Linear Algebra 47 | Rule of Sarrus
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The determinant definition uses the Leibniz formula: a sum over all permutations of column indices, with each term weighted by a sign determined by permutation parity.
Briefing
For 3×3 matrices, the determinant can be computed quickly using the Rule of Sarrus—an efficient shortcut that reproduces exactly the six terms from the full Leibniz (permutation) formula, without missing any contributions. The method matters because it turns a permutation-sum problem (which grows explosively with matrix size) into a straightforward “diagonal products” recipe that humans can carry out fast on paper.
The starting point is the general determinant definition: for an n×n matrix, the determinant is a sum over all permutations of the column indices, pairing each permutation with a sign (±1) determined by whether the permutation is even or odd. Each term is a product of n matrix entries, with every row and every column used exactly once. Counting permutations explains why shortcuts are rare: there are n! permutations total. For n=2 there are only 2 permutations, and for n=3 there are 6—small enough that a special rule can list them all. For n=4, the count jumps to 24, making the diagonal shortcut impractical.
In the 3×3 case, the Rule of Sarrus organizes the six permutation terms as products along diagonals. One set comes from the “main” orientation: multiplying the three diagonals in the downward direction yields the three even-permutation terms, all added with a plus sign. The other set comes from the “flipped” orientation: multiplying the three diagonals in the opposite direction yields the three odd-permutation terms, which are subtracted. This mirrors the Leibniz signs: even permutations contribute positively, odd permutations contribute negatively. The rule works precisely because those six diagonal products correspond one-to-one with the six permutations of {1,2,3}.
A worked example demonstrates the mechanics. After writing the 3×3 matrix, the calculation proceeds by multiplying the three positive diagonals (each product uses three entries). Then the three products from the flipped diagonals are computed. When the positive and negative groups are combined, many terms cancel, leaving the final determinant value. The example’s cancellations underscore the practical benefit: the arithmetic is organized so the final result emerges with minimal bookkeeping.
Finally, the rule’s limits are made explicit. The Rule of Sarrus is tailored to n=3 and does not extend to n=4 or higher, because the diagonal products no longer cover all permutations. For larger matrices, the next step is the Laplace expansion, which provides a general method that can handle bigger n even though it is more computationally involved than the 3×3 diagonal trick.
Cornell Notes
The determinant of a 3×3 matrix can be computed using the Rule of Sarrus, a shortcut that reproduces the six terms from the full Leibniz permutation formula. Even permutations correspond to three “downward” diagonal products and are added; odd permutations correspond to three “upward/flipped” diagonal products and are subtracted. This works because a 3×3 matrix has exactly 3! = 6 permutations, so the diagonal products cover every required term with the correct sign. For n=4 and higher, the number of permutations becomes too large (n! grows quickly), so Sarrus no longer accounts for all terms. For larger sizes, Laplace expansion is used instead.
Why does the Rule of Sarrus work specifically for 3×3 determinants?
How are the signs determined in the Rule of Sarrus?
What goes wrong if someone tries to use Sarrus for a 4×4 determinant?
How does the Rule of Sarrus relate to the Leibniz formula?
In a worked example, why do cancellations often occur?
Review Questions
- How many terms does the Leibniz formula require for a 3×3 determinant, and how does that number connect to the Rule of Sarrus?
- Describe which diagonal products are added and which are subtracted in the Rule of Sarrus, and explain how this matches even/odd permutations.
- Why is the Rule of Sarrus not applicable to 4×4 determinants? Use the permutation-count argument (n!).
Key Points
- 1
The determinant definition uses the Leibniz formula: a sum over all permutations of column indices, with each term weighted by a sign determined by permutation parity.
- 2
For an n×n matrix, the number of permutation terms is n!, which grows rapidly as n increases.
- 3
A 3×3 determinant has exactly 3! = 6 terms, making a complete shortcut possible.
- 4
The Rule of Sarrus computes the 3×3 determinant by multiplying three diagonals in one direction (added) and three diagonals in the opposite direction (subtracted).
- 5
The three “main-direction” diagonal products correspond to even permutations (+), while the three “flipped-direction” diagonal products correspond to odd permutations (−).
- 6
Sarrus fails for n=4 because 4! = 24 permutation terms cannot be captured by only six diagonal products.
- 7
For larger determinants where diagonal shortcuts don’t work, Laplace expansion is the next general method discussed.