Augmented matrices and Gaussian elimination
Master augmented matrices and Gaussian elimination for Year 12 Specialist Mathematics in Queensland (QCAA). A system of linear equations is packed into an augmented matrix \([A\,|\,\mathbf{b}]\) and solved with elementary row operations that reduce it to a solvable staircase.
You will learn to build the augmented matrix, reach row-echelon form, back-substitute for \(x\), \(y\) and \(z\), and read off whether a system has a unique, no, or infinitely many solutions — with the geometry of three planes meeting at a point, a line, or nowhere.
Theory
An augmented matrix \([A\,|\,\mathbf{b}]\) packs a system of linear equations into one array so it can be solved by Gaussian elimination. In Year 12 Specialist Mathematics (QCAA, Queensland) you apply elementary row operations to reach row-echelon form, then back-substitute, and read off whether the system has a unique, no, or infinitely many solutions.
A system of linear equations in several variables can be recorded as an augmented matrix. The coefficients fill the left block \(A\), a vertical bar marks the divide, and the constants form the right column \(\mathbf{b}\), written \([A\,|\,\mathbf{b}]\). Every variable keeps a column, so a missing variable is entered with a coefficient of \(0\).
Three elementary row operations change the matrix without changing its solution set: swap two rows (\(R_i\leftrightarrow R_j\)), multiply a row by a non-zero constant (\(R_i\to kR_i\)), and add a multiple of one row to another (\(R_i\to R_i+kR_j\)). The last of these is the workhorse for creating zeros below a leading entry.
Gaussian elimination uses row operations to reach row-echelon form: a staircase of leading entries (the first non-zero in each row), each strictly to the right of the one above, with zeros beneath every leading entry and any all-zero rows at the bottom. Back-substitution then works up from the last row to recover each variable.
The final form reveals the number of solutions. A full staircase down to a single \(z\) gives a unique solution; a row reading \(0=k\) with \(k\neq0\) means no solution (the system is inconsistent); an all-zero row \(0=0\) leaves a free variable, so there are infinitely many solutions. Geometrically the three equations are three planes: they meet at a point, or share no common point, or meet in a common line.
A \(3\times3\) system and its augmented matrix:
The three elementary row operations:
Row-echelon form is upper-triangular in the coefficient block, ready for back-substitution:
How to solve a system by Gaussian elimination
- Build the augmented matrix. Write each equation in the general form \(ax+by+cz=d\) (a missing variable gets coefficient \(0\)), then load the coefficients into \(A\) and the constants into \(\mathbf{b}\) as \([A\,|\,\mathbf{b}]\).
- Clear the first column. Use \(R_2\to R_2+kR_1\) and \(R_3\to R_3+kR_1\) to make the entries below the leading \(1\) of row \(1\) equal to \(0\); swap rows if needed to get a non-zero pivot.
- Clear the second column below its leading entry with \(R_3\to R_3+kR_2\), reaching row-echelon form (the staircase).
- Back-substitute and classify. Solve the last row for \(z\), then work upward for \(y\) and \(x\); or, if a row reads \(0=k\) or \(0=0\), report no solution or infinitely many.
Load each coefficient into the left block and each constant into the right column; every variable keeps its column:
| \(x-2y+3z\) | \(=\) | \(5\) |
| \(2x+y-z\) | \(=\) | \(3\) |
| \(-x+4y+2z\) | \(=\) | \(1\) |
| \([A\,|\,\mathbf{b}]\) | \(=\) | \(\left(\begin{array}{ccc|c} 1&-2&3&5\\ 2&1&-1&3\\ -1&4&2&1 \end{array}\right)\) |
\([A\,|\,\mathbf{b}]=\left(\begin{array}{ccc|c} 1&-2&3&5\\ 2&1&-1&3\\ -1&4&2&1 \end{array}\right)\).
Clear the first column below the leading \(1\):
| \(R_2\to R_2-2R_1\) | \(:\) | \((0,\,-3,\,-1\,|\,-5)\) |
| \(R_3\to R_3-R_1\) | \(:\) | \((0,\,1,\,-2\,|\,4)\) |
Swap to get a leading \(1\) in row \(2\), then clear the second column:
| \(R_2\leftrightarrow R_3\) | \(:\) | \((0,\,1,\,-2\,|\,4)\text{ on top}\) |
| \(R_3\to R_3+3R_2\) | \(:\) | \((0,\,0,\,-7\,|\,7)\) |
Back-substitute up the staircase:
| \(-7z\) | \(=\) | \(7\) |
| \(z\) | \(=\) | \(-1\) |
| \(y-2z=4\) | \(\Rightarrow\) | \(y=4+2(-1)=2\) |
| \(x+y+z=2\) | \(\Rightarrow\) | \(x=2-2-(-1)=1\) |
\((x,y,z)=(1,2,-1)\).
Clear the first column below row \(1\):
| \(R_2\to R_2-2R_1\) | \(:\) | \((0,\,0,\,0\,|\,1)\) |
| \(R_3\to R_3-R_1\) | \(:\) | \((0,\,-2,\,1\,|\,1)\) |
Read the impossible row:
| \(\text{row 2}\) | \(:\) | \(0x+0y+0z=1\) |
| \(0\) | \(=\) | \(1\quad(\text{impossible})\) |
No solution; rows \(1\) and \(2\) are parallel planes with no common point.
Clear the first column, then the second:
| \(R_2\to R_2-R_1\) | \(:\) | \((0,\,1,\,2\,|\,8)\) |
| \(R_3\to R_3-2R_1\) | \(:\) | \((0,\,1,\,2\,|\,8)\) |
| \(R_3\to R_3-R_2\) | \(:\) | \((0,\,0,\,0\,|\,0)\) |
An all-zero row leaves a free variable; set \(z=\lambda\) and back-substitute:
| \(y+2z=8\) | \(\Rightarrow\) | \(y=8-2\lambda\) |
| \(x+y+z=6\) | \(\Rightarrow\) | \(x=6-(8-2\lambda)-\lambda\) |
| \(x\) | \(=\) | \(-2+\lambda\) |
\((x,y,z)=(-2+\lambda,\;8-2\lambda,\;\lambda)\); the planes meet in a common line.
Common pitfalls
Frequently asked questions
What is an augmented matrix?
It is a system of linear equations written as one array \([A\,|\,\mathbf{b}]\): the coefficients fill the left block \(A\) and the constants form the right column \(\mathbf{b}\), separated by a vertical bar.
What are the three elementary row operations?
Swap two rows \((R_i\leftrightarrow R_j)\), multiply a row by a non-zero constant \((R_i\to kR_i)\), and add a multiple of one row to another \((R_i\to R_i+kR_j)\). None of them changes the solution set.
What is row-echelon form?
A matrix is in row-echelon form when each row's leading entry sits strictly to the right of the one above, every entry below a leading entry is \(0\), and any all-zero rows are at the bottom — a staircase ready for back-substitution.
How do you tell if a system has no solution or infinitely many?
Reduce to row-echelon form and read the bottom row. A row \((0\;0\;0\,|\,k)\) with \(k\neq0\) means \(0=k\), so there is no solution; a row \((0\;0\;0\,|\,0)\) leaves a free variable, so there are infinitely many.
What is back-substitution?
Once the matrix is in row-echelon form, solve the last row for its single variable, then substitute upward row by row to find the remaining variables.
How does Gaussian elimination link to three planes?
Each equation is a plane. A unique solution means the three planes meet at one point; no solution means they share no common point; infinitely many means they meet along a common line.