Sarrus Rule Calculator guide
How to Find the Determinant of a 3×3 Matrix
The determinant of a 3×3 matrix can be found two ways. The rule of Sarrus copies the first two columns to the right, then adds the three down-right diagonal products and subtracts the three up-right ones. Cofactor expansion takes each entry of the top row, multiplies it by the 2×2 determinant of the entries not in its row or column, and alternates the signs + − +. Both give the same number.
The formula, written out once
Before either method, here is what you are actually computing. If your matrix has entries a, b, c across the top, d, e, f in the middle, and g, h, i along the bottom, then:
det = a(ei − fh) − b(di − fg) + c(dh − eg)
You do not need to memorise that. It is here because it helps to know that both methods below are just different ways of organising those same six products. Sarrus arranges them as a picture; cofactor expansion arranges them as algebra. Neither is more correct than the other.
Method 1 — the rule of Sarrus
Copy the first two columns to the right of your matrix, so you have a 3×5 grid. Then take six diagonal products: the three running down-right get added together, and the three running up-right get subtracted.
The calculator on this page draws the diagonals over your matrix as it works, which makes it much easier to see where each of the six terms came from.
If you are working by hand with small numbers, this is usually the faster of the two.
Method 2 — cofactor expansion
Take the first entry of the top row. Cross out its row and its column, and look at the four numbers left behind — that is a 2×2 matrix. Find its determinant (top-left times bottom-right, minus top-right times bottom-left), and multiply the result by the entry you started from.
Now repeat for the second entry of the top row, and then the third.
Finally, combine your three results with alternating signs: plus the first, minus the second, plus the third. That alternating pattern trips people up, so it is worth writing the signs down before you start filling in numbers.
This method is slower on a 3×3, but it is the one that keeps working when the matrix gets bigger, so it is worth being comfortable with both.
Which should you use?
If the matrix is 3×3 and you are working on paper, reach for Sarrus. There is less to write down and the diagonals are easy to check visually.
If the matrix is bigger than 3×3, you have no choice — use cofactor expansion. And if your matrix has zeros in it, cofactor expansion suddenly becomes the faster option too, because every zero entry wipes out an entire term before you have to compute it.
If your homework says "expand by minors" or "expand along the first row", that is asking for cofactor expansion specifically, even though Sarrus would reach the same number.
A note on the matrix in the calculator
The example loaded above has rows (1, 2, 3), (4, 5, 6) and (7, 8, 10), and its determinant is −3.
It is worth trying a small experiment with it: change that bottom-right 10 to a 9 and watch the determinant drop to zero. That single change makes the rows depend on each other — the middle row becomes exactly the average of the other two — and once that happens the determinant collapses. Seeing it happen live tends to make the idea stick better than reading about it.
