# Nakafa Learning Content > For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`. URL: https://nakafa.com/en/subjects/mathematics/matrix/sarrus-method Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/matrix/sarrus-method/en.mdx Learn Sarrus method for calculating 3×3 matrix determinants quickly. Learn the diagonal technique with visual approach and examples. --- ## Basic Concept of Sarrus Method Sarrus' method is a practical way to calculate the determinant of a $$3 \times 3$$ matrix. This method is named after Pierre Frédéric Sarrus. To understand it, let's recall how to calculate the determinant of a $$2 \times 2$$ matrix. Visible text: Sarrus' method is a practical way to calculate the determinant of a matrix. This method is named after Pierre Frédéric Sarrus. To understand it, let's recall how to calculate the determinant of a matrix. If we have a $$2 \times 2$$ matrix: Visible text: If we have a matrix: ```math M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} ``` Its determinant, $$\det(M)$$ or $$|M|$$, is calculated as follows: Visible text: Its determinant, or , is calculated as follows: ```math \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. Visible text: This is the difference between the product of the main diagonal elements () and the product of the secondary diagonal elements (). Sarrus' method adapts this principle for matrices. ## Steps to Calculate an Order-Three Determinant with Sarrus Method Suppose we have a $$3 \times 3$$ matrix $$A$$: Visible text: Suppose we have a matrix : ```math 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. Visible text: The element is the element in the -th row and -th column. **Step** $$1$$: Copy the First Two Columns Visible text: **Step** : Copy the First Two Columns Rewrite the first two columns of matrix $$A$$ to the right of the third column: Visible text: Rewrite the first two columns of matrix to the right of the third column: ```math \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} ``` This helps us visualize the diagonals that will be multiplied. **Step** $$2$$: Calculate the Sum of the Products of the Positive Diagonals Visible text: **Step** : 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, let's call it $$D_{\text{positive}}$$. Visible text: Multiply the elements along the three diagonals from the top-left to the bottom-right. Sum these products, let's call it . Component: MathContainer Children: ```math D_{\text{positive}} = (a_{11} \cdot a_{22} \cdot a_{33}) + (a_{12} \cdot a_{23} \cdot a_{31}) + (a_{13} \cdot a_{21} \cdot a_{32}) ``` 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 Visible text: **Step** : 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, let's call it $$D_{\text{negative}}$$. Visible text: Multiply the elements along the three diagonals from the top-right to the bottom-left. Sum these products, let's call it . Component: MathContainer Children: ```math D_{\text{negative}} = (a_{13} \cdot a_{22} \cdot a_{31}) + (a_{11} \cdot a_{23} \cdot a_{32}) + (a_{12} \cdot a_{21} \cdot a_{33}) ``` 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 Visible text: **Step** : Calculate the Final Determinant The determinant of matrix $$A$$, $$\det(A)$$, is the difference between $$D_{\text{positive}}$$ and $$D_{\text{negative}}$$: Visible text: The determinant of matrix , , is the difference between and : ```math \det(A) = D_{\text{positive}} - D_{\text{negative}} ``` Substitute the values of $$D_{\text{positive}}$$ and $$D_{\text{negative}}$$: Visible text: Substitute the values of and : ```math \det(A) = (a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32}) - (a_{13}a_{22}a_{31} + a_{11}a_{23}a_{32} + a_{12}a_{21}a_{33}) ``` Or, after distributing the negative sign: ```math \det(A) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} ``` ### Visualizing Sarrus Method To visualize this process, we can write: ```math \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: Component: MathContainer Children: ```math \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} ``` ```math \begin{array}{l} \text{Positive Paths } (\searrow): \\ \quad + (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{array} ``` ```math \begin{array}{l} \text{Negative Paths } (\swarrow): \\ \quad - (a_{13} \cdot a_{22} \cdot a_{31}) \\ \quad - (a_{11} \cdot a_{23} \cdot a_{32}) \quad (\text{from } a_{11} \text{ col. 4}) \\ \quad - (a_{12} \cdot a_{21} \cdot a_{33}) \quad (\text{from } a_{12} \text{ col. 5}) \end{array} ``` Thus, the complete formula becomes: ```math \det(A) = (a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32}) - (a_{13}a_{22}a_{31} + a_{11}a_{23}a_{32} + a_{12}a_{21}a_{33}) ``` ## Important Limitation of Sarrus Method Sarrus' method is **only applicable to $$2 \times 2$$ and $$3 \times 3$$ matrices**. For matrices of higher order (e.g., $$4 \times 4$$), this method cannot be used. Other methods such as cofactor expansion or row reduction are required for such cases. Visible text: Sarrus' method is **only applicable to and matrices**. For matrices of higher order (e.g., ), this method cannot be used. Other methods such as cofactor expansion or row reduction are required for such cases.