Solution
- $x$: —
- $y$: —
- $z$: —
Step-by-step
Explanation:
3x3 System of Equations Solver
This tool solves a system of three linear equations with three variables and shows the work step by step using Cramer’s rule. Enter the coefficients for $a_1x+b_1y+c_1z=d_1,\quad a_2x+b_2y+c_2z=d_2,\quad a_3x+b_3y+c_3z=d_3$ and the solver will compute the determinants $D$, $D_x$, $D_y$, and $D_z$ to determine whether the system has a unique solution, no solution, or infinitely many solutions. When a unique solution exists, it returns the exact values of $x$, $y$, and $z$. For more algebra tools, explore Algebra.
Solve a 3x3 system in the form:
$$a_1x + b_1y + c_1z = d_1$$
$$a_2x + b_2y + c_2z = d_2$$
$$a_3x + b_3y + c_3z = d_3$$
What Is a 3x3 System of Equations?
A 3x3 system is three linear equations written in the same three variables, typically $x$, $y$, and $z$. A solution is a triple $(x,y,z)$ that satisfies all three equations at the same time. Some systems have exactly one solution, some have no solution (because the equations conflict), and some have infinitely many solutions (because at least one equation is redundant).
How This Solver Works
This solver uses Cramer’s rule, which is based on determinants. First it computes the main determinant $D$ from the coefficient matrix. Then it computes three additional determinants $D_x$, $D_y$, and $D_z$ by replacing one column at a time with the constants. These values determine both the solution type and, when it exists, the exact values of $x$, $y$, and $z$.
$$D=\begin{vmatrix}
a_1 & b_1 & c_1\\
a_2 & b_2 & c_2\\
a_3 & b_3 & c_3
\end{vmatrix},\quad
D_x=\begin{vmatrix}
d_1 & b_1 & c_1\\
d_2 & b_2 & c_2\\
d_3 & b_3 & c_3
\end{vmatrix},\quad
D_y=\begin{vmatrix}
a_1 & d_1 & c_1\\
a_2 & d_2 & c_2\\
a_3 & d_3 & c_3
\end{vmatrix},\quad
D_z=\begin{vmatrix}
a_1 & b_1 & d_1\\
a_2 & b_2 & d_2\\
a_3 & b_3 & d_3
\end{vmatrix}$$
$$D\ne 0 \Rightarrow x=\frac{D_x}{D},\; y=\frac{D_y}{D},\; z=\frac{D_z}{D}$$
How to Interpret the Determinants
Cramer’s rule also gives a clear classification:
- Unique solution: if $D \ne 0$, the system has exactly one solution $(x,y,z)$.
- Infinitely many solutions: if $D=0$ and $D_x=0$ and $D_y=0$ and $D_z=0$, the equations are consistent but dependent, so there are infinitely many solutions.
- No solution: if $D=0$ but at least one of $D_x$, $D_y$, or $D_z$ is nonzero, the equations are inconsistent, so no triple $(x,y,z)$ satisfies all of them.
If you only need a two-variable system, use the 2x2 System of Equations Solver. For a single equation in one variable, use the Linear Equation Solver. If your equation includes an $x^2$ term, use the Quadratic Solver.
Input Notes and Numerical Stability
Coefficients can be integers or decimals, including negative values. The tool formats outputs using the configured decimal precision. In real-world inputs, determinants can be extremely close to zero due to rounding, even when the exact value is zero. To keep the classification stable, the solver treats values that are very close to zero as zero. If your result is near the boundary between cases, try increasing precision or using simpler exact fractions where possible.
Worked Examples With Steps
Example 1: Unique solution
A standard system with one clear solution.
Solve: $$\begin{cases} x + y + z = 6\\ 2x - y + z = 5\\ x + 2y - z = 3 \end{cases}$$
Compute the main determinant: $$D=\begin{vmatrix} 1 & 1 & 1\\ 2 & -1 & 1\\ 1 & 2 & -1 \end{vmatrix}=-9$$
Replace columns to compute: $$D_x=\begin{vmatrix} 6 & 1 & 1\\ 5 & -1 & 1\\ 3 & 2 & -1 \end{vmatrix}=-18,\quad D_y=\begin{vmatrix} 1 & 6 & 1\\ 2 & 5 & 1\\ 1 & 3 & -1 \end{vmatrix}=-9,\quad D_z=\begin{vmatrix} 1 & 1 & 6\\ 2 & -1 & 5\\ 1 & 2 & 3 \end{vmatrix}=-27$$
Since $D\ne 0$: $$x=\frac{D_x}{D}=2,\quad y=\frac{D_y}{D}=1,\quad z=\frac{D_z}{D}=3$$
Example 2: Infinitely many solutions
The equations describe the same constraint, so there are infinitely many solutions.
Solve: $$\begin{cases} x + y + z = 3\\ 2x + 2y + 2z = 6\\ 3x + 3y + 3z = 9 \end{cases}$$
Here each equation is a multiple of the first, so the coefficient matrix is dependent and: $$D=0,\quad D_x=0,\quad D_y=0,\quad D_z=0$$
Since all determinants are zero, the system has infinitely many solutions. For example, any $(x,y,z)$ with $x+y+z=3$ works.
Example 3: No solution
The equations conflict, so no triple $(x,y,z)$ can satisfy all of them.
Solve: $$\begin{cases} x + y + z = 1\\ x + y + z = 2\\ x - y + z = 0 \end{cases}$$
The first two equations already contradict each other. In determinant terms, the coefficient rows are dependent so $D=0$, but the constants are inconsistent so at least one of $D_x$, $D_y$, or $D_z$ is nonzero.
Therefore, the system has no solution.
Keep exploring Mathematics or jump back to Calculators to browse more tools.
Questions About the 3x3 System Solver
Quick answers about determinants, Cramer's rule, and solution types.
It solves three linear equations in three variables written as $a_1x+b_1y+c_1z=d_1$, $a_2x+b_2y+c_2z=d_2$, and $a_3x+b_3y+c_3z=d_3$, and reports whether the system has a unique solution, no solution, or infinitely many solutions.
It uses Cramer’s rule. The solver computes the determinants $D$, $D_x$, $D_y$, and $D_z$ from the coefficient matrix and the constants to determine the solution type and, when $D\ne 0$, the values of $x$, $y$, and $z$.
The system has a unique solution when the main determinant $D\ne 0$. In that case, $x=\frac{D_x}{D}$, $y=\frac{D_y}{D}$, and $z=\frac{D_z}{D}$.
It has infinitely many solutions when $D=0$ and $D_x=0$ and $D_y=0$ and $D_z=0$. This means the equations are consistent but dependent, so there are infinitely many solutions.
It has no solution when $D=0$ but at least one of $D_x$, $D_y$, or $D_z$ is nonzero. This indicates the equations are inconsistent, so no triple $(x,y,z)$ satisfies all of them.
Yes. The solver accepts decimal and negative coefficients for all inputs and formats the results using the tool’s configured decimal precision.
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