Minors and cofactors

To compute determinants of $n \times n$ matrices recursively, we extract smaller matrices by deleting one row and one column. The determinant of that smaller matrix is the minor, and with a sign attached it becomes the cofactor. The collection of cofactors organised into a matrix is called the cofactor matrix, and its transpose is the adjoint , used to invert matrices in the next subtopic.

Definition of minor

For a square matrix $A = (a_{ij})$ of order $n$, the minor $M_{ij}$ of the entry $a_{ij}$ is the determinant of the $(n - 1) \times (n - 1)$ matrix obtained by deleting row $i$ and column $j$ of $A$.

Example: for $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$,

$M_{11} = \det\begin{pmatrix} 5 & 6 \\ 8 & 9 \end{pmatrix} = -3$.

$M_{23} = \det\begin{pmatrix} 1 & 2 \\ 7 & 8 \end{pmatrix} = -6$.

Definition of cofactor

The cofactor $C_{ij}$ is

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

The sign pattern $(-1)^{i+j}$ alternates checkerboard-style starting with $+$ in the top-left:

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

In the example above: $C_{11} = +M_{11} = -3$, $C_{23} = -M_{23} = +6$.

Cofactor expansion

The determinant of $A$ can be computed by expansion along any row or column:

$\det A = \sum_{j = 1}^n a_{ij} C_{ij} \quad \text{(expansion along row } i \text{)}.$

Similarly along column $j$:

$\det A = \sum_{i = 1}^n a_{ij} C_{ij}.$

This recursive procedure reduces an $n \times n$ determinant to a linear combination of $(n - 1) \times (n - 1)$ determinants.

Orthogonality of cofactors

A subtle but important fact: the sum $\sum_j a_{ij} C_{kj}$ equals $\det A$ when $i = k$ and equals $0$ when $i \neq k$. The "wrong-row" expansion always vanishes.

Reason: the wrong-row expansion equals the determinant of the matrix obtained by replacing row $k$ of $A$ with row $i$ of $A$ , which has two equal rows.

This orthogonality is the basis for the adjoint formula in the next subtopic.

Cofactor matrix

The cofactor matrix of $A$ is the matrix $\text{cof}(A) = (C_{ij})$. Its transpose is the adjoint (next subtopic).

For our example:

$C_{11} = -3, C_{12} = -(36 - 42) = 6, C_{13} = 32 - 35 = -3$.

$C_{21} = -(18 - 24) = 6, C_{22} = 9 - 21 = -12, C_{23} = -(8 - 14) = 6$.

$C_{31} = 12 - 15 = -3, C_{32} = -(6 - 12) = 6, C_{33} = 5 - 8 = -3$.

$\text{cof}(A) = \begin{pmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{pmatrix}$.

Worked examples

Example 1. Find $M_{12}$ for $A = \begin{pmatrix} 2 & 3 & 1 \\ 0 & 4 & -1 \\ 5 & 2 & 1 \end{pmatrix}$.

Delete row $1$, column $2$: $\det\begin{pmatrix} 0 & -1 \\ 5 & 1 \end{pmatrix} = 0 + 5 = 5$. So $M_{12} = 5$, $C_{12} = -5$.

Example 2. Find $\det A$ in the above example by expansion along row 1.

$a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13} = 2 \cdot (4 + 2) + 3 \cdot (-5) + 1 \cdot (0 - 20)$ $= 12 - 15 - 20 = -23$.

Let me double-check $C_{11}$: $M_{11} = \det\begin{pmatrix} 4 & -1 \\ 2 & 1 \end{pmatrix} = 4 + 2 = 6$, so $C_{11} = +6$.

$C_{13}$: $M_{13} = \det\begin{pmatrix} 0 & 4 \\ 5 & 2 \end{pmatrix} = 0 - 20 = -20$, so $C_{13} = +(-20) = -20$.

Det $= 2 \cdot 6 + 3 \cdot (-5) + 1 \cdot (-20) = 12 - 15 - 20 = -23$.

Example 3. Show that the cofactor matrix of the identity $I_3$ is itself the identity.

Each $C_{ii}$ is the determinant of an $(n - 1) \times (n - 1)$ identity, which is $1$. Each $C_{ij}$ for $i \neq j$ has a zero row in the minor, so is $0$. So $\text{cof}(I) = I$.

Example 4. Find $C_{32}$ for $A = \begin{pmatrix} 1 & 0 & 2 \\ 3 & -1 & 1 \\ 4 & 5 & 6 \end{pmatrix}$.

Delete row $3$, column $2$: $\det\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} = 1 - 6 = -5$. Sign $(-1)^{3 + 2} = -1$. So $C_{32} = 5$.

Example 5. Use orthogonality to verify: for the matrix in Example 4, $a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} = 0$.

Compute: $C_{21} = (-1)^3 \det\begin{pmatrix} 0 & 2 \\ 5 & 6 \end{pmatrix} = -(-10) = 10$.

$C_{22} = +\det\begin{pmatrix} 1 & 2 \\ 4 & 6 \end{pmatrix} = -2$.

$C_{23} = -\det\begin{pmatrix} 1 & 0 \\ 4 & 5 \end{pmatrix} = -5$.

Sum: $1 \cdot 10 + 0 \cdot (-2) + 2 \cdot (-5) = 10 - 10 = 0$. $\checkmark$

Example 6. Find the cofactor matrix of $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

$C_{11} = 4, C_{12} = -3, C_{21} = -2, C_{22} = 1$. So $\text{cof} = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}$.

Try it yourself

1. Find $M_{11}, M_{12}, M_{13}$ for $\begin{pmatrix} 2 & 1 & 3 \\ 0 & 4 & 5 \\ -1 & 2 & 6 \end{pmatrix}$.
2. Find the corresponding cofactors.
3. Expand the determinant of the matrix in question 1 along row 1.
4. Expand the same determinant along column 1, verify same answer.
5. Compute the cofactor matrix of $\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$.
6. Compute the cofactor matrix of $I_3$.
7. Verify the orthogonality $\sum_j a_{1j} C_{2j} = 0$ for $\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$.
8. Find $C_{31}$ for the matrix in question 7.
9. For a $2 \times 2$ matrix $A$, show that the cofactor matrix is $\begin{pmatrix} d & -c \\ -b & a \end{pmatrix}$. Notice it's the transpose of $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.
10. Show that for a diagonal matrix the cofactor matrix is also diagonal.
11. Compute the cofactor matrix of $\begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & 1 \\ 2 & 0 & 4 \end{pmatrix}$.
12. Find all entries of the cofactor matrix of the rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$.
13. Prove: if $A$ is symmetric, so is its cofactor matrix.
14. Prove: $\det(\text{cof}(A)) = (\det A)^{n - 1}$ for an $n \times n$ matrix.

Pitfalls / Tricks

• Always include the sign $(-1)^{i+j}$ , the most common error is to confuse minor and cofactor.
• The sign pattern starts with $+$ at the $(1, 1)$ entry and alternates.
• Orthogonality of cofactors ($\sum a_{ij}C_{kj} = 0$ for $i \neq k$) is what powers the adjoint inverse formula.
• For large matrices, the cofactor expansion is inefficient ($O(n!)$). Row reduction is faster ($O(n^3)$). The cofactor approach is more useful in proofs than in practical computation.

Next, the adjoint and the explicit inverse formula.