From Minors to Cofactor Expansion

From Smaller Determinants to Cofactors

A \(3\times3\) determinant can be organized around its first row. Each entry in that row is paired with a \(2\times2\) determinant: delete the entry’s row and column, then take the determinant of what remains. That smaller determinant is the entry’s minor. To combine the three minors correctly, we also need the cofactor signs. For the first row, those signs are \(+,-,+\).

This tutorial focuses on expressing a \(3\times3\) determinant as a sum using those three first-row entries, minors, and signs. The calculations use the \(2\times2\) formula \(\det\begin{bmatrix}r&s\\t&u\end{bmatrix}=ru-st\). To keep the derivation grounded in explicit calculations, we begin with the six-term formula for a \(3\times3\) determinant. We will group its terms to obtain the cofactor expansion.

The minor and cofactor are related, but they are not interchangeable: \(M_{ij}\) is the smaller determinant, while \(C_{ij}\) includes the sign \((-1)^{i+j}\). In the first row, the signs for \(j=1,2,3\) are \((-1)^2=+1\), \((-1)^3=-1\), and \((-1)^4=+1\). Thus \(C_{11}=M_{11}\), \(C_{12}=-M_{12}\), and \(C_{13}=M_{13}\).

Why the First-Row Signs Are \(+,-,+\)

Write a general matrix as \(\mathbf{A}=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\). Deleting row 1 and column 1 leaves \(\begin{bmatrix}e&f\\h&i\end{bmatrix}\), whose determinant is \(ei-fh\). Deleting row 1 and column 2 leaves \(\begin{bmatrix}d&f\\g&i\end{bmatrix}\), with determinant \(di-fg\). Deleting row 1 and column 3 leaves \(\begin{bmatrix}d&e\\g&h\end{bmatrix}\), with determinant \(dh-eg\).

Proof. Begin with the six-term determinant formula and collect terms according to the first-row entry they contain:

By the \(2\times2\) determinant formula, \(ei-fh=M_{11}\), \(di-fg=M_{12}\), and \(dh-eg=M_{13}\). The cofactor definition gives \(C_{11}=M_{11}\), \(C_{12}=-M_{12}\), and \(C_{13}=M_{13}\). Substituting these identities into the last line yields both stated forms of the expansion. \(\square\)

This derivation accounts for every sign: the middle term is negative because its grouped products are \(b(fg-di)=-b(di-fg)\). The two outer terms have positive signs and are exactly the first and third minors multiplied by their entries.

Computing with the Expansion

A useful way to organize the work is to keep three roles separate: the first-row entry, the minor obtained by deleting its row and column, and the sign that converts that minor into a cofactor. The cofactor expansion pairs each entry only with its own cofactor; it does not pair an entry with a minor from a different column.

What the Formula Says About the First Row

Proof. Let the fixed lower rows be \((d,e,f)\) and \((g,h,i)\). The expansion already proved is \(a(ei-fh)-b(di-fg)+c(dh-eg)\). Each coefficient \(ei-fh\), \(-(di-fg)\), and \(dh-eg\) depends only on the lower rows, not on \(a,b,c\). The expression is therefore linear in the first-row entries: replacing \((a,b,c)\) by \((a_1+a_2,b_1+b_2,c_1+c_2)\) adds the corresponding two expressions, and multiplying all three entries by a scalar multiplies the determinant by that scalar. This proves the proposition. \(\square\)

For example, keep the last two rows \((1,2,0)\) and \((3,-1,4)\) fixed, and let the first row be \((x,y,z)\). Its minors are \(8\), \(4\), and \(-7\), so the first-row cofactors are \(8\), \(-4\), and \(-7\). The expansion gives \(8x-4y-7z\). The coefficients do not change as \(x,y,z\) change, because they come entirely from the lower rows.

Common Sign Pitfall

A frequent error is to compute the middle minor correctly and then use it as though it were already the middle cofactor. In Worked Example 1, \(M_{12}=9\), but \(C_{12}=-9\). Using \(M_{12}\) directly would give the incorrect calculation \(2(-20)+(-1)(9)+3(5)=-40-9+15=-34\). The correct middle contribution is the entry \(-1\) multiplied by the cofactor \(-9\), giving \(+9\), and the determinant is \(-40+9+15=-16\).

The reliable correction is to write down all three signs before combining terms: \(+M_{11}\), \(-M_{12}\), \(+M_{13}\). Then multiply each signed minor by the entry in the same first-row column. A minor is not signed just because it was computed from a determinant; the cofactor sign is a separate part of the calculation.

For this tutorial, the expansion is specifically along the first row. The sign and minor in each term are tied to a particular matrix position. In the next calculation, first verify the minor by deleting the correct row and column, then apply the checkerboard sign, and only then multiply by the first-row entry. This sequence makes the calculation auditable and helps locate an error if the final result does not match a direct six-term check.