Every matrix skill and trap for IB Mathematics Applications & Interpretation HL — multiplication, determinants, inverses, solving AX=B on the GDC, and the AI applications of transition matrices & network walks. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Matrices are the AHL workhorse of IB Mathematics Applications & Interpretation HL. AI treats a matrix less as abstract algebra and more as a modelling container: one object that packs an entire system of equations, a network, or the month-to-month behaviour of a population into a single symbol you can add, multiply and invert. Because a calculator is always allowed, the marks live in setting the problem up as a matrix and reading the GDC's answer correctly — not in grinding the arithmetic by hand.
This cheatsheet condenses the whole of AHL 1.14 — order and notation, the algebra of matrices, determinants and inverses (2×2 by hand, 3×3 on the GDC), solving $AX=B$, and the AI applications of powers, transition matrices and networks — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Order, notation & special matrices AHL 1.14
A matrix is a rectangular array of numbers. Its order is stated rows-by-columns, and every operation below depends on that order matching up.
Order:an $m\times n$ matrix has $m$ rows and $n$ columns; the entry in row $i$, column $j$ is $a_{ij}$
Square matrix:$n\times n$ — only square matrices have a determinant, an inverse and powers
Identity:$I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ leaves any matrix unchanged: $AI=IA=A$
Zero matrix:$O$ has every entry $0$, so $A+O=A$
NoteTwo matrices are equal only when they have the same order and every corresponding entry matches — that is how "find the value of $x$" entry-matching questions are solved.
TrapOrder is always (rows × columns), in that order. A $2\times3$ matrix is a different object from a $3\times2$ one, and swapping them makes a product undefined.
§2 — Adding, subtracting & scaling AHL 1.14
Addition, subtraction and scalar multiplication all act entry by entry, so they are the easy operations — the only catch is that the orders must agree.
Add / subtract:$(A\pm B)_{ij}=a_{ij}\pm b_{ij}$ — both matrices must have the same order
Scalar multiple:$kA$ multiplies every entry of $A$ by $k$
TrickLine the two matrices up and work one position at a time, top-left to bottom-right, so entries never get mixed up.
TrapYou can only add or subtract matrices of the same order. A $2\times2$ and a $2\times3$ cannot be combined — the operation is simply undefined, not zero.
§3 — Matrix multiplication AHL 1.14
Multiplication is the operation that makes matrices powerful — and the one that trips students up, because it pairs whole rows with whole columns.
Conformable:$A_{m\times n}\,B_{n\times p}$ needs the inner dimensions to match; the product is $m\times p$
Entry $(i,j)$:$(AB)_{ij}=\sum_{k}a_{ik}b_{kj}$ — row $i$ of $A$ paired term-by-term with column $j$ of $B$
Powers:for a square $A$: $A^2=AA,\ A^3=AAA,\ \ldots,\ A^0=I$
TrickRun one finger along the row of $A$ and another down the column of $B$ together, multiplying the pairs as you go — that keeps every term matched.
TrapOrder matters: in general $AB\neq BA$, so you may never swap the factors. The identity is the one exception, $AI=IA=A$; multiplication is associative and distributive but not commutative.
§4 — Determinants: 2×2 by hand, 3×3 by GDC AHL 1.14
Every square matrix carries a single number, the determinant, that measures how it scales area (2D) or volume (3D) — and whether it can be inverted at all.
3×3:enter the matrix and read $\det$ straight from the GDC — no expansion by hand needed in AI
Meaning:$|\det M|$ is the area/volume scale factor; $\det M=0\Rightarrow$ the matrix is singular (no inverse)
TrapMind the signs in $ad-bc$: a negative times a negative inside $bc$ can flip your answer. Write $ad$ and $bc$ out separately before subtracting.
Note$\det M=0$ is the single test for "no inverse / no unique solution". Compute it first — if it is zero, every method that relies on $M^{-1}$ stops here.
§5 — The inverse matrix AHL 1.14
The inverse undoes a matrix, $AA^{-1}=A^{-1}A=I$, and it exists exactly when $\det A\neq0$. Learn the 2×2 formula; let the GDC handle 3×3.
2×2 inverse:$M^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ — swap the leading diagonal, negate the other two, divide by $\det M$
3×3 inverse:enter $M$ on the GDC and compute $M^{-1}$ directly
TrickAlways find $\det M$ first. If it is $0$ the matrix is singular — stop, because no inverse exists and $\tfrac{1}{0}$ is the tell.
TrapIn the 2×2 formula you swap $a$ and $d$ but only negate $b$ and $c$ — do not put a minus on $a$ or $d$, and do not forget to divide the whole matrix by $ad-bc$.
§6 — Solving linear systems $AX=B$ AHL 1.14
This is the headline use of matrices: a whole system of linear equations collapses into one statement, which the inverse then solves in a single step.
Encode:coefficients $\to A$, unknowns $\to X$, constants $\to B$, so the system becomes $AX=B$
Solve:if $\det A\neq0$ then $X=A^{-1}B$ — one GDC step for a 2×2 or 3×3 system
TrickStore $A$ and $B$ as matrices on the GDC and evaluate $A^{-1}B$ in one keystroke — quicker and less error-prone than the simultaneous-equation app, and it works for three unknowns just as easily as two.
TrapIf $\det A=0$ there is no unique solution (either none or infinitely many), and the GDC will refuse to invert. That error is information about the system, not a typing mistake — though it is worth re-checking your entries too.
§7 — Transformations & the area scale factor AHL 1.14
A $2\times2$ matrix maps the plane to itself, and its determinant tells you exactly what it does to area — the geometric face of §4.
Area scale:a region of area $A_0$ becomes area $|\det M|\,A_0$ under the transformation $M$
Two-way stretch:$\begin{pmatrix}a&0\\0&b\end{pmatrix}$ has $\det=ab$, stretching by $a$ horizontally and $b$ vertically
TrapAreas multiply by $|\det M|$, never by the linear factor alone. A negative determinant is fine — its magnitude still gives the area, and the sign just means the shape's orientation has been reversed.
NoteThe same idea runs one dimension up: for a $3\times3$ matrix $|\det M|$ scales volume, and $\det M=0$ squashes the solid flat — which is exactly why a singular matrix has no inverse.
§8 — Powers & transition matrices AHL 1.14
Raising a matrix to a power is how AI steps a system forward in time. A transition matrix $T$ moves a population between states each step; here we use the column convention, so each column sums to $1$.
Transition matrix:$T_{ij}$ is the probability of moving from state $j$ to state $i$; every column of $T$ sums to $1$
Step forward:$\mathbf{s}_n=T^{\,n}\mathbf{s}_0$ — apply $T$ once per time step and read the power off the GDC
Steady state:solve $T\mathbf{p}=\mathbf{p}$; for $T=\begin{pmatrix}1-a&b\\a&1-b\end{pmatrix}$ the long-run share in state 1 is $\dfrac{b}{a+b}$
A two-state transition diagram: the arrows leaving each state show the transition probabilities, and the options out of a state sum to 1 — so every column of T totals 1.
TrickThe steady state does not depend on where you start: whatever the initial split $\mathbf{s}_0$, the long-run distribution settles to the same $\mathbf{p}$. That is why the answer is just $\tfrac{b}{a+b}$.
NoteThe long-run growth or decay of $T^{\,n}$ is governed by its dominant eigenvalue (the largest solution of $\lambda^2-(\operatorname{tr}T)\lambda+\det T=0$), and the steady state is the eigenvector for $\lambda=1$. The full eigenvalue and Markov toolkit is the AHL 1.15 topic.
§9 — Adjacency matrices: counting walks AHL 1.14
Matrices meet graph theory here. Store a network in its adjacency matrix $A$, where $A_{ij}$ counts the edges joining vertex $i$ to vertex $j$ — then powers of $A$ count paths through the network.
Walks:$\left(A^{\,n}\right)_{ij}$ = the number of walks of length $n$ from vertex $i$ to vertex $j$
Whole network:total walks of length 2 = the sum of all entries of $A^2$ $\left(=\sum_i \deg(i)^2\right)$
TrapA walk of length $n$ uses $n$ edges, so raise $A$ to the power $n$ — not $n-1$. Read the single entry in row $i$, column $j$ for one pair of vertices; sum all the entries only when the question asks about the whole network.
NoteThis is an AHL 3.16 graph-theory synthesis: square or cube $A$ on the GDC, then pick out one entry. No walk-tracing by hand is ever needed.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
"Find the determinant" (3×3)
Enter the matrix on the GDC and read $\det$
One wrong sign or entry changes everything
"Find the inverse"
2×2 by formula ($\tfrac{1}{ad-bc}$, swap & negate); 3×3 on the GDC
No inverse when $\det=0$
"Solve the system"
Write $AX=B$, compute $X=A^{-1}B$
$\det A=0\Rightarrow$ no unique solution
"Find the product $AB$"
Row $i$ × column $j$; check inner dimensions match
$AB\neq BA$; wrong sizes = undefined
"State / population after $n$ steps"
Compute $\mathbf{s}_n=T^{\,n}\mathbf{s}_0$ on the GDC
Columns of $T$ sum to 1; do not transpose
"Long-run / steady state"
Solve $T\mathbf{p}=\mathbf{p}$ (or use $\tfrac{b}{a+b}$)
Independent of the start; totals are conserved
"Area after a transformation"
Multiply the area by $|\det M|$
Not the linear factor; $\det$ may be negative
"Walks of length $n$"
Compute $A^{\,n}$ and read entry $(i,j)$
Use power $n$, not $n-1$; one entry, not the row sum
Worked Example — AI HL-Style Transition Matrix
Question (AI HL Paper 2 style — 6 marks)
A streaming service reviews its members each month. Each month $80\%$ of current subscribers renew and $20\%$ cancel; of those not currently subscribed, $30\%$ take out a subscription and $70\%$ stay unsubscribed. The service begins with $2000$ subscribers and $3000$ non-subscribers ($5000$ people in total). Let state 1 be "subscribed" and state 2 be "not subscribed". (a) Write down the transition matrix $T$ (columns summing to 1) and the initial state vector $\mathbf{s}_0$. (b) Find the number of subscribers after 2 months. (c) Find the long-run (steady-state) number of subscribers.
Solution
(a) Each column lists where one state goes. From subscribed: $0.8$ stay, $0.2$ leave; from not-subscribed: $0.3$ join, $0.7$ stay. So $T=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}$, $\mathbf{s}_0=\begin{pmatrix}2000\\3000\end{pmatrix}$. (A1)(A1)(each column of $T$ sums to 1 — a quick self-check.)
(b) After two months $\mathbf{s}_2=T^{2}\mathbf{s}_0$. On the GDC $T^{2}=\begin{pmatrix}0.70&0.45\\0.30&0.55\end{pmatrix}$, so $\mathbf{s}_2=\begin{pmatrix}0.70&0.45\\0.30&0.55\end{pmatrix}\begin{pmatrix}2000\\3000\end{pmatrix}=\begin{pmatrix}2750\\2250\end{pmatrix}$. (M1)(A1) There are 2750 subscribers after 2 months.
(c) The steady state solves $T\mathbf{p}=\mathbf{p}$. With $p$ subscribers out of $5000$: $0.8p+0.3(5000-p)=p\Rightarrow 0.5p=1500\Rightarrow p=3000$. (M1)(A1) In the long run there are 3000 subscribers (and 2000 non-subscribers). (Equivalently $\tfrac{b}{a+b}=\tfrac{0.3}{0.5}=0.6$, and $0.6\times5000=3000$.)
Examiner's note: state the convention first — here columns sum to 1, so $\mathbf{s}_{n+1}=T\mathbf{s}_n$ — and do not transpose $T$; swapping its rows and columns is the classic slip. The total must stay at $5000$ at every step, and the steady state is independent of the starting numbers. Give whole people and keep exact GDC values along the way.
Common Student Questions
Why is $AB$ usually not the same as $BA$?
Because each entry of a product pairs a row of the first matrix with a column of the second. Swap the order and different rows meet different columns, so the numbers change — and the sizes may not even be conformable any more, making one product undefined. Multiplication order is part of the operation, so always keep the factors exactly as the question gives them.
My GDC gives an error when I try to invert a matrix — what went wrong?
Almost always the determinant is zero, so the matrix is singular and has no inverse. If it came from a system of equations, that means there is no unique solution — the equations are either inconsistent or dependent. Check $\det=0$: the "error" is telling you the geometry of the problem, not that you mistyped (though it is still worth re-checking every entry).
Should I use the inverse-matrix method or the GDC's simultaneous-equation solver?
They give the same answer, so either is fine for a unique solution. Storing $A$ and $B$ and computing $A^{-1}B$ is usually quickest and it scales straight from two unknowns to three. The one advantage of the inverse route is diagnostic: if $\det A=0$ it flags "no unique solution" immediately, which is often exactly what a harder question is testing.
Do the rows or the columns of a transition matrix add up to 1?
It depends on the convention. With state vectors written as columns and updated by $\mathbf{s}_{n+1}=T\mathbf{s}_n$ — the convention used on this sheet — the columns sum to 1, because each column lists where everything in one state goes. Some sources use row vectors with $\mathbf{s}_{n+1}=\mathbf{s}_nT$, and then the rows sum to 1. Decide which you are using at the start and never transpose $T$ midway.
For counting walks, do I raise the adjacency matrix to the power $n$ or $n-1$?
To the power $n$. A walk of length $n$ uses $n$ edges, and $\left(A^n\right)_{ij}$ counts precisely those walks from vertex $i$ to vertex $j$. Read the single entry in row $i$, column $j$ for one pair of vertices, or sum every entry of $A^n$ for the whole network. Using $n-1$ is the classic off-by-one error on this synthesis question.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every matrix rule, determinant and inverse method in full, with worked edge cases and the exact GDC keystrokes.
Tutorial booklet — 25+ AI HL-style questions from basic products to full transition-matrix and network applications.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every matrix method.
Predicted-paper questions — the exact matrix and transition-matrix question types most likely in the next session.
