QCE Specialist Mathematics - Unit 3 - Further matrices

Matrix Algebra and Systems | QCE Specialist Mathematics

Learn Specialist Mathematics matrix algebra, determinants, inverse matrices, Gaussian elimination and systems with unique or non-unique solutions.

Updated 2026-05-18 - 5 min read

Matrices give you a compact way to store a system of linear equations. Once the equations are in matrix form, row operations and inverse matrices can do the solving more systematically than ordinary substitution.

Matrix form of a system

The system

$ \begin{aligned} x-2y+3z&=11\\ 2x-y+3z&=10\\ 5x-3y-4z&=15 \end{aligned} $

can be written as:

$ \begin{pmatrix} 1&-2&3\\ 2&-1&3\\ 5&-3&-4 \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} 11\\10\\15 \end{pmatrix}. $

In short:

$ A\mathbf x=\mathbf b. $

If $A$ is invertible, then:

$ \mathbf x=A^{-1}\mathbf b. $

This is a very efficient technology-active method, but it depends on $A^{-1}$ existing.

Specialist also asks you to solve matrix equations beyond $2\times2$, usually with technology. The order of multiplication matters. For example:

$ AX=B \quad\Rightarrow\quad X=A^{-1}B $

but

$ XA=B \quad\Rightarrow\quad X=BA^{-1}. $

For an equation like $AX+BX=C$, factor $X$ on the correct side:

$ (A+B)X=C $

so, if $A+B$ is invertible,

$ X=(A+B)^{-1}C. $

Gaussian elimination

Gaussian elimination uses row operations on an augmented matrix:

$ \left[ \begin{array}{ccc|c} 1&-2&3&11\\ 2&-1&3&10\\ 5&-3&-4&15 \end{array} \right]. $

The goal is to create zeros below the main diagonal, then back-substitute. Legal row operations include:

• swapping two rows
• multiplying a row by a non-zero scalar
• adding a multiple of one row to another row

You are not changing the solution set; you are replacing the system with an equivalent one that is easier to read.

Row echelon form has a staircase shape: each leading entry is to the right of the leading entry in the row above, and zero rows sit at the bottom. Reduced row echelon form goes further by making each leading entry $1$ and clearing the entries above it. Either form can solve a system, but reduced form is usually easier to read.

Types of solutions

A row echelon form tells you the nature of the solution.

If every variable column has a leading entry and the final column is not contradictory, there is a unique solution.

If a row becomes

$ \begin{pmatrix} 0&0&0&|&5 \end{pmatrix}, $

then it represents $0=5$, so there is no solution.

If at least one variable is free and there is no contradiction, there are infinitely many solutions. You introduce a parameter for the free variable and express the others in terms of it.

Geometric interpretation

For three variables, each linear equation represents a plane. A unique solution means the planes meet at one point. No solution means the planes do not share a common point. Infinitely many solutions can mean the planes share a line or are effectively the same plane after simplification.

Technology can calculate determinants, inverses and row-reduced forms quickly, but the interpretation is still mathematical. If a calculator returns a row like:

$ \begin{pmatrix} 0&0&0&|&0 \end{pmatrix}, $

that row is not a contradiction; it says one equation gave no new information. If there is also a free variable, expect infinitely many solutions.

Gaussian elimination by hand

Gaussian elimination is a controlled way of creating zeros below the main diagonal. Start with the first pivot, clear the entries below it, then move to the next column and repeat. If the pivot is inconvenient, you may swap rows or multiply a row by a non-zero constant.

The legal row operations are:

• $R_i\leftrightarrow R_j$
• $R_i\to kR_i$, where $k\ne0$
• $R_i\to R_i+kR_j$

These operations keep the system equivalent. What you should avoid is changing a row using a version of itself that has already been changed in the same step. It is fine to use one untouched row to update two other rows, but do not let the row operations blur together.

Row-echelon form is not unique. Different valid row operations can produce different-looking echelon matrices, but the solution set should be the same.

Here is a compact hand walkthrough. Solve:

$ \begin{aligned} x+y+z&=6\\ 2x+y-z&=3\\ -x+2y+3z&=14. \end{aligned} $

Start with:

$ \left[ \begin{array}{ccc|c} 1&1&1&6\\ 2&1&-1&3\\ -1&2&3&14 \end{array} \right]. $

Clear below the first pivot:

$ R_2\to R_2-2R_1,\quad R_3\to R_3+R_1. $

This gives:

$ \left[ \begin{array}{ccc|c} 1&1&1&6\\ 0&-1&-3&-9\\ 0&3&4&20 \end{array} \right]. $

Scale the second row by $-1$ and clear below the second pivot:

$ R_2\to -R_2,\quad R_3\to R_3-3R_2. $

Then:

$ \left[ \begin{array}{ccc|c} 1&1&1&6\\ 0&1&3&9\\ 0&0&-5&-7 \end{array} \right]. $

Back-substitution gives:

$ z=\frac75,\quad y=9-3z=\frac{24}{5},\quad x=6-y-z=-\frac15. $

This is the level of detail you should be able to reproduce when technology is not being used.

Leading columns

In row-echelon form, the first non-zero entry in a row is called a leading entry. The columns containing leading entries tell you the solution type:

• every variable column leading and no contradiction: unique solution
• right-hand column leading in a row like $0=5$: no solution
• at least one non-leading variable column and no contradiction: infinitely many solutions

For infinitely many solutions, choose a parameter for the variable in a non-leading column. Back-substitute to express the other variables in terms of that parameter.

For a no-solution example, row reduction might give:

$ \left[ \begin{array}{ccc|c} 1&0&2&4\\ 0&1&-1&3\\ 0&0&0&5 \end{array} \right]. $

The last row says $0=5$, so the planes have no common point.

For an infinite-solution example:

$ \left[ \begin{array}{ccc|c} 1&0&2&7\\ 0&1&-1&3\\ 0&0&0&0 \end{array} \right] $

gives:

$ x=7-2t,\quad y=3+t,\quad z=t. $

In vector form:

$ \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}7\\3\\0\end{pmatrix} +t\begin{pmatrix}-2\\1\\1\end{pmatrix}. $

Geometrically, that is a line of intersection.

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax College Algebra 2e: Solving Systems with Gaussian Elimination
• OpenStax College Algebra 2e: Matrices and Matrix Operations