Sarrus Method

The Diagonal Products in a Two by Two Determinant

Sarrus' method is a practical way to calculate the determinant of a $3 \times 3$ matrix. To understand its sign pattern, first recall how to calculate the determinant of a $2 \times 2$ matrix.

If we have a $2 \times 2$ matrix:

M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}

Its determinant, $\det(M)$ or $|M|$ , is calculated as follows:

\det(M) = ad - bc

This is the difference between the product of the main diagonal elements ( $ad$ ) and the product of the secondary diagonal elements ( $bc$ ). Sarrus' method adapts this principle for $3 \times 3$ matrices.

Calculating a Three by Three Determinant with the Sarrus Method

Suppose we have a $3 \times 3$ matrix $A$ :

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

The element $a_{ij}$ is the element in the $i$ -th row and $j$ -th column.

Step 1: Copy the First Two Columns

Rewrite the first two columns of matrix $A$ to the right of the third column:

\begin{array}{|ccc|cc} a_{11} & a_{12} & a_{13} & a_{11} & a_{12} \\ a_{21} & a_{22} & a_{23} & a_{21} & a_{22} \\ a_{31} & a_{32} & a_{33} & a_{31} & a_{32} \end{array}

The copied columns form the three downward and three upward diagonals used in the products.

Step 2: Calculate the Sum of the Products of the Positive Diagonals

Multiply the elements along the three diagonals from the top-left to the bottom-right. Sum these products, call it $D_{\text{positive}}$ .

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

The first term is the product of the main diagonal. The second and third terms are products of parallel diagonals involving elements from the copied columns.

Step 3: Calculate the Sum of the Products of the Negative Diagonals

Multiply the elements along the three diagonals from the top-right to the bottom-left. Sum these products, call it $D_{\text{negative}}$ .

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

The first term is the product of the secondary diagonal (anti-diagonal). The second and third terms are products of parallel diagonals involving elements from the copied columns, moving towards the bottom-left.

Step 4: Calculate the Final Determinant

The determinant of matrix $A$ , $\det(A)$ , is the difference between $D_{\text{positive}}$ and $D_{\text{negative}}$ :

\det(A) = D_{\text{positive}} - D_{\text{negative}}

Substitute the values of $D_{\text{positive}}$ and $D_{\text{negative}}$ :

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

Or, after distributing the negative sign:

\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}

Numerical Worked Example

Calculate the determinant of matrix $P$ :

P=\begin{bmatrix}1&3&2\\2&6&2\\5&9&4\end{bmatrix}

Keep the two diagonal sums separate until the final subtraction. This makes the sign easier to check:

\begin{aligned} D_+ &=(1)(6)(4)+(3)(2)(5) \\ &\quad+(2)(2)(9) \\ &=24+30+36 \\ &=90 \end{aligned}

The determinant is $-12$ . Expanding the same matrix by cofactors gives the same value, which is a useful independent check.

Six Diagonal Products in the Sarrus Method

Begin with the matrix:

\det(A) = \left| \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right|

Then, using Sarrus' Method, we expand the matrix and identify the multiplication paths:

\begin{array}{ccc|cc} a_{11} & a_{12} & a_{13} & a_{11} & a_{12} \\ a_{21} & a_{22} & a_{23} & a_{21} & a_{22} \\ a_{31} & a_{32} & a_{33} & a_{31} & a_{32} \end{array}

Follow the negative paths $\swarrow$ . The second starts from $a_{11}$ in copied column 4. The third starts from $a_{12}$ in copied column 5:

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

The complete formula becomes:

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

Sarrus Method Applies Only to Three by Three Matrices

Sarrus' method is only applicable to $3 \times 3$ matrices. The $2 \times 2$ calculation above simply reminds us of the diagonal sign pattern. For larger matrices, such as a $4 \times 4$ matrix, use another method such as cofactor expansion or row reduction.

Practice with a Worked Solution

Use Sarrus' method to calculate the determinant of:

B=\begin{bmatrix}2&-1&3\\0&4&5\\1&2&-2\end{bmatrix}

First calculate the two diagonal sums, then subtract the negative-diagonal sum from the positive-diagonal sum:

\begin{aligned} D_+ &=(2)(4)(-2)+(-1)(5)(1) \\ &\quad+(3)(0)(2) \\ &=-16-5+0 \\ &=-21 \end{aligned}

$\det(B)=-53$ .