Laplace Expansion Theorem

Cofactors and the Two Expansion Formulas

Laplace expansion writes a determinant as a sum of entries multiplied by their cofactors $along any one row or column$ . Each cofactor contains the determinant of a matrix that is one size smaller.

Choose a row or column with many zero entries. Its zero terms vanish, so fewer smaller determinants need to be calculated.

Expanding a Determinant Along a Row or Column

For matrix $A \in \mathbb{R}^{n \times n}$ , the determinant can be calculated in two ways:

Let $A_{ij}$ be the $(n-1) \times (n-1)$ matrix obtained by deleting row $i$ and column $j$ from $A$ . The minor determinant with its $alternating sign$ is the cofactor

C_{ij}=(-1)^{i+j}\det A_{ij}

This sign pattern for three rows and columns shows that each step right or down reverses the sign.

	j = 1	j = 2	j = 3
i = 1	+	−	+
i = 2	−	+	−
i = 3	+	−	+

Expanding along any chosen column $j$ gives

\det A = \sum_{i=1}^{n} (-1)^{i+j} \cdot a_{ij} \cdot \det A_{ij}

The column-expansion formula holds for every $j \in \{1, 2, \ldots, n\}$ .

The determinant can also be calculated based on row $i$ :

\det A = \sum_{j=1}^{n} (-1)^{i+j} \cdot a_{ij} \cdot \det A_{ij}

The row-expansion formula holds for every $i \in \{1, 2, \ldots, n\}$ .

Cofactor Expansion by Matrix Size

The expansion rule works the same way at every size, but the $number of terms grows quickly$ , so the cases below show how the work changes as the matrix gets larger.

Two by Two Matrix

For a matrix of size $n = 2$ :

A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}

$Expansion based on the first row$ :

\det A = a_{11} \cdot \det(a_{22}) - a_{12} \cdot \det(a_{21})

Three by Three Matrix

For a matrix of size $n = 3$ :

A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

$Expansion based on the first row$ :

\begin{aligned} \det A &= a_{11} \cdot \det \begin{pmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} \\ &\quad{}- a_{12} \cdot \det \begin{pmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{pmatrix} \\ &\quad{}+ a_{13} \cdot \det \begin{pmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} \end{aligned}

After calculating the $2 \times 2$ determinants:

\begin{aligned} &= a_{11} \cdot (a_{22} \cdot a_{33} - a_{23} \cdot a_{32}) \\ &\quad{}- a_{12} \cdot (a_{21} \cdot a_{33} - a_{23} \cdot a_{31}) \\ &\quad{}+ a_{13} \cdot (a_{21} \cdot a_{32} - a_{22} \cdot a_{31}) \end{aligned}

If we expand it fully:

\begin{aligned} &= a_{11} \cdot a_{22} \cdot a_{33} - a_{11} \cdot a_{23} \cdot a_{32} \\ &\quad{}- a_{12} \cdot a_{21} \cdot a_{33} + a_{12} \cdot a_{23} \cdot a_{31} \\ &\quad{}+ a_{13} \cdot a_{21} \cdot a_{32} - a_{13} \cdot a_{22} \cdot a_{31} \end{aligned}

The expansion above gives the same expression as $Sarrus Rule$ . For a $3 \times 3$ matrix, the rule adds the products along three downward diagonals and subtracts the products along three upward diagonals.

Sarrus formula for $3 \times 3$ matrices:

\begin{aligned} \det A &= a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31} \\ &\quad{}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31} \\ &\quad{}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33} \end{aligned}

Sarrus rule uses diagonal patterns to determine which terms are added and which are subtracted.

Sarrus rule applies only to $3 \times 3$ matrices. For larger matrices, use cofactor expansion or a factorization-based method. Extending the diagonal pattern gives an invalid rule.

Computational Complexity

For a dense $n \times n$ matrix, a direct recursive implementation evaluates $n$ determinants of size $n-1$ at each level. Its work therefore follows the recurrence

T(n)=nT(n-1)+O(n)

The $factorial growth$ makes recursive Laplace expansion impractical for large dense matrices. Gaussian elimination or an $\mathrm{LU}$ factorization computes a determinant in $O(n^3)$ arithmetic operations.

Using Zero Entries

When a matrix has many zero entries, choose a row or column containing $as many zeros as possible$ . Every term whose selected coefficient $a_{ij}$ is zero vanishes, so its minor never needs to be evaluated.

Choosing a Row with Zero Entries

Suppose we have a matrix:

A = \begin{pmatrix} 0 & 3 & 3 \\ 2 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix}

$Expansion based on the first row$ :

\begin{aligned} \det A &= 0 \cdot \det(\ldots) - 3 \cdot \det \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \\ &\quad{}+ 3 \cdot \det \begin{pmatrix} 2 & 2 \\ 1 & 0 \end{pmatrix} \end{aligned}

Since the $first element is zero$ , the calculation becomes:

\begin{aligned} &= -3 \cdot (2 \cdot 1 - 0 \cdot 1) \\ &\quad{}+ 3 \cdot (2 \cdot 0 - 2 \cdot 1) \end{aligned}

By choosing rows or columns that have many zeros, we can save calculations.

Determinant of the Transpose

Laplace expansion also gives the following property:

\det A^T = \det A

This means the determinant of a matrix equals the $determinant of its transpose$ .

Consequences for Rows and Columns

Due to this $transpose property$ , all determinant properties that apply to rows of matrix $A$ also apply to columns of matrix $A$ .

For example:

If two rows are identical then the determinant is zero, likewise if two columns are identical
Swapping two rows changes the sign of the determinant, so does swapping two columns
The corresponding elementary row and column operations obey analogous determinant rules

Cofactors and the Two Expansion Formulas

Laplace expansion writes a determinant as a sum of entries multiplied by their cofactors $along any one row or column$ . Each cofactor contains the determinant of a matrix that is one size smaller.

Choose a row or column with many zero entries. Its zero terms vanish, so fewer smaller determinants need to be calculated.

Expanding a Determinant Along a Row or Column

For matrix $A \in \mathbb{R}^{n \times n}$ , the determinant can be calculated in two ways:

Let $A_{ij}$ be the $(n-1) \times (n-1)$ matrix obtained by deleting row $i$ and column $j$ from $A$ . The minor determinant with its $alternating sign$ is the cofactor

C_{ij}=(-1)^{i+j}\det A_{ij}

This sign pattern for three rows and columns shows that each step right or down reverses the sign.

	j = 1	j = 2	j = 3
i = 1	+	−	+
i = 2	−	+	−
i = 3	+	−	+

Expanding along any chosen column $j$ gives

\det A = \sum_{i=1}^{n} (-1)^{i+j} \cdot a_{ij} \cdot \det A_{ij}

The column-expansion formula holds for every $j \in \{1, 2, \ldots, n\}$ .

The determinant can also be calculated based on row $i$ :

\det A = \sum_{j=1}^{n} (-1)^{i+j} \cdot a_{ij} \cdot \det A_{ij}

The row-expansion formula holds for every $i \in \{1, 2, \ldots, n\}$ .

Cofactor Expansion by Matrix Size

The expansion rule works the same way at every size, but the $number of terms grows quickly$ , so the cases below show how the work changes as the matrix gets larger.

Two by Two Matrix

For a matrix of size $n = 2$ :

A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}

$Expansion based on the first row$ :

\det A = a_{11} \cdot \det(a_{22}) - a_{12} \cdot \det(a_{21})

Three by Three Matrix

For a matrix of size $n = 3$ :

A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

$Expansion based on the first row$ :

\begin{aligned} \det A &= a_{11} \cdot \det \begin{pmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} \\ &\quad{}- a_{12} \cdot \det \begin{pmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{pmatrix} \\ &\quad{}+ a_{13} \cdot \det \begin{pmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} \end{aligned}

After calculating the $2 \times 2$ determinants:

\begin{aligned} &= a_{11} \cdot (a_{22} \cdot a_{33} - a_{23} \cdot a_{32}) \\ &\quad{}- a_{12} \cdot (a_{21} \cdot a_{33} - a_{23} \cdot a_{31}) \\ &\quad{}+ a_{13} \cdot (a_{21} \cdot a_{32} - a_{22} \cdot a_{31}) \end{aligned}

If we expand it fully:

\begin{aligned} &= a_{11} \cdot a_{22} \cdot a_{33} - a_{11} \cdot a_{23} \cdot a_{32} \\ &\quad{}- a_{12} \cdot a_{21} \cdot a_{33} + a_{12} \cdot a_{23} \cdot a_{31} \\ &\quad{}+ a_{13} \cdot a_{21} \cdot a_{32} - a_{13} \cdot a_{22} \cdot a_{31} \end{aligned}

Sarrus Rule

Sarrus formula for $3 \times 3$ matrices:

\begin{aligned} \det A &= a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31} \\ &\quad{}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31} \\ &\quad{}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33} \end{aligned}

Sarrus rule uses diagonal patterns to determine which terms are added and which are subtracted.

Sarrus rule applies only to $3 \times 3$ matrices. For larger matrices, use cofactor expansion or a factorization-based method. Extending the diagonal pattern gives an invalid rule.

Computational Complexity

For a dense $n \times n$ matrix, a direct recursive implementation evaluates $n$ determinants of size $n-1$ at each level. Its work therefore follows the recurrence

T(n)=nT(n-1)+O(n)

Using Zero Entries

Choosing a Row with Zero Entries

Suppose we have a matrix:

A = \begin{pmatrix} 0 & 3 & 3 \\ 2 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix}

$Expansion based on the first row$ :

\begin{aligned} \det A &= 0 \cdot \det(\ldots) - 3 \cdot \det \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \\ &\quad{}+ 3 \cdot \det \begin{pmatrix} 2 & 2 \\ 1 & 0 \end{pmatrix} \end{aligned}

Since the $first element is zero$ , the calculation becomes:

\begin{aligned} &= -3 \cdot (2 \cdot 1 - 0 \cdot 1) \\ &\quad{}+ 3 \cdot (2 \cdot 0 - 2 \cdot 1) \end{aligned}

By choosing rows or columns that have many zeros, we can save calculations.

Determinant of the Transpose

Laplace expansion also gives the following property:

\det A^T = \det A

This means the determinant of a matrix equals the $determinant of its transpose$ .

Consequences for Rows and Columns

Due to this $transpose property$ , all determinant properties that apply to rows of matrix $A$ also apply to columns of matrix $A$ .

For example:

If two rows are identical then the determinant is zero, likewise if two columns are identical
Swapping two rows changes the sign of the determinant, so does swapping two columns
The corresponding elementary row and column operations obey analogous determinant rules