Simultaneous equations (three or more variables)
Master simultaneous equations in three or more variables for Year 12 Specialist Mathematics in Queensland (QCAA). A system of linear equations is written compactly as the matrix equation \(A\mathbf{x}=\mathbf{b}\), and its solution is an ordered triple \((x,y,z)\) that satisfies every equation at once.
You will learn to set up the coefficient matrix, solve \(3\times3\) systems by elimination or with technology, and classify each system as having a unique solution, no solution, or infinitely many solutions — reading the result as three planes that meet at a point, share a line, or never all meet.
Theory
A system of three or more linear equations can be written compactly as the matrix equation \(A\mathbf{x}=\mathbf{b}\). In Year 12 Specialist Mathematics (QCAA, Queensland) you solve \(3\times3\) systems for the ordered triple \((x,y,z)\), and classify every system as having a unique solution, no solution, or infinitely many solutions — each with a picture of three planes.
A linear equation in three variables such as \(ax+by+cz=d\) describes a plane in space. A system is a set of such equations that must hold at the same time, and a solution is an ordered triple \((x,y,z)\) that satisfies every equation.
Collecting the coefficients into a coefficient matrix \(A\), the unknowns into a column \(\mathbf{x}\), and the constants into a column \(\mathbf{b}\), the whole system becomes the single matrix equation \(A\mathbf{x}=\mathbf{b}\). The rows of \(A\) are just the coefficients of each equation read in the order \(x,y,z\).
There are exactly three cases. If \(A\) is non-singular (\(\det A\neq0\)) the system has a unique solution. If \(A\) is singular (\(\det A=0\)) the system has either no solution (the equations contradict each other) or infinitely many solutions (an equation carries no new information).
Geometrically each equation is a plane. A unique solution is three planes meeting at a single point; no solution is planes with no common point (some are parallel, or they form a triangular prism); infinitely many solutions is three planes sharing a common line.
A system of three equations in \(x,y,z\) is written as the matrix equation \(A\mathbf{x}=\mathbf{b}\):
The determinant of \(A\) decides which case the system is in:
When \(A\) is non-singular the solution can be found by the inverse (or with technology):
How to solve and classify a 3-variable system
- Write it as \(A\mathbf{x}=\mathbf{b}\). Read each equation's coefficients (in the order \(x,y,z\)) into a row of \(A\), the unknowns into \(\mathbf{x}\), and the constants into \(\mathbf{b}\).
- Test the determinant. If \(\det A\neq0\) the system has a unique solution; if \(\det A=0\) it is either no solution or infinitely many, so go on to check consistency.
- Solve for the triple. Use elimination (add or subtract equations to remove one variable, then back-substitute) or technology to find \((x,y,z)\).
- Interpret geometrically. State the case as three planes: meeting at a point, having no common point, or sharing a common line.
Read each equation's coefficients (in the order \(x,y,z\)) into a row of \(A\); the constants form \(\mathbf{b}\):
| \(3x-y+2z=5\) | \(\Rightarrow\) | \((3,\ -1,\ 2)\) |
| \(x+2y-z=3\) | \(\Rightarrow\) | \((1,\ 2,\ -1)\) |
| \(2x+y+z=6\) | \(\Rightarrow\) | \((2,\ 1,\ 1)\) |
\(\begin{pmatrix}3 & -1 & 2\\ 1 & 2 & -1\\ 2 & 1 & 1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}5\\ 3\\ 6\end{pmatrix}\).
Add equations in pairs to eliminate \(z\) (the \(+z\) and \(-z\) terms cancel):
| \((x+y+z)+(x+2y-z)\) | \(=\) | \(6+1\) |
| \(2x+3y\) | \(=\) | \(7\) |
| \((x+2y-z)+(3x-y+z)\) | \(=\) | \(1+8\) |
| \(4x+y\) | \(=\) | \(9\) |
Solve the two-variable pair, then back-substitute for \(z\):
| \(y\) | \(=\) | \(9-4x\) |
| \(2x+3(9-4x)\) | \(=\) | \(7\) |
| \(-10x\) | \(=\) | \(-20 \;\Rightarrow\; x=2\) |
| \(y\) | \(=\) | \(9-4(2)=1\) |
| \(z\) | \(=\) | \(6-x-y=6-2-1=3\) |
\((x,y,z)=(2,1,3)\).
Add the first two equations and compare the result with the third:
| \((x+y+z)+(x+2y+3z)\) | \(=\) | \(3+4\) |
| \(2x+3y+4z\) | \(=\) | \(7\) |
| \(\text{but equation 3 says }2x+3y+4z\) | \(=\) | \(10\) |
| \(7\) | \(\neq\) | \(10 \;\Rightarrow\; \text{contradiction}\) |
The equations are inconsistent, so there is no solution — the three planes have no common point.
Add the first two equations; check against the third:
| \((x+y+z)+(x+2y+3z)\) | \(=\) | \(4+9\) |
| \(2x+3y+4z\) | \(=\) | \(13\) |
| \(\text{equation 3 also says }2x+3y+4z\) | \(=\) | \(13\) |
| \(\Rightarrow\) | \(\text{no new information}\) |
Two independent equations remain for three unknowns, so one variable is free:
| \(\text{rank}\) | \(=\) | \(2 < 3 \text{ unknowns}\) |
| \(\Rightarrow\) | \(\text{a free variable}\) |
There are infinitely many solutions; the three planes share a common line.
Common pitfalls
Frequently asked questions
How do you write a system of equations as \(A\mathbf{x}=\mathbf{b}\)?
Put each equation's coefficients (in the order \(x,y,z\)) into a row of the coefficient matrix \(A\), the unknowns into the column \(\mathbf{x}=(x,y,z)^{\!\top}\), and the constants into the column \(\mathbf{b}\). A variable that is missing from an equation contributes a \(0\).
What are the three cases for the solutions of a system?
A system in three variables has either a unique solution, no solution, or infinitely many solutions. The determinant of \(A\) is non-zero for a unique solution; if it is zero, the system has no solution or infinitely many.
How does the determinant tell you the number of solutions?
If \(\det A\neq0\) the coefficient matrix is non-singular and the system has exactly one solution. If \(\det A=0\) the matrix is singular, so there is no unique solution — check consistency to decide between no solution and infinitely many.
What is the geometric interpretation of a solution in three variables?
Each equation is a plane. A unique solution is three planes meeting at a single point, no solution is three planes with no common point, and infinitely many solutions is three planes sharing a common line.
How do you tell no solution apart from infinitely many?
Both have \(\det A=0\). Combine the equations: if you reach a contradiction such as \(0=1\) there is no solution; if an equation just repeats information already given (dependent and consistent), there are infinitely many solutions.
How do you solve a \(3\times3\) system by elimination?
Add or subtract equations to cancel one variable, leaving two equations in two unknowns; solve that pair, then back-substitute to find the third unknown and state the ordered triple \((x,y,z)\).