System of Equations (3×3)
This solver handles three linear equations a·x + b·y + c·z = d in the unknowns x, y, z. It computes the coefficient determinant D and the three replaced-column determinants Dₓ, Dᵧ, D_z, then reports x = Dₓ/D, y = Dᵧ/D, z = D_z/D.
The same rule, one dimension up
Three linear equations in three unknowns describe three planes in space. A unique solution is the single point all three pass through — and it is easy to underestimate how special that is, since three planes can also meet along a whole line, or in no common point at all.
This solver extends Cramer's rule to that size. Four determinants are computed instead of three, and each unknown is one of them divided by the first.
The structure is identical to the two-variable case: build a determinant from the coefficients of the unknowns, then replace one column at a time with the constants. What changes is the size of each determinant and therefore the amount of arithmetic behind each one.
That growth is why the method stops being practical shortly after this. Each determinant here expands into three smaller ones, and the count multiplies with every additional variable — which is why larger systems are solved by elimination instead.
How to use this calculator
- Rearrange each equation into standard form Three coefficients and a constant per equation, with all variable terms on the left. Twelve numbers in total.
- Enter the coefficients row by row Four fields per equation, in the order of the three unknowns and then the right-hand side. A variable absent from an equation takes a coefficient of zero.
- Read the coefficient determinant Computed first. A zero value ends the calculation, since every unknown would require dividing by it.
- Read the three replaced determinants and the ratios One per unknown, each shown before its division. Comparing them against your own working localises an error to a single determinant.
The formula, and where it comes from
D = det of the coefficients Dₓ, Dᵧ, D_z = D with that column replaced by the constants x = Dₓ/D, y = Dᵧ/D, z = D_z/D
Each determinant is computed by expansion along the first row: every entry is multiplied by the determinant of the smaller matrix left when its row and column are removed, and the three products are combined with alternating signs. That middle minus sign is where hand-computed determinants most often go wrong.
The replacement pattern is the same as at the smaller size. To solve for the first unknown, the first column is swapped for the constants; for the second, the second column; and so on. Only one column changes each time, and the rest of the matrix is untouched.
All four determinants are computed from the same twelve numbers, so an error in reading a single coefficient corrupts several of them at once. Checking the echoed equations before reading the results is worth the moment it takes.
The three unknowns are found independently, each by its own division. Nothing is substituted back, which means an error in one value does not disturb the other two — and also that a solution has to be verified against all three original equations rather than just one.
What each input means
- a₁…d₁ First equation — form field “a₁ (eq 1, x)”
- Coefficients of the three unknowns followed by the constant. An absent variable has coefficient zero.
- a₂…d₂ Second equation — form field “a₂ (eq 2, x)”
- The same four values for the second equation.
- a₃…d₃ Third equation — form field “a₃ (eq 3, x)”
- And for the third.
- D Coefficient determinant
- Built from the nine coefficients of the unknowns. Non-zero exactly when the three planes meet at one point.
Worked examples
Every number below is produced by the same calculation engine the tool above runs. Nothing here is typed by hand, so the walkthrough cannot drift from what you get when you enter the same values yourself.
A system with a unique solution
Three equations whose coefficient determinant is non-zero, so the three planes intersect at a single point.
Inputs a₁ (eq 1, x) = 1, b₁ (eq 1, y) = 1, c₁ (eq 1, z) = 1, d₁ (eq 1, =) = 6, a₂ (eq 2, x) = 2, b₂ (eq 2, y) = -1, c₂ (eq 2, z) = 1, d₂ (eq 2, =) = 3, a₃ (eq 3, x) = 1, b₃ (eq 3, y) = 2, c₃ (eq 3, z) = -1, d₃ (eq 3, =) = 2
- Equation 1 1x + 1y + 1z = 6
- Equation 2 2x − 1y + 1z = 3
- Equation 3 1x + 2y − 1z = 2
- Determinant D = 7
- Dₓ 7
- Dᵧ 14
- D_z 21
- x Dₓ / D = 1
- y Dᵧ / D = 2
- z D_z / D = 3
Result x = 1, y = 2, z = 3
The coefficient determinant is non-zero, so the three planes are in genuinely independent orientations and cross at one point. Each unknown then follows from its own replaced determinant.
Substituting the answer into all three equations is the necessary check. Because the three values are computed independently, verifying only one equation could leave an error in another undetected.
A system with zeros among the coefficients
Each equation omits one of the three unknowns, so a third of the coefficient entries are zero.
Inputs a₁ (eq 1, x) = 1, b₁ (eq 1, y) = 1, c₁ (eq 1, z) = 0, d₁ (eq 1, =) = 5, a₂ (eq 2, x) = 0, b₂ (eq 2, y) = 1, c₂ (eq 2, z) = 1, d₂ (eq 2, =) = 7, a₃ (eq 3, x) = 1, b₃ (eq 3, y) = 0, c₃ (eq 3, z) = 1, d₃ (eq 3, =) = 6
- Equation 1 1x + 1y + 0z = 5
- Equation 2 0x + 1y + 1z = 7
- Equation 3 1x + 0y + 1z = 6
- Determinant D = 2
- Dₓ 4
- Dᵧ 6
- D_z 8
- x Dₓ / D = 2
- y Dᵧ / D = 3
- z D_z / D = 4
Result x = 2, y = 3, z = 4
The zeros are entries the determinants need, not omissions. They make the expansion shorter, since any term multiplied by zero drops out, but the method is otherwise unchanged.
The determinant is non-zero, so the three planes still meet at one point. A sparse coefficient matrix says nothing about whether a system is degenerate — only the determinant does.
Reading the result
What three planes can do
Meet at one point, meet along a line, coincide entirely, or have no common point at all. Only the first gives a unique solution, and only that case has a non-zero coefficient determinant.
A zero determinant is a stopping point, not a classification
It rules out a unique solution without saying whether there are none or infinitely many. The general linear-system solver reduces the augmented matrix and distinguishes the two, reporting free variables where they exist.
Verify against all three equations
The unknowns are computed independently rather than by back-substitution, so satisfying one equation says nothing about the others. A complete check means substituting into all three.
When you would use this
Three constraints in three quantities
A mixture problem with three components meeting a total, a cost and a proportion is exactly this shape once each condition is written as a linear equation.
Fitting a curve through three points
Finding the quadratic through three given points means solving for its three coefficients, and substituting each point produces one linear equation in them.
Assumptions and limitations
What this calculator assumes
- All three equations are linear in the same three unknowns and are entered in standard form.
- Coefficients are finite real numbers; an absent variable has coefficient zero.
- A unique solution exists exactly when the coefficient determinant is non-zero.
- Each unknown is computed independently as a ratio of determinants, with no back-substitution.
Where it stops being the right tool
- Exactly three equations in three unknowns; larger systems need the general solver.
- A zero determinant ends the calculation without saying which degenerate case applies.
- No parametric form is produced when the solution set is a line or a plane.
- Coefficients are entered rather than parsed from typed equations.
Common mistakes
Losing the alternating sign in a determinant
Why it happens. The expansion has a plus, a minus and a plus, and the pattern depends on position rather than on the numbers. Adding all three terms produces a plausible determinant that is simply wrong.
How to avoid it. Compare your determinant against the reported one before checking the ratios. An error there corrupts every unknown at once.
Entering the constants in the coefficient columns
Why it happens. Four fields per row look interchangeable, and the constant is just another number in the sequence.
How to avoid it. The fourth field of each row is the right-hand side. The echoed equations show how the twelve numbers were interpreted.
Verifying against only one equation
Why it happens. One successful substitution feels like confirmation, and it would be under back-substitution — where later values depend on earlier ones.
How to avoid it. Check all three. Cramer's rule computes the unknowns independently, so each equation is a separate test.
Frequently asked questions
What method does it use?
Cramer's rule at three variables. The coefficient determinant is computed first, then three more with one column each replaced by the constants, and every unknown is its own determinant divided by the first.
What does a zero determinant mean?
That the three planes are not independently oriented — they may share a line, coincide, or simply have no common point. Either infinitely many solutions or none, and the division Cramer's rule needs is impossible in both cases.
How do I enter my equations?
Rearrange each into the form with three variable terms on the left and a constant on the right, then enter four numbers per row. A variable missing from an equation has a coefficient of zero rather than a blank field.
Why not use this method for larger systems?
Because the work grows very quickly. Each determinant at this size expands into three smaller ones, and that multiplication compounds with every extra variable. Elimination on the augmented matrix scales far better and also classifies the degenerate cases.