Cramer's Rule: Solving 3 Equations with Determinants

What you need before you start

Cramer's rule works when you have the same number of equations as unknowns, and the determinant of the coefficient matrix is not zero. For a 3×3 system that means three equations, three unknowns, and D ≠ 0.

Check D first, before anything else. If it turns out to be zero, stop — the rule genuinely cannot help you here, and carrying on will only produce a division by zero. A zero D means the system has either no solution or infinitely many, and you will need elimination or row reduction to sort out which.

The nice thing about this method is that it is completely mechanical. There are no choices to make and no moments where you have to spot something clever, which is exactly what you want when you are tired or under time pressure.

Step 1 — find D

Take the coefficients of x, y and z, arrange them as a 3×3 matrix in the same order they appear in your equations, and find its determinant. This is D.

D is the denominator that all three of your answers will share, so it is worth double-checking before you go further. The calculator on this page is loaded with the coefficient matrix from our worked example, whose determinant is −5.

Steps 2 to 4 — replace one column at a time

Now you build three more matrices, each one a small edit of the original.

For Dx, take the coefficient matrix and replace the first column — the x coefficients — with the constants from the right-hand side of your equations. Everything else stays exactly as it was. Take the determinant of that.

For Dy, go back to the *original* matrix and replace the second column with those same constants. For Dz, replace the third.

The thing to watch here is that each replacement starts from the original matrix, not from the one you just edited. It is an easy slip to make when you are working quickly down a page.

Each of these is an ordinary 3×3 determinant, so the rule of Sarrus works on all of them. Four Sarrus calculations and you have everything you need.

Step 5 — divide

x = Dx ÷ D, y = Dy ÷ D, z = Dz ÷ D. That is your answer.

Substitute all three values back into one of your original equations and check that it balances. With four determinants behind you, an arithmetic slip somewhere is genuinely likely, and this check takes about twenty seconds.

Is it worth using?

For a 3×3 system by hand, yes. It is competitive with elimination, and its being mechanical is a real advantage — there is no point at which you can get stuck wondering what to do next.

For anything larger it falls apart quickly. A 4×4 system needs five 4×4 determinants, and each of those is a serious piece of work on its own. Gaussian elimination is the method that scales, and it is what any software would use.