverse are built from cofactors.

Minor. The minor $M_{ij}$ of the element $a_{ij}$ is the determinant left after deleting row $i$ and column $j$. In a determinant of order $n$ (with $n \geq 2$), every minor is a determinant of order $n - 1$.

Cofactor. The cofactor attaches the chessboard sign to the minor:

$$\boxed{\;A_{ij} = (-1)^{i+j} M_{ij}\;}$$

When $i + j$ is even, $A_{ij} = M_{ij}$; when $i + j$ is odd, $A_{ij} = -M_{ij}$. Minor and cofactor differ only by that sign — mixing them up is the classic half-mark leak of this section.

The expansion, restated

With cofactors, every expansion of Section 1 compresses into one sentence:

$$\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$$

— the sum of the products of the elements of any one row (or column) with their own cofactors. All six versions (three rows, three columns) give the same $\Delta$, and the signs are already inside the cofactors, so there is nothing extra to remember.

The alien-cofactor rule

What if a row's elements are multiplied by the cofactors of a different row? The sum is always zero:

$$a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$$

Why: this sum is exactly the expansion of a determinant whose first two rows are identical (row 1's elements sitting in both positions) — and such a determinant vanishes. So:

1. elements $\times$ own cofactors $= \Delta$,
2. elements $\times$ alien cofactors $= 0$.

This own/alien dichotomy is precisely what will make $A(\operatorname{adj} A) = \vert A \vert I$ work in the next section — the diagonal entries of the product are case 1, and every off-diagonal entry is case 2.

Solved Examples

Example 1 — One minor, quickly

Find the minor of the element $6$ in $\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$.

Step 1 — locate: $6$ sits in row $2$, column $3$, so we need $M_{23}$.

Step 2 — delete $R_2$ and $C_3$: $$M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = 8 - 14 = -6$$

Answer: $M_{23} = -6$. (Its cofactor would be $A_{23} = (-1)^{2+3}(-6) = 6$ — sign flipped, since $2 + 3$ is odd.)

Example 2 — All four, for a $2 \times 2$

Find the minors and cofactors of all elements of $\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}$.

Step 1 — minors: deleting the row and column of each element leaves a single entry: $$M_{11} = 3, \quad M_{12} = 4, \quad M_{21} = -2, \quad M_{22} = 1$$

Step 2 — attach signs: $$A_{11} = +3, \quad A_{12} = -4, \quad A_{21} = +2, \quad A_{22} = +1$$

Answer: as above — note the diagonal cofactors kept their minors' values ($i + j$ even) while the off-diagonal pair flipped sign.

Example 3 — A full $3 \times 3$ cofactor table, plus the alien check

For $\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$, find the cofactors of the first row and verify that $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} = 0$.

Step 1 — first-row cofactors: $$A_{11} = +\begin{vmatrix} 0 & 4 \\ 5 & -7 \end{vmatrix} = -20, \quad A_{12} = -\begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = 46, \quad A_{13} = +\begin{vmatrix} 6 & 0 \\ 1 & 5 \end{vmatrix} = 30$$

Step 2 — third-row cofactors (needed for the check): $$A_{31} = +\begin{vmatrix} -3 & 5 \\ 0 & 4 \end{vmatrix} = -12, \quad A_{32} = -\begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = 22, \quad A_{33} = +\begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} = 18$$

Step 3 — alien-cofactor sum: $2(-12) + (-3)(22) + 5(18) = -24 - 66 + 90 = 0$. ✓

Answer: the row-1 elements against row-3 cofactors cancel exactly, as the alien-cofactor rule promises. (For contrast, against their own cofactors: $2(-20) + (-3)(46) + 5(30) = -40 - 138 + 150 = -28 = \Delta$.)

Example 4 — Evaluating with a chosen row's cofactors

Using the cofactors of the second row, evaluate $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$.

Step 1 — second-row cofactors: $$A_{21} = -\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = -(9 - 16) = 7, \quad A_{22} = +\begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 7, \quad A_{23} = -\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -7$$

Step 2 — own-cofactor sum: $\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2(7) + 0(7) + 1(-7)$.

Answer: $\Delta = 14 - 7 = 7$. The zero in the second row was exactly why this row was worth choosing.