This guide covers the matrix side of Unit 3. You solve matrix equations with an inverse, write a system of linear equations as a matrix equation, and use Gaussian elimination on an augmented matrix. Then you decide how many solutions a system has and describe what its three planes look like. Most of these questions are simple familiar, so they are marks you should expect to get. Work through it in order, and tick off each skill once you can do it without help.
Skills in this topic
Only tick a skill when you could do it in an exam.
Find the determinant and inverse of a $2\times2$ matrix by hand, and use the determinant to decide whether a matrix is singular. Part 1
Solve matrix equations such as $AX=B$, $XA=B$ and $XA-XC=B$, multiplying by the inverse on the correct side. Part 2
Write a system of linear equations as $AX=B$ and solve it with $X=A^{-1}B$.Part 3
Use row operations on an augmented matrix to reach row echelon form, then back-substitute. Part 3
Decide whether a system has a unique solution, no solution or infinitely many solutions, and describe the planes. Part 4
Find the value of a constant that gives infinitely many solutions, and write the solutions as a vector equation of a line. Part 4
Part 1
The determinant and the inverse
For a square matrix $A$, the determinant is a single number, written $\det(A)$ or $|A|$. On a Technology-active paper you find it with your calculator. In this course it answers one question: does $A$ have an inverse?
Non-singular
$$\det(A)\ne0$$
The inverse $A^{-1}$ exists, and $AA^{-1}=A^{-1}A=I$.
Singular
$$\det(A)=0$$
There is no inverse. If $A$ is the coefficient matrix of a system of equations, the system does not have a unique solution.
The $2\times2$ case by hand
Technology-free questions only ask for the inverse of a $2\times2$ matrix. Anything bigger is done with technology.
Check by multiplying. $MM^{-1}$ should come out as $I$. It only takes a moment and it catches most sign slips.
Part 2
Matrix equations
Matrix multiplication is not commutative. In general $AB\ne BA$. So when you multiply both sides of a matrix equation by an inverse, you have to multiply on the same side of both sides. Look at which side of $X$ the matrix sits on, and put its inverse on that side.
$A^{-1}B$ and $BA^{-1}$ are usually different matrices. 2022 Paper 2, Question 9 puts both of them in the options, so getting the side wrong costs you the mark.
When $X$ appears more than once, factorise first and keep $X$ on the side it is already on.
Check: $\begin{pmatrix}2&-4\\4&-7\end{pmatrix}\begin{pmatrix}2&-1\\1&-1\end{pmatrix}=\begin{pmatrix}0&2\\1&3\end{pmatrix}=B$. Writing $(A-C)X=B$ is the slip that loses most of the marks here.
Answer
$X=\begin{pmatrix}2&-4\\4&-7\end{pmatrix}$
2021 Paper 2, Question 16
Short response · Technology-active · Complex familiar · [6 marks]
Any three entries will do, as long as they give three different equations. Row 1 is what the marking guide uses. The question says to use matrix algebra, so solve the three equations as a matrix equation and show it.
Any system of linear equations can be written as $AX=B$. $A$ is the matrix of coefficients, $X$ is the column of variables and $B$ is the column of constants. Here is the system from 2024 Paper 2, Question 12.
Tidy each equation before you read off the coefficients. Put the variables on the left in the same order, the constant on the right, and write a zero for any variable that is missing. An equation given as $3x+3z=y$ has to become $3x-y+3z=0$ first.
Choosing the route
The question saysWhat to do
“Express as $AX=B$”, “use matrix algebra”, or a plain solve on a Technology-active paperInverse route. Write $A$, $X$ and $B$ separately, then find $X=A^{-1}B$ with technology.
“Use a Gaussian technique”, an augmented matrix in the stem, or a stated row operationGaussian route. Write the augmented matrix and reduce it to row echelon form.
How many solutions, or for what value of a constantGaussian route. If $\det(A)=0$ there is no unique solution, but the determinant cannot tell you whether there are none or infinitely many.
Gaussian elimination on an augmented matrix
The augmented matrix is the coefficient matrix with the constants attached as an extra column. Rule a vertical line before the constants. This is the matrix from 2025 Paper 1, Question 11.
You are allowed three row operations, and only these three:
swap two rows
multiply a row by a non-zero number
add a multiple of one row to another row.
The aim is row echelon form, with zeros below the leading diagonal. The bottom row then gives you one variable, and you back-substitute upwards. QCAA writes row operations as $R_2'=R_2+2R_1$, which means the new row 2 is the old row 2 plus twice row 1. Apply it to every entry in the row, including the constant after the line.
The idea
Clear one column at a time
Game plan: systems of equations
1Tidy the system: the same variable order in every equation, constants on the right, and zeros for missing variables.
2Choose the inverse route or the Gaussian route from the wording of the question.
3Inverse route: write $A$, $X$ and $B$ as three separate matrices, then calculate $X=A^{-1}B$ with technology.
4Gaussian route: write the augmented matrix and reduce it using the row operations you are given or allowed.
5Read the bottom row, then finish for that case. Part 4 covers the cases.
Watch it on an exam question
One row operation, then back-substitution
Practise
Exam questions: solving systems
2024 Paper 2, Question 12
Short response · Technology-active · Simple familiar · [4 marks]
a)Express the system of equations as a matrix equation of the form $AX=B$, where $A$ is a $3\times3$ matrix and both $X$ and $B$ are $3\times1$ column vectors.[1 mark]
b)Use matrix algebra to express $X$ in terms of $A$ and $B$.[1 mark]
c)Use your result from Question 12b) to determine the solution of the system of equations.[1 mark]
d)Verify your result from Question 12c) using one of the given linear equations.[1 mark]
Key: $R_2'=R_2+2R_1$ indicates that the new row 2 values are equal to the sum of the existing row 2 and twice row 1 values.
b)Given the row 1 values represent the equation $x-y-z=-6$, use your result from Question 11a) to determine the solution of the system of linear equations.[3 marks]
The system of linear equations is geometrically represented by three planes.
c)Use your result from Question 11b) to describe a geometrical interpretation of your solution of the system of linear equations.[1 mark]
Given that row 1 values of the matrix represent $x+4y+2z=-10$, the unique solution for $y$ is
Row 2 is the equation $0x+2y+0z=5$. It only contains $y$, so you don't need rows 1 or 3.
$$2y=5\ \Rightarrow\ y=\frac{5}{2}$$
(A) comes from dividing the wrong way. (B) is the value of $z$, from row 3.
Answer
(D) $\dfrac{5}{2}$
Part 4
The three cases and their geometry
Each linear equation in $x$, $y$ and $z$ is a plane. Solving the system means finding the points that lie on all three planes at once. Once the augmented matrix is in row echelon form, the bottom row tells you which case you are in.
Bottom rowSolutionsThe planes
A leading entry in every row, like $0\ \ 0\ \ 4\mid 4$Unique. Here $\det(A)\ne0$.Meet at a single point
$0\ \ 0\ \ 0\mid k$ with $k\ne0$None. The row says $0=k$, which is impossible.No common point: two or three parallel planes, or a triangular prism
$0\ \ 0\ \ 0\mid 0$Infinitely many. One variable is free.Share a common line (or are all the same plane)
Unique solution. The three planes meet at one point.No solution. Two parallel planes, cut by the third.No solution. Three parallel planes.No solution. Each pair meets in a line, but the three lines are parallel. This is a triangular prism.Infinitely many solutions. The three planes share a common line.
To tell parallel planes from a prism, compare the normal vectors, which are the coefficients of $x$, $y$ and $z$. If one normal is a multiple of another, those two planes are parallel. If none are, and there is no solution, the planes form a prism.
Infinitely many solutions as a line
A row of zeros leaves one variable free. Set it equal to a parameter, such as $z=\mu$, then back-substitute to write $x$ and $y$ in terms of $\mu$. The solutions form a line, so write them as $\boldsymbol{r}=\boldsymbol{a}+\mu\boldsymbol{d}$, just like the vector equations of lines in Topic 3.2.
When the system has a constant in it, such as $\lambda$ or $k$, reduce until the constant is alone in the bottom row. The value that makes that row all zeros gives infinitely many solutions. Every other value gives no solution.
Game plan: the three cases
1Reduce the augmented matrix to row echelon form.
2Read the bottom row: a leading entry, $0\ \ 0\ \ 0\mid k$ with $k\ne0$, or $0\ \ 0\ \ 0\mid 0$.
3If there is a constant in the system, choose its value so the bottom row is all zeros.
4For infinitely many solutions, let the free variable be a parameter, back-substitute, and write a vector equation of a line.
5For the geometry, compare the normal vectors to decide between parallel planes and a prism.
Try it yourself
Three planes
Pick a system from the exam questions. For the two with a constant, change its value and watch the bottom row of the echelon form. The solution set is drawn in orange whenever there is one. Turn the view to see how the planes sit against each other.
Row echelon form
The bottom row says
Solution
Practise
Exam questions: the three cases
2023 Paper 1, Question 9
Multiple choice · Technology-free · [1 mark]
The geometric interpretation of a certain system of three equations with no solution is shown.
Given two of the equations are $x+y-z=0.5$ and $x-y-z=0.5$, the third equation could be
The diagram shows two parallel planes cut by a third plane. The given planes have normals $(1,1,-1)$ and $(1,-1,-1)$. These are not parallel, so the given planes are two that cross. The third plane has to be parallel to one of them without being the same plane.
(A) is $x-y-z=0.5$ doubled, and (B) is $x+y-z=0.5$ doubled. Each is the same plane as one already given. (C) has normal $(1,-1,1)$, which is not parallel to either.
$$2x+2y-2z=3\ \Rightarrow\ x+y-z=1.5$$
(D) has the same normal as the first plane and a different constant, so it is parallel to it and distinct.
Answer
(D) $2x+2y-2z=3$
2021 Paper 1, Question 5
Multiple choice · Technology-free · [1 mark]
The augmented matrix shown is produced when a Gaussian elimination technique is used to solve a certain system of equations with three variables.
A unique solution means three planes meeting at a single point, which is (D). (A) shows three planes sharing a common line, (B) shows a triangular prism and (C) shows two parallel planes cut by a third.
Answer
(D)
2022 Paper 1, Question 16
Short response · Technology-free · Complex familiar · [7 marks]
Consider this system of equations that corresponds to three planes.
a)Use a Gaussian technique to determine the value of $\lambda$ for which this system of equations has infinitely many solutions.[4 marks]
b)Use the result from Question 16a) to determine the infinitely many solutions. Express your answer in the form of a vector equation of a line.[3 marks]
Part a)
Rearrange each equation with the variables on the left, in the order $x$,$y$,$z$.
1 markDetermines the infinite solutions expressed in the form of a vector equation of a line.
Any multiple of the direction vector is fine, such as $(1,3,8)$. Use a new letter for the parameter, because $\lambda$ already has a job in this question. Try $\lambda=-2$ in the interactive above, then any other value, which turns the line into a prism.
Answer
a) $\lambda=-2$b) $\boldsymbol{r}=\begin{pmatrix}\frac{9}{4}\\-\frac{1}{4}\\0\end{pmatrix}+\mu\begin{pmatrix}\frac{1}{8}\\\frac{3}{8}\\1\end{pmatrix}$
1 markDetermines the reduced augmented matrix for $k=3$.
The bottom row says $0=-5$, which is impossible, so the system has no solution.
1 markStates there is no solution and justifies it from the bottom row.
Part c)
The third plane, $2x+4y+6z=3$, has normal $(2,4,6)=2(1,2,3)$, so it is parallel to the first plane. It is a different plane, because doubling the first equation gives $2x+4y+6z=8$.
1 markIdentifies that the first and third planes are parallel and distinct.
The second plane is not parallel to either, so it cuts each of them in a line. The two lines are parallel, and there is no point on all three planes.
1 markDescribes how the second plane intersects the parallel planes.
The sample paper has no marking guide, so this solution and its marks are my own. This is the third system in the interactive above.
Answer
a) $k=8$b) no solutionc) two parallel planes, both cut by the second plane
1Matrix equation: factorise with $X$ kept on its side, then multiply both sides by the inverse on that same side.
2Check the determinant is not zero before you use an inverse.
3System of equations: tidy the equations, then choose the inverse route or the Gaussian route from the wording.
4Apply each row operation to the whole row, including the constant after the line.
5Read the bottom row to name the case, then describe the planes.
6For infinitely many solutions, let the free variable be a parameter and write a vector equation of a line.
Mistakes I see a lot
writing $A^{-1}B$ when the equation was $XA=B$
factorising $XA-XC$ as $(A-C)X$
forgetting to apply a row operation to the constant after the line
leaving out the zero when a variable is missing from an equation
saying there is no solution because $\det(A)=0$, without checking for infinitely many
calling the parameter $\lambda$ when $\lambda$ is already in the question
Practice
Practice questions
Ten questions of my own, worth 53 marks in total. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Check whether each one is Technology-free or Technology-active before you start.
Question 1 of 10
Matrix equation $AX-BX=C$; factorising and a $2\times2$ inverse · Technology-free · Simple familiar · [5 marks]
Multiplying in the wrong order gives $C(A-B)^{-1}=\begin{bmatrix}3&-1\\1&2\end{bmatrix}$, which loses the last two marks. Factorising as $X(A-B)$ is also wrong, because $X$ is on the right of both terms.
Answer
$X=\begin{bmatrix}2&-1\\1&3\end{bmatrix}$
Question 2 of 10
System as $AX=B$; determinant; geometric meaning · Technology-active · Simple familiar · [5 marks]
$A^{-1}$ must be post-multiplied, since matrix multiplication is not commutative. Also $A^{-1}B$ is undefined: $(3\times3)(2\times3)$ is not conformable.
1 markExplains the order of multiplication.
Part b)
Using technology: $\det(A)=5\neq0$, so $A^{-1}$ exists.
1 markUses the determinant to show $A$ is non-singular.
Part c)
$X=\begin{bmatrix}5&2&2\\-1&3&1\end{bmatrix}A^{-1}$, with $A$ and $B$ entered with technology
1 markSets up the correct product.
$X=\begin{bmatrix}1&2&0\\-1&0&3\end{bmatrix}$
1 markDetermines $X$.
Part d)
$X$ is a $2\times3$ matrix.
1 markStates the order.
Answer
a) $A^{-1}B$ is undefinedb) $\det(A)=5\ne0$c) $X=\begin{bmatrix}1&2&0\\-1&0&3\end{bmatrix}$d) $2\times3$
Question 6 of 10
Singular matrix with a constant; exact values · Technology-active · Simple familiar · [5 marks]
a)Show that $\pi_1$ and $\pi_2$ are parallel but not coincident.[2 marks]
b)Use Gaussian elimination to show that the system has no solution, and describe the geometric interpretation.[2 marks]
c)The equation of $\pi_2$ is changed to $2x+4y-2z=6$. Describe how this changes the solution of the system.[1 mark]
Part a)
$\boldsymbol{n}_1=\begin{bmatrix}1\\2\\-1\end{bmatrix}$, $\boldsymbol{n}_2=\begin{bmatrix}2\\4\\-2\end{bmatrix}=2\boldsymbol{n}_1$, so the planes are parallel.
1 markShows the normal vectors are parallel.
$2\times\pi_1$: $2x+4y-2z=6\neq5$, so the planes are not coincident.
Row 2 states $0=-1$, which is impossible, so there is no solution.
1 markUses a row operation to show the system is inconsistent.
$\pi_1$ and $\pi_2$ are parallel, so they have no common point. $\pi_3$ cuts each of them in a line, and these two lines are parallel, so no point lies on all three planes.
1 markDescribes the geometric interpretation.
Part c)
$\pi_1$ and $\pi_2$ are now the same plane. This plane meets $\pi_3$ (not parallel, as $\boldsymbol{n}_3$ is not a multiple of $\boldsymbol{n}_1$) in a line, so there are infinitely many solutions.
1 markDescribes the change in the solution.
Answer
a) $\boldsymbol{n}_2=2\boldsymbol{n}_1$ and $6\ne5$b) the bottom row says $0=-1$c) infinitely many solutions, on a line
Question 8 of 10
Inverse from a matrix identity · Technology-free · Simple familiar · [5 marks]
Consider the matrix
$A=\begin{bmatrix}2&1\\1&3\end{bmatrix}$
and let $I$ be the $2\times2$ identity matrix.
a)Show that $A^2=5A-5I$.[2 marks]
b)Use your result from Question 8a) to determine $A^{-1}$.[2 marks]
c)Verify your answer to Question 8b) using the determinant of $A$.[1 mark]
Determine the number of solutions of the system for all real values of $\lambda$. Where the system has infinitely many solutions, express the solution as a vector equation of a line.
1 markDetermines the vector equation of the line for infinitely many solutions.
Unique solution: no value of $\lambda$
No solution: $\lambda\neq8$
Infinitely many solutions: $\lambda=8$
1 markShows logical organisation, having fully attempted the question.
Saying there is a unique solution when $\lambda\ne8$ loses the third and fourth marks. The geometry is not needed here, but for $\lambda=8$ the three planes share a common line.
Answer
never a unique solution$\lambda\ne8$: no solution$\lambda=8$: $\boldsymbol{r}=\begin{pmatrix}-1\\3\\0\end{pmatrix}+t\begin{pmatrix}4\\-3\\1\end{pmatrix}$
Question 10 of 10
Cubic through four points; a $4\times4$ system; evaluating a claim · Technology-active · Complex familiar · [6 marks]
The graph of the cubic function $y=ax^3+bx^2+cx+d$ passes through the points $(-1,4)$, $(1,2)$, $(2,10)$ and $(3,40)$.
A student claims that the graph has a local minimum between $x=0$ and $x=1$.
Use matrix methods to determine the function. Evaluate the reasonableness of the student's claim. Justify your decision using mathematical reasoning.
1 markWrites the system as a matrix equation and states the method of solution.
$a=2$, $b=-1$, $c=-3$, $d=4$
$y=2x^3-x^2-3x+4$
1 markDetermines the function.
$\dfrac{dy}{dx}=6x^2-2x-3=0$
Using technology: $x\approx-0.560$ or $x\approx0.893$
1 markDetermines the $x$-values of the stationary points.
$\dfrac{d^2y}{dx^2}=12x-2$; at $x\approx0.893$, $\dfrac{d^2y}{dx^2}\approx8.72>0$, so this is a local minimum, and $0<0.893<1$.
1 markUses mathematical reasoning to justify the decision.
The student's claim is reasonable.
1 markStates an appropriate evaluation of the reasonableness of the claim.
A sign diagram of $\frac{dy}{dx}$, or reading the minimum off a graph of the function on your calculator, also justifies the decision. The evaluation mark needs that justification.
Answer
$y=2x^3-x^2-3x+4$local minimum at $x\approx0.893$, so the claim is reasonable
Keep practising
Ready for something harder?
The complex familiar questions in this topic, 2021 Paper 2, Question 16 and 2022 Paper 1, Question 16, are built from the same simple skills. Once these feel comfortable, try the complex unfamiliar questions in the collection.
