Source codeVideos

Command Palette

Search for a command to run...

Matrix

Sarrus Method

Basic Concept of Sarrus Method

Sarrus' method is a practical way to calculate the determinant of a 3x3 matrix. This method is named after Pierre Frédéric Sarrus. To understand it, let's recall how to calculate the determinant of a 2x2 matrix.

If we have a 2x2 matrix:

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

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

det(M)=adbc\det(M) = ad - bc

This is the difference between the product of the main diagonal elements (adad) and the product of the secondary diagonal elements (bcbc). Sarrus' method adapts this principle for 3x3 matrices.

Steps to Calculate the Determinant of a 3x3 Matrix using Sarrus Method

Suppose we have a 3x3 matrix A:

A=[a11a12a13a21a22a23a31a32a33]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 aija_{ij} is the element in the ii-th row and jj-th column.

Step 1: Copy the First Two Columns

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

a11a12a13a11a12a21a22a23a21a22a31a32a33a31a32\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

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

Dpositive=(a11a22a33)+(a12a23a31)+(a13a21a32)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

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

Dnegative=(a13a22a31)+(a11a23a32)+(a12a21a33)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

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

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

Substitute the values of DpositiveD_{\text{positive}} and DnegativeD_{\text{negative}}:

det(A)=(a11a22a33+a12a23a31+a13a21a32)(a13a22a31+a11a23a32+a12a21a33)\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:

det(A)=a11a22a33+a12a23a31+a13a21a32a13a22a31a11a23a32a12a21a33\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:

det(A)=a11a12a13a21a22a23a31a32a33\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:

a11a12a13a11a12a21a22a23a21a22a31a32a33a31a32\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}
Positive Paths ():+(a11a22a33)+(a12a23a31)+(a13a21a32)\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}
Negative Paths ():(a13a22a31)(a11a23a32)(from a11 col. 4)(a12a21a33)(from a12 col. 5)\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:

det(A)=(a11a22a33+a12a23a31+a13a21a32)(a13a22a31+a11a23a32+a12a21a33)\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 2x2 and 3x3 matrices. For matrices of higher order (e.g., 4x4), this method cannot be used. Other methods such as cofactor expansion or row reduction are required for such cases.