MIT 18.06 Lecture 19: Determinant Formulas and Cofactors

linear algebra

MIT 18.06

determinants

cofactors

permutations

Three computational methods for determinants: pivots, the big formula, and cofactor expansion

Author

Chao Ma

Published

November 4, 2025

2 by 2 Matrix

\[ \begin{vmatrix}a&b\\c&d\end{vmatrix}=\begin{vmatrix}a&0\\c&d\end{vmatrix}+\begin{vmatrix}0&b\\c&d\end{vmatrix}=\\ \begin{vmatrix}a&0\\0&d\end{vmatrix}+\begin{vmatrix}a&0\\c&0\end{vmatrix}+\begin{vmatrix}0&b\\c&0\end{vmatrix}+\begin{vmatrix}0&b\\0&d\end{vmatrix}=\\ \begin{vmatrix}a&0\\0&d\end{vmatrix}+0+\begin{vmatrix}0&b\\c&0\end{vmatrix}+0 \]

$\begin{vmatrix}a&0\\0&d\end{vmatrix}=ad$ is the product of the diagonal
$\begin{vmatrix}0&b\\c&0\end{vmatrix}=-bc$ requires exchanging rows, then taking the product of the diagonal

The Method

Keep rows 2, 3, …, n unchanged
Split the first row into n pieces, one nonzero element per part

\[ |A|=\begin{bmatrix}a&0&0\\.&\text{row}_2&.\\.&\text{row}_n&.\end{bmatrix}+ \begin{bmatrix}0&b&0\\.&\text{row}_2&.\\.&\text{row}_n&.\end{bmatrix}+ \begin{bmatrix}0&0&c\\.&\text{row}_2&.\\.&\text{row}_n&.\end{bmatrix}+ \ldots \]

Now we have n smaller matrices, each matrix keeps all lower rows the same and has only 1 active element entry in the first row
Repeat this step recursively
We can apply the same splitting process to the second row, then the third, and so on — each time expanding the matrix into smaller pieces
Eventually, we will obtain many ($n^n$) matrices, each with only 1 active (non-zero) entry per row

3 by 3 Matrix

\[ \begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix} \]

The Survivors

The survivors have one entry in each row and each column. Because if one column is all zeros, the matrix is singular and the determinant is 0.

Proof: - For an $n \times n$ matrix, after the processing, there should be $n^n$ matrices - For each matrix, each row has only 1 non-zero entry, so in total there are n non-zero entries - If one column has >1 non-zero entry, there must be at least one other column that is all zero - Reason: we have n columns and n non-zero entries - If one column has >1 non-zero entries, the other columns (n-1 in total) have <n-1 non-zeros to share

Survivors of 3×3:

\[ \begin{vmatrix}a_{11}&0&0\\0&a_{22}&0\\0&0&a_{33}\end{vmatrix}+\begin{vmatrix}a_{11}&0&0\\0&0&a_{23}\\0&a_{32}&0\end{vmatrix}+\begin{vmatrix}0&a_{12}&0\\a_{21}&0&0\\0&0&a_{33}\end{vmatrix}+\\ \begin{vmatrix}0&a_{12}&0\\0&0&a_{23}\\a_{31}&0&0\end{vmatrix}+\begin{vmatrix}0&0&a_{13}\\a_{21}&0&0\\0&a_{32}&0 \end{vmatrix}+\begin{vmatrix}0&0&a_{13}\\0&a_{22}&0\\a_{31}&0&0 \end{vmatrix} \]

The determinants:

\[ a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31} \]

Each part is the product of the diagonal
The sign is determined by how many row exchanges are needed to make it diagonal

The Big Formula

For an n by n matrix, we have $n!$ (n factorial) terms of survivors and the determinant is:

\[ \det A=\sum_{n!} \pm a_{1\alpha}a_{2\beta}a_{3\gamma}\ldots a_{n\omega} \]

where $(\alpha, \beta, \gamma,\ldots,\omega)$ is a permutation of $(1,\ldots,n)$.

Why n! Terms?

It is because of permutations: - Row 1 can be chosen in n ways - Row 2 can be chosen in n-1 ways - … - Row n can be chosen in 1 way

Therefore: $n \times (n-1) \times \ldots \times 1 = n!$

$\det A=0$ and Singularity

Consider this 4×4 matrix:

\[ \begin{vmatrix}0&0&1&1\\0&1&1&0\\1&1&0&0\\1&0&0&1\end{vmatrix} \]

Try the big formula, we can find 2 permutations that each make every row and column have a 1:

$D_1=\begin{vmatrix}0&0&1&\underline{1}\\0&1&\underline{1}&0\\1&\underline{1}&0&0\\\underline{1}&0&0&1\end{vmatrix}$
$D_2=\begin{vmatrix}0&0&\underline{1}&1\\0&\underline{1}&1&0\\\underline{1}&1&0&0\\1&0&0&\underline{1}\end{vmatrix}$

\[ \det D_1+\det D_2=1-1=0 \]

The determinant of the 4×4 matrix is 0, and there are dependent rows: row₁ + row₃ = row₂ + row₄, which proves the finding.

Cofactor

3×3 Example

\[ \det A=a_{11}(a_{22}a_{33}-a_{23}a_{32})+\\ a_{12}(\ldots)+\\ a_{13}(\ldots) \]

The picture for $a_{11}(a_{22}a_{33}-a_{23}a_{32})$:

\[ \begin{vmatrix}a_{11}&0&0\\0&a_{22}&a_{23}\\0&a_{32}&a_{33}\end{vmatrix} \]

This is just $a_{11} \det \begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}$.

Cofactor Definition

The cofactor of $a_{ij}=C_{ij}$ is $\pm \det$ of the (n-1) × (n-1) matrix with $\text{row}_i$ and $\text{col}_j$ erased.

\[ C_{ij} = \begin{cases}+ \det M_{ij} & \text{ if } i+j \text{ is even}\\ - \det M_{ij} & \text{ if } i+j \text{ is odd} \end{cases} \]

where $M_{ij}$ is the matrix with row i and column j removed.

Cofactor Formula

\[ \det A=a_{11}C_{11}+a_{12}C_{12}+\ldots+a_{1n}C_{1n} \]

This formula allows us to expand along any row (or column) to compute the determinant.

Three Formulas for Computing Determinants

Product of pivots (from elimination)
Big formula (sum over all n! permutations)
Cofactor formula (recursive expansion along a row or column)

Source: MIT 18.06SC Linear Algebra - Lecture 19

--- title: "MIT 18.06 Lecture 19: Determinant Formulas and Cofactors" author: "Chao Ma" date: "2025-11-04" categories: [linear algebra, MIT 18.06, determinants, cofactors, permutations] description: "Three computational methods for determinants: pivots, the big formula, and cofactor expansion" --- {{< video https://www.youtube.com/watch?v=23LLB9mNJvc >}} ## 2 by 2 Matrix $$ \begin{vmatrix}a&b\\c&d\end{vmatrix}=\begin{vmatrix}a&0\\c&d\end{vmatrix}+\begin{vmatrix}0&b\\c&d\end{vmatrix}=\\ \begin{vmatrix}a&0\\0&d\end{vmatrix}+\begin{vmatrix}a&0\\c&0\end{vmatrix}+\begin{vmatrix}0&b\\c&0\end{vmatrix}+\begin{vmatrix}0&b\\0&d\end{vmatrix}=\\ \begin{vmatrix}a&0\\0&d\end{vmatrix}+0+\begin{vmatrix}0&b\\c&0\end{vmatrix}+0 $$ - $\begin{vmatrix}a&0\\0&d\end{vmatrix}=ad$ is the product of the diagonal - $\begin{vmatrix}0&b\\c&0\end{vmatrix}=-bc$ requires exchanging rows, then taking the product of the diagonal ### The Method - Keep rows 2, 3, …, n unchanged - Split the first row into n pieces, one nonzero element per part $$ |A|=\begin{bmatrix}a&0&0\\.&\text{row}_2&.\\.&\text{row}_n&.\end{bmatrix}+ \begin{bmatrix}0&b&0\\.&\text{row}_2&.\\.&\text{row}_n&.\end{bmatrix}+ \begin{bmatrix}0&0&c\\.&\text{row}_2&.\\.&\text{row}_n&.\end{bmatrix}+ \ldots $$ - Now we have n smaller matrices, each matrix keeps all lower rows the same and has only 1 active element entry in the first row - Repeat this step recursively - We can apply the same splitting process to the second row, then the third, and so on — each time expanding the matrix into smaller pieces - **Eventually, we will obtain many ($n^n$) matrices, each with only 1 active (non-zero) entry per row** ## 3 by 3 Matrix $$ \begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix} $$ ### The Survivors The survivors have one entry in each row and each column. Because if one column is all zeros, the matrix is singular and the determinant is 0. **Proof**: - For an $n \times n$ matrix, after the processing, there should be $n^n$ matrices - For each matrix, each row has only 1 non-zero entry, so in total there are n non-zero entries - If one column has >1 non-zero entry, there must be at least one other column that is all zero - Reason: we have n columns and n non-zero entries - If one column has >1 non-zero entries, the other columns (n-1 in total) have <n-1 non-zeros to share **Survivors of 3×3**: $$ \begin{vmatrix}a_{11}&0&0\\0&a_{22}&0\\0&0&a_{33}\end{vmatrix}+\begin{vmatrix}a_{11}&0&0\\0&0&a_{23}\\0&a_{32}&0\end{vmatrix}+\begin{vmatrix}0&a_{12}&0\\a_{21}&0&0\\0&0&a_{33}\end{vmatrix}+\\ \begin{vmatrix}0&a_{12}&0\\0&0&a_{23}\\a_{31}&0&0\end{vmatrix}+\begin{vmatrix}0&0&a_{13}\\a_{21}&0&0\\0&a_{32}&0 \end{vmatrix}+\begin{vmatrix}0&0&a_{13}\\0&a_{22}&0\\a_{31}&0&0 \end{vmatrix} $$ The determinants: $$ a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31} $$ - Each part is the product of the diagonal - The sign is determined by how many row exchanges are needed to make it diagonal ## The Big Formula For an n by n matrix, we have $n!$ (n factorial) terms of survivors and the determinant is: $$ \det A=\sum_{n!} \pm a_{1\alpha}a_{2\beta}a_{3\gamma}\ldots a_{n\omega} $$ where $(\alpha, \beta, \gamma,\ldots,\omega)$ is a permutation of $(1,\ldots,n)$. ### Why n! Terms? It is because of permutations: - Row 1 can be chosen in n ways - Row 2 can be chosen in n-1 ways - … - Row n can be chosen in 1 way Therefore: $n \times (n-1) \times \ldots \times 1 = n!$ ## $\det A=0$ and Singularity Consider this 4×4 matrix: $$ \begin{vmatrix}0&0&1&1\\0&1&1&0\\1&1&0&0\\1&0&0&1\end{vmatrix} $$ Try the big formula, we can find 2 permutations that each make every row and column have a 1: - $D_1=\begin{vmatrix}0&0&1&\underline{1}\\0&1&\underline{1}&0\\1&\underline{1}&0&0\\\underline{1}&0&0&1\end{vmatrix}$ - $D_2=\begin{vmatrix}0&0&\underline{1}&1\\0&\underline{1}&1&0\\\underline{1}&1&0&0\\1&0&0&\underline{1}\end{vmatrix}$ $$ \det D_1+\det D_2=1-1=0 $$ The determinant of the 4×4 matrix is 0, and there are dependent rows: row₁ + row₃ = row₂ + row₄, which proves the finding. ## Cofactor ### 3×3 Example $$ \det A=a_{11}(a_{22}a_{33}-a_{23}a_{32})+\\ a_{12}(\ldots)+\\ a_{13}(\ldots) $$ The picture for $a_{11}(a_{22}a_{33}-a_{23}a_{32})$: $$ \begin{vmatrix}a_{11}&0&0\\0&a_{22}&a_{23}\\0&a_{32}&a_{33}\end{vmatrix} $$ This is just $a_{11} \det \begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}$. ### Cofactor Definition The cofactor of $a_{ij}=C_{ij}$ is $\pm \det$ of the (n-1) × (n-1) matrix with $\text{row}_i$ and $\text{col}_j$ erased. $$ C_{ij} = \begin{cases}+ \det M_{ij} & \text{ if } i+j \text{ is even}\\ - \det M_{ij} & \text{ if } i+j \text{ is odd} \end{cases} $$ where $M_{ij}$ is the matrix with row i and column j removed. ### Cofactor Formula $$ \det A=a_{11}C_{11}+a_{12}C_{12}+\ldots+a_{1n}C_{1n} $$ This formula allows us to expand along any row (or column) to compute the determinant. ## Three Formulas for Computing Determinants 1. **Product of pivots** (from elimination) 2. **Big formula** (sum over all n! permutations) 3. **Cofactor formula** (recursive expansion along a row or column) --- *Source: MIT 18.06SC Linear Algebra - Lecture 19*