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Exercise 449Linear algebra › DeterminantsHigh schoolAdded 3 Nov 2026

Determinant of a 3×3 matrix by Sarrus' rule

#determinant#Sarrus rule#3x3 matrices

Statement

Compute the determinant of
A=(1234567810).A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix}.

Hint

Use Sarrus' rule: add the three products along downward diagonals and subtract the three products along upward diagonals.

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Solution

Sarrus' rule.
det⁡A=1⋅5⋅10+2⋅6⋅7+3⋅4⋅8−3⋅5⋅7−1⋅6⋅8−2⋅4⋅10.\det A = 1\cdot5\cdot10 + 2\cdot6\cdot7 + 3\cdot4\cdot8 - 3\cdot5\cdot7 - 1\cdot6\cdot8 - 2\cdot4\cdot10.
Compute each term:
=50+84+96−105−48−80=230−233=−3.= 50 + 84 + 96 - 105 - 48 - 80 = 230 - 233 = -3.

Conclusion. det⁡A=−3\det A = -3, so AA is invertible.

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