# Cofactor Matrix — Formula, Signs, Examples

TL;DR

The cofactor matrix of a square matrix is built by replacing each entry with its cofactor — the determinant of the smaller matrix left after deleting that entry's row and column, signed by $(-1)^{i+j}$. This article covers minors versus cofactors, the checkerboard sign pattern, the formula, how the cofactor matrix builds the adjugate, determinant, and inverse, and six worked examples.

## What Is a Cofactor Matrix?

The **cofactor matrix** of a square matrix $A$ is the matrix formed by replacing every entry $a_{ij}$ with its **cofactor** $C_{ij}$. To get there, two definitions stack:

- The **minor** $M_{ij}$ of entry $a_{ij}$ is the determinant of the submatrix that remains after deleting **row $i$ and column $j$** from $A$.

- The **cofactor** $C_{ij}$ is that minor multiplied by a sign:

$$C_{ij} = (-1)^{i+j}M_{ij}.$$

So a minor measures the _size_ of the leftover determinant, and the cofactor adds the _sign_ that the entry's position demands. The distinction matters: a minor is always just a determinant, while a cofactor can be a minor or its negative, depending on where the entry sits.

## What Is the Cofactor Sign Pattern?

The factor $(-1)^{i+j}$ produces a **checkerboard of signs**. When $i + j$ is even the sign is $+$; when odd, the sign is $-$. Laid out, it never changes:

$$\begin{bmatrix} + & - & + \ - & + & - \ + & - & + \end{bmatrix}.$$

The top-left is always $+$, and the signs alternate from there in both directions. You do not need to recompute $(-1)^{i+j}$ each time — once the pattern is memorised, you read the sign straight off the position. This same pattern shows up in the determinant of a matrix, because cofactor expansion _is_ how a determinant is computed.

## What Is the Cofactor Matrix Formula?

For each entry, the cofactor is

$$C_{ij} = (-1)^{i+j}M_{ij},$$

and the cofactor matrix collects them all:

$$\text{cof}(A) = \begin{bmatrix} C_{11} & C_{12} & C_{13} \ C_{21} & C_{22} & C_{23} \ C_{31} & C_{32} & C_{33} \end{bmatrix}.$$

For a 2x2 matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, the minors are single entries, so the cofactor matrix is simply

$$\text{cof}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}.$$

The procedure for any size is fixed: for each position, delete its row and column, take the determinant of the rest, then apply the checkerboard sign.

## How Does the Cofactor Matrix Build the Adjugate, Determinant, and Inverse?

The cofactor matrix is rarely the final goal — it is the step that powers three bigger results:

- **Determinant.** Expanding along any row, the determinant is the sum of each entry times its cofactor: $\det A = a_{i1}C_{i1} + a_{i2}C_{i2} + a_{i3}C_{i3}$. This is cofactor expansion.

- **Adjugate.** The [adjugate (adjoint)](/content/math/algebra/adjoint-of-a-matrix/index.html) is the **transpose** of the cofactor matrix: $\text{adj}(A) = \text{cof}(A)^T$.

- **Inverse.** The [inverse of the matrix](/content/math/algebra/inverse-of-a-matrix/index.html) is the adjugate over the determinant: $A^{-1} = \frac{1}{\det A}\text{adj}(A)$, valid whenever $\det A \neq 0$.

So the cofactor matrix sits one transpose away from the adjugate, and one division away from the inverse. Get the cofactors right and everything downstream follows.

## Examples of Cofactor Matrix

The set runs from a single minor, through the most common sign mistake, to a full 2x2 and 3x3 cofactor matrix, a cofactor expansion of the determinant, and building the adjugate.

### Example 1

**Find the minor $M_{11}$ of $A = \begin{bmatrix} 4 & 3 & 2 \ 1 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix}$.**

Delete row 1 and column 1, leaving $\begin{bmatrix} 5 & 6 \ 8 & 9 \end{bmatrix}$, and take its determinant:

$$M_{11} = (5)(9) - (6)(8) = 45 - 48 = -3.$$

**Final answer:** $M_{11} = -3$. Since position $(1,1)$ has sign $+$, here the cofactor equals the minor: $C_{11} = -3$.

### Example 2

**Find the cofactor $C_{21}$ of $A = \begin{bmatrix} 4 & 3 & 2 \ 1 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix}$.**

_Wrong attempt._ A student deletes row 2 and column 1, gets the minor $M_{21} = \det\begin{bmatrix} 3 & 2 \ 8 & 9 \end{bmatrix} = 27 - 16 = 11$, and writes $C_{21} = 11$. The minor is computed correctly — but the position's sign was never applied.

Check the position. Entry $(2,1)$ has $i + j = 3$, which is odd, so $(-1)^{2+1} = -1$. The cofactor cannot match the minor here.

_Correct._ Apply the sign:

$$C_{21} = (-1)^{2+1} M_{21} = -11.$$

**Final answer:** $C_{21} = -11$. The minor and the cofactor agree only when the position's sign is $+$.

### Example 3

**Find the cofactor matrix of $A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$.**

For a 2x2, each minor is the single remaining entry. Applying the signs:

$$\text{cof}(A) = \begin{bmatrix} 4 & -3 \ -2 & 1 \end{bmatrix}.$$

**Final answer:** $\begin{bmatrix} 4 & -3 \ -2 & 1 \end{bmatrix}$. Notice this is the "swap-and-negate" pattern from the 2x2 inverse — the same cofactors, before transposing and dividing.

### Example 4

**Find the cofactor matrix of $A = \begin{bmatrix} 1 & 0 & 2 \ 3 & 1 & 0 \ 0 & 2 & 1 \end{bmatrix}$.**

Compute all nine signed minors. For example, $C_{11} = +\det\begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} = 1$ and $C_{12} = -\det\begin{bmatrix} 3 & 0 \ 0 & 1 \end{bmatrix} = -3$. Continuing through every entry:

$$\text{cof}(A) = \begin{bmatrix} 1 & -3 & 6 \ 4 & 1 & -2 \ -2 & 6 & 1 \end{bmatrix}.$$

**Final answer:** the matrix above. Each entry was a signed 2x2 determinant — the checkerboard handles the alternation automatically.

### Example 5

**Use cofactor expansion along the first row to find $\det A$ for the matrix in Example 4.**

The determinant is each first-row entry times its cofactor: $\det A = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$. Using the cofactors $C_{11} = 1$, $C_{12} = -3$, $C_{13} = 6$ from Example 4:

$$\det A = (1)(1) + (0)(-3) + (2)(6) = 1 + 0 + 12 = 13.$$

**Final answer:** $\det A = 13$. The zero in position $(1,2)$ wiped out a whole cofactor term — choosing a row with zeros saves work.

### Example 6

**Build the adjugate of the matrix in Example 4 from its cofactor matrix.**

The adjugate is the transpose of the cofactor matrix. Transposing $\text{cof}(A)$ from Example 4:

$$\text{adj}(A) = \text{cof}(A)^T = \begin{bmatrix} 1 & 4 & -2 \ -3 & 1 & 6 \ 6 & -2 & 1 \end{bmatrix}.$$

**Final answer:** the matrix above. Dividing this by $\det A = 13$ would give $A^{-1}$ — the cofactor matrix is two short steps from the inverse.

## Why the Cofactor Matrix Earns Its Place

Cofactors come from the theory of determinants, shaped through the 1800s by figures including [James Joseph Sylvester](https://mathshistory.st-andrews.ac.uk/Biographies/Sylvester/) (1814–1897, England), who coined the word "matrix" in 1850 and worked on the minors that underlie cofactor expansion. The idea solved a real problem: a large determinant is impossible to compute directly, but cofactor expansion reduces it to a chain of small, manageable determinants.

Where the cofactor matrix does real work:

- **Computing determinants by hand.** Cofactor (Laplace) expansion is the standard textbook method for $3 \times 3$ and larger determinants — the only practical hand method beyond the 2x2.
- **Finding inverses.** The adjugate-over-determinant formula is built entirely on the cofactor matrix; it is how most students first compute a 3x3 inverse.
- **Cramer's rule.** Solving systems by ratios of determinants relies on cofactors to evaluate each determinant.
- **Symbolic computation.** When entries are variables rather than numbers, cofactor expansion is one of the few methods that stays exact rather than approximate.

## Where Students Trip Up on the Cofactor Matrix

### Mistake 1: Confusing the minor with the cofactor

**Where it slips in:** Filling the cofactor matrix straight from the minors.

**Don't do this:** Copy each minor in as-is, ignoring the $(-1)^{i+j}$ sign.

**The correct way:** A cofactor is a _signed_ minor. Apply the checkerboard sign to every position before placing the value.

### Mistake 2: Deleting the wrong row or column

**Where it slips in:** Computing the minor $M_{ij}$.

**Don't do this:** Delete row $j$ and column $i$ (swapping the indices), or delete by value instead of by position.

**The correct way:** For $M_{ij}$, delete **row $i$ and column $j$** — row first, then column, matching the subscript order. The rusher who works fast often transposes the indices and computes the wrong submatrix entirely.

### Mistake 3: Forgetting to transpose when building the adjugate

**Where it slips in:** Going from the cofactor matrix to the adjugate (and then the inverse).

**Don't do this:** Use the cofactor matrix directly as the adjugate.

**The correct way:** The adjugate is the **transpose** of the cofactor matrix. Skip the transpose and the inverse comes out wrong everywhere except on the diagonal. The second-guesser who knows the rule still sometimes "un-transposes" on a recheck — trusting the one transpose keeps it right.

## Key Takeaways

- The **cofactor matrix** replaces each entry with its cofactor $C_{ij} = (-1)^{i+j}M_{ij}$ — a signed minor.
- A minor is the leftover determinant after deleting a row and column; the cofactor adds the position's sign.
- The signs follow a fixed checkerboard, top-left always $+$: $\begin{bmatrix} + & - & + \ - & + & - \ + & - & + \end{bmatrix}$.
- Transposing the cofactor matrix gives the adjugate; dividing the adjugate by the determinant gives the inverse.
- The most common mistake is treating the minor as the cofactor — always apply the sign.
- Cofactor expansion is the standard way to compute determinants of $3 \times 3$ and larger matrices.