In linear algebra, the Rule of Sarrus is a mnemonic device for computing the determinant of a \( 3\times 3 \) matrix named after the French mathematician Pierre Frédéric Sarrus.[1]
Consider a \( 3\times 3 \) matrix
\( {\displaystyle M={\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}},} \)
then its determinant can be computed by the following scheme.
Write out the first two columns of the matrix to the right of the third column, giving five columns in a row. Then add the products of the diagonals going from top to bottom (solid) and subtract the products of the diagonals going from bottom to top (dashed). This yields[1][2]
\( {\displaystyle {\begin{aligned}\det(M)&=\det {\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}}\\[6pt]& =a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{31}a_{22}a_{13}-a_{32}a_{23}a_{11}-a_{33}a_{21}a_{12}.\end{aligned}}} \)
Alternative vertical arrangement
A similar scheme based on diagonals works for \( {\displaystyle 2\times 2} \) matrices:[1]
\( {\displaystyle \det(M)=\det {\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}=a_{11}a_{22}-a_{21}a_{12}.} \)
Both are special cases of the Leibniz formula, which however does not yield similar memorization schemes for larger matrices. Sarrus's rule can also be derived using the Laplace expansion of a \( 3\times 3 \) matrix.[1]
Another way of thinking of Sarrus' rule is to imagine that the matrix is wrapped around a cylinder, such that the right and left edges are joined.
References
Fischer, Gerd (1985). Analytische Geometrie (in German) (4th ed.). Wiesbaden: Vieweg. p. 145. ISBN 3-528-37235-4.
Paul Cohn: Elements of Linear Algebra. CRC Press, 1994, ISBN 9780412552809, p. 69
Undergraduate Texts in Mathematics
Graduate Studies in Mathematics
Hellenica World - Scientific Library
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