When do I solve by hand and when do I use the GDC?

For two equations in two unknowns you can eliminate or substitute by hand, and you should be ready to show at least one algebraic step. For three unknowns the course expects you to use technology — the GDC's system solver or the inverse-matrix method. Either way, write down the equations (or the matrix you entered) so the method is visible to the examiner.

How do I know if two matrices can be multiplied?

Write their orders side by side, for example $(2\times3)$ and $(3\times1)$. The product exists only if the two inner numbers match — here both are 3 — and the result takes the outer numbers, giving a $(2\times1)$ matrix. If the inner numbers differ, the product is undefined, and you should say so rather than force an answer.

Why is the order of multiplication so important?

Matrix multiplication is not commutative: $AB$ and $BA$ are usually different, and sometimes only one of them even exists. That is why you must left-multiply $A\mathbf{x}=\mathbf{b}$ by $A^{-1}$ to reach $\mathbf{x}=A^{-1}\mathbf{b}$. Writing $\mathbf{b}A^{-1}$ reverses the order and is both conceptually wrong and, for a column $\mathbf{b}$, not conformable.

What does it mean when the determinant is zero?

A zero determinant means the matrix is singular and has no inverse. For a $2\times2$ system the two lines are then either parallel (no solution) or the same line (infinitely many solutions). You cannot use $\mathbf{x}=A^{-1}\mathbf{b}$; the GDC will report no unique solution.

Do I have to memorise the inverse formula, or can I just use the GDC?

Both are examinable. The $2\times2$ inverse formula $A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ can be asked for directly, so know it. For solving a system quickly and accurately, the GDC's matrix or system function is usually faster and less error-prone — use it, but show the set-up.

Systems of equations and matrix operations —…

Revise Systems of equations and matrix operations for IB Mathematics: Applications and Interpretation (1.5)

Looking for free IB Mathematics: Applications and Interpretation notes on Systems of equations and matrix operations? This Higher Level lesson covers official syllabus point 1.5 on Maths AI HL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Systems of equations and matrix operations. Built for students searching "IB Maths AI 1.5", "Systems of equations and matrix operations IB revision", or a free alternative to paid IB note sites.

Systems of equations and matrix operations

Many real problems fix several unknown quantities at once through several linked conditions: ticket sales that must match both a headcount and a takings figure, a diet that must hit targets for two or three nutrients, currents in a circuit that obey several rules together. Each condition is a linear equation, and together they form a system. This lesson shows you how to solve such systems — by hand for two unknowns, and on the GDC for up to three — and how the language of matrices packages a whole system into a single equation, $A\mathbf{x}=\mathbf{b}$ , that you can solve with an inverse or with technology. Along the way you will master the SL matrix operations: addition, subtraction, scalar multiplication, multiplication with its conformability rule, the identity matrix, and the determinant and inverse of a 2x2 matrix. As in every calculation topic, the worked examples below are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1Solve a system of two linear equations in two unknowns algebraically, and a system of up to three unknowns using the GDC.
2Translate a worded context into a system of linear equations, defining variables clearly.
3Add, subtract and scalar-multiply matrices, and multiply matrices, applying the conformability rule to decide when a product exists.
4State the role of the identity matrix, and find the determinant and inverse of a 2x2 matrix.
5Write a system as a matrix equation $A\mathbf{x}=\mathbf{b}$ and solve it using the inverse $\mathbf{x}=A^{-1}\mathbf{b}$ or the GDC, presenting working so method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Many clues, one tidy grid

A system of equations is several clues about the same unknowns. A matrix is a tidy grid that packages those clues so the whole system becomes a single equation, $A\mathbf{x}=\mathbf{b}$ , that you can undo in one move with an inverse or hand straight to your GDC.

THINK OF IT LIKE…

Imagine a ticket office that sells adult and child tickets. You are told two facts: how many tickets sold in total, and how much money came in. Each fact is a clue that mixes the two unknown numbers together. Written as a grid, the ticket prices become the matrix $A$ , the unknown counts become the column $\mathbf{x}$ , and the totals become $\mathbf{b}$ . Multiplying $A$ by $A^{-1}$ is like having a master key that unlocks $\mathbf{x}$ in a single turn — or you can just let the calculator turn the key for you.

1
Translate the context into equations. Define each variable in words, then write every equation in standard form, for example $ax+by=c$ , with the variables in the same order.
2
Choose your route. Two unknowns can be solved algebraically; three unknowns are solved on the GDC. Either way you can package the coefficients into a matrix $A$ , the unknowns into a column $\mathbf{x}$ , and the constants into a column $\mathbf{b}$ .
3
Solve. For a 2x2 system find $\det A$ and the inverse $A^{-1}$ , then compute $\mathbf{x}=A^{-1}\mathbf{b}$ ; or type the system straight into the GDC.
4
Interpret and check. Read the solution back into the context, and substitute into an original equation to confirm it works.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$k\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}ka&kb\\kc&kd\end{pmatrix}$

$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$

Full topic notes

Formal explanation with the rigour you need for the exam.

Solving systems of linear equations

A system of linear equations is a set of equations that must all hold at the same time. A solution is a set of values for the unknowns that satisfies every equation simultaneously. Geometrically, two equations in two unknowns are two straight lines: a unique solution is the point where they cross, no solution means they are parallel, and infinitely many solutions means they are the same line.

For example, to solve $2x+3y=13$ and $x-y=1$ by elimination, rearrange the second equation to $x=y+1$ and substitute: $2(y+1)+3y=13$ , so $5y+2=13$ , giving $y=2.2$ and then $x=3.2$ . The same pair of values is exactly what the GDC returns from its simultaneous-equation solver — the technology is doing the elimination for you.

Two unknowns: solve by elimination or substitution by hand, or with the GDC's simultaneous-equation solver. Show at least one algebraic step for the method mark.
Up to three unknowns: use the GDC — the system solver, or entering the coefficients and constants as matrices. Write down what you entered.
Standard form first: arrange every equation as $ax+by=c$ (or $ax+by+cz=d$ ) with the variables in the same order before reading off coefficients.
Interpret the outcome: a unique solution, no solution (inconsistent) or infinitely many solutions (dependent).

Setting up a system from a context

Most exam marks are lost not in the solving but in the setting up. The reliable routine is: define each unknown in words, turn each sentence of the problem into one equation, and line the variables up in the same order in every equation. A missing variable is written with a coefficient of zero so the columns still align.

Matrix operations at SL

A matrix is a rectangular array of numbers; its order is 'rows x columns'. Matrices give you a compact language for handling many numbers at once, and four operations are examinable at SL: addition, subtraction, scalar multiplication and matrix multiplication.

Scalar multiple: $k\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}ka&kb\\kc&kd\end{pmatrix}$ . \n Product (2x2): $\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}e&f\\g&h\end{pmatrix}=\begin{pmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{pmatrix}$ .

Two features of multiplication trap students. First, conformability: $(2\times3)(3\times1)$ works and gives a $(2\times1)$ matrix, but $(2\times3)(2\times3)$ is undefined because the inner numbers, 3 and 2, disagree. Second, multiplication is not commutative: even when both $AB$ and $BA$ exist, they are usually different matrices, so you must never swap the order.

Addition and subtraction: only for matrices of the SAME order; work element by element.
Scalar multiplication: multiply EVERY element by the scalar.
Matrix multiplication: row-by-column. The product $AB$ exists only when the columns of $A$ match the rows of $B$ .
Order of a product: an $(m\times n)$ times an $(n\times p)$ gives an $(m\times p)$ matrix — inner numbers must match, outer numbers give the size.

The identity, determinant and inverse of a 2x2 matrix

Among square matrices, the identity matrix $I$ plays the role of the number 1: for the 2x2 case $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ , and $AI=IA=A$ for every 2x2 matrix $A$ . The inverse of $A$ , written $A^{-1}$ , plays the role of a reciprocal: it satisfies $AA^{-1}=A^{-1}A=I$ , so multiplying by it undoes $A$ .

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ : determinant $\det A=ad-bc$ , and inverse $A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ , defined only when $\det A\neq0$ .

To build the inverse: swap the two entries on the leading diagonal, put a minus sign on the other two, and divide the whole matrix by the determinant. If $\det A=0$ the matrix is singular — it has no inverse, and the associated system has no unique solution. The determinant is therefore a quick test: compute $ad-bc$ before attempting an inverse.

The matrix equation $A\mathbf{x}=\mathbf{b}$

Any linear system can be packaged as a single matrix equation. Collect the coefficients into a matrix $A$ , the unknowns into a column $\mathbf{x}$ , and the constants into a column $\mathbf{b}$ . Then the whole system reads $A\mathbf{x}=\mathbf{b}$ . To solve, left-multiply both sides by $A^{-1}$ : since $A^{-1}A=I$ , the left side collapses to $\mathbf{x}$ , giving $\mathbf{x}=A^{-1}\mathbf{b}$ . The order is essential — because matrices do not commute, you must multiply on the left, and $\mathbf{b}A^{-1}$ is both wrong and, for a column $\mathbf{b}$ , not even conformable.

Package it: $A$ = coefficients, $\mathbf{x}$ = unknowns (a column), $\mathbf{b}$ = constants (a column).
Left-multiply by $A^{-1}$ : $A^{-1}A\mathbf{x}=A^{-1}\mathbf{b}$ gives $\mathbf{x}=A^{-1}\mathbf{b}$ .
Order matters: it is $A^{-1}\mathbf{b}$ , never $\mathbf{b}A^{-1}$ .
Or use the GDC: enter $A$ and $\mathbf{b}$ and compute $A^{-1}\mathbf{b}$ , or use the system solver directly. Both are accepted; show the set-up.

EXAM TIP

On this course you may solve $A\mathbf{x}=\mathbf{b}$ entirely on the GDC, but you still write down the matrix equation you set up. That single line — $A$ , $\mathbf{x}$ and $\mathbf{b}$ correctly identified — is where the method marks live, and it is also your evidence if a keying slip spoils the final numbers.

Common mistakes examiners penalise

Adding or subtracting matrices of different orders — these operations only exist for matrices of the SAME order; check the orders before you start.
Forcing a product that is not conformable — $AB$ exists only if the columns of $A$ equal the rows of $B$ . If the inner numbers disagree, write 'not defined' rather than inventing entries.
Swapping the order of a product — $AB\neq BA$ in general. Copy the factors down in the order the question gives them and never reverse them.
Mis-signing the determinant — $\det A=ad-bc$ , so watch the subtraction and the signs of $b$ and $c$ ; a sign slip here corrupts the whole inverse.
Forgetting to divide by the determinant — the inverse is $\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ ; writing only the adjusted matrix without the $\frac{1}{\det A}$ factor is a classic lost accuracy mark.
Solving as $\mathbf{b}A^{-1}$ instead of $A^{-1}\mathbf{b}$ — you must left-multiply; the reversed product is wrong and usually not conformable.
Trying to invert a singular matrix — if $\det A=0$ there is no inverse and no unique solution; say so instead of dividing by zero.
Setting up equations with misaligned variables — line up $x$ , $y$ , $z$ in the same order in every equation and use a coefficient of 0 for any missing variable before reading them into $A$ .

Model answer — marked the way our engine marks it

On Paper 1 the marks are analytic: each is tied to a specific line of working. A method mark (M) rewards a correct approach; an accuracy mark (A) rewards a correct result and depends on the method mark it follows. Follow-through (FT) means an earlier slip need not cost the marks that depend on it, provided the later step is carried out correctly on your own figures. The engine also accepts any equivalent form and any correctly-rounded value, and it accepts a valid GDC route in place of by-hand working — but only if the set-up is written down. Study how each mark below is earned by a specific line.

Where this leads

The matrix machinery you have built here is the entry point to the wider linear-algebra strand of the course. The same idea of collecting coefficients into a matrix underlies solving three-variable systems on the GDC, and $A\mathbf{x}=\mathbf{b}$ reappears whenever a model has several unknowns tied together. In HL, determinants and inverses extend to larger matrices, and the determinant becomes the gateway to eigenvalues and eigenvectors used in modelling change over time. Master the SL essentials — define your variables, keep the order straight, test the determinant before inverting, and solve on the correct side — and every later system, however large, becomes a variation on the equation $A\mathbf{x}=\mathbf{b}$ you already know how to unlock.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A cinema sells adult tickets at $12 and child tickets at $8. One evening it sells 200 tickets in total and takes $2040. Let$ a$ be the number of adult tickets and $c$ the number of child tickets sold. Set up and solve a system to find $a$ and $c $. [5]$

1
Define the variables (given): $a=$ adult tickets, $c=$ child tickets.\n\nSet up the system. 'Total tickets' and 'total takings' give one equation each:\n $a+c=200$ \n $12a+8c=2040$ [M1: two correct equations from the context]\n\nSolve. From the first equation $a=200-c$ . Substitute into the second:\n $12(200-c)+8c=2040$ \n $2400-4c=2040$ \n $-4c=-360$ , so $c=90$ . [M1: a valid elimination/substitution step] [A1: $c=90$ ]\n\nThen $a=200-90=110$ . [A1: $a=110$ ]\n\nInterpret. 110 adult tickets and 90 child tickets were sold. Check: $12(110)+8(90)=1320+720=2040$ . Correct. [A1: both values stated in context, with a check]

✓ FULLY WORKED

EXAMPLE 2

Let $A=\begin{pmatrix}1&2\\3&0\end{pmatrix}$ and $B=\begin{pmatrix}4&-1\\2&5\end{pmatrix}$ . Find (a) $2A-B$ , and (b) the product $AB$ , and (c) state whether $BA$ equals $AB$ . [5]

1
(a) $2A-B$ . Scalar-multiply first, then subtract element by element:\n $2A=\begin{pmatrix}2&4\\6&0\end{pmatrix}$ , so $2A-B=\begin{pmatrix}2-4&4-(-1)\\6-2&0-5\end{pmatrix}=\begin{pmatrix}-2&5\\4&-5\end{pmatrix}$ . [M1: scalar multiple and subtraction of like elements] [A1]\n\n**(b) $AB$ .** Both are $2\times2$ , so the product is $2\times2$ . Multiply row by column:\n $AB=\begin{pmatrix}1(4)+2(2)&1(-1)+2(5)\\3(4)+0(2)&3(-1)+0(5)\end{pmatrix}=\begin{pmatrix}8&9\\12&-3\end{pmatrix}$ . [M1: row-by-column method] [A1]\n\n**(c) Does $BA=AB$ ?** Computing $BA=\begin{pmatrix}4(1)+(-1)(3)&4(2)+(-1)(0)\\2(1)+5(3)&2(2)+5(0)\end{pmatrix}=\begin{pmatrix}1&8\\17&4\end{pmatrix}$ , which is different from $AB$ . So $BA\neq AB$ : matrix multiplication is not commutative. [A1: correct comparison and conclusion]

✓ FULLY WORKED

EXAMPLE 3

Let $A=\begin{pmatrix}4&3\\2&1\end{pmatrix}$ . (a) Find $\det A$ . (b) Hence find $A^{-1}$ . (c) Verify that $AA^{-1}=I$ . [5]

1
(a) Determinant. $\det A=ad-bc=(4)(1)-(3)(2)=4-6=-2$ . [A1]\n\n**(b) Inverse.** Since $\det A=-2\neq0$ , the inverse exists. Swap the leading diagonal, negate the other diagonal, and divide by $-2$ :\n $A^{-1}=\dfrac{1}{-2}\begin{pmatrix}1&-3\\-2&4\end{pmatrix}=\begin{pmatrix}-0.5&1.5\\1&-2\end{pmatrix}$ . [M1: correct inverse structure] [A1]\n\n**(c) Verify.** $AA^{-1}=\begin{pmatrix}4&3\\2&1\end{pmatrix}\begin{pmatrix}-0.5&1.5\\1&-2\end{pmatrix}=\begin{pmatrix}4(-0.5)+3(1)&4(1.5)+3(-2)\\2(-0.5)+1(1)&2(1.5)+1(-2)\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ . [M1: product attempted] [A1: shown equal to $I$ ]

✓ FULLY WORKED

EXAMPLE 4

Solve the system $2x+3y=13$ , $x-y=1$ using matrices. [5]

1
Model answer — full working.\n\nWrite the system as a matrix equation. Collect the coefficients, unknowns and constants:\n $A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad \mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}13\\1\end{pmatrix},\qquad A\mathbf{x}=\mathbf{b}.$ \n\nDeterminant and inverse. $\det A=(2)(-1)-(3)(1)=-2-3=-5\neq0$ , so $A^{-1}$ exists:\n $A^{-1}=\dfrac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.$ \n\nSolve. Left-multiply by $A^{-1}$ : $\mathbf{x}=A^{-1}\mathbf{b}=\dfrac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}\begin{pmatrix}13\\1\end{pmatrix}=\dfrac{1}{-5}\begin{pmatrix}-16\\-11\end{pmatrix}=\begin{pmatrix}3.2\\2.2\end{pmatrix}.$ \n\nConclusion: $x=3.2,\ y=2.2$ . (Equivalently, entering the two equations in the GDC's simultaneous-equation solver returns $x=3.2$ , $y=2.2$ directly.) Check: $2(3.2)+3(2.2)=6.4+6.6=13$ . Correct.\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — set up the matrix equation. Awarded for correctly packaging the system as $A\mathbf{x}=\mathbf{b}$ with $A=\begin{pmatrix}2&3\\1&-1\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}13\\1\end{pmatrix}$ . It is the correct set-up that is rewarded, so it survives an arithmetic slip further down.\n- M1 — determinant/inverse or GDC solve. A method mark for finding $\det A=-5$ and forming $A^{-1}$ , or equivalently for entering the system into the GDC's solver. A valid technology route earns this mark just as the by-hand inverse does.\n- A1 — value of $x$ . Awarded for $x=3.2$ . This accuracy mark depends on the two method marks above and is protected by follow-through: a candidate whose inverse differed but who multiplied correctly still earns it on their own figures.\n- A1 — value of $y$ . Awarded for $y=2.2$ , again dependent and FT-protected.\n- A1 — method/answer presentation. Awarded for showing the solution reached through $\mathbf{x}=A^{-1}\mathbf{b}$ (or a clearly evidenced GDC solve) and stating both values; the engine checks the inverse was applied on the LEFT, not as $\mathbf{b}A^{-1}$ .\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $x=3.2$ written as $\frac{16}{5}$ and $y=2.2$ as $\frac{11}{5}$ , and accepts the answer whether reached by the inverse or by the GDC solver. Once the correct pair appears, subsequent working is ignored (ISW).\n\nBottom line: two of the five marks are method marks that survive an arithmetic slip, and the accuracy marks for $x$ and $y$ are shielded by follow-through — but only because the matrix equation and the inverse (or GDC set-up) are on the page. A student who writes only ' $x=3.2$ , $y=2.2$ ' with no set-up risks losing the three method-bearing marks.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEASystems of equations and matrix operations

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 14

System of linear equations

TAP OR SPACE TO FLIP

ANSWER

Two or more linear equations sharing the same unknowns, for example $2x+3y=13$ and $x-y=1$ . A solution is a set of values that satisfies every equation at once.

TAP OR SPACE TO FLIP BACK

1 / 14 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Two unknowns: solve by elimination or substitution by hand, or with the GDC's simultaneous-equation solver. Show at least one algebraic step for the method mark.
✓
Up to three unknowns: use the GDC — the system solver, or entering the coefficients and constants as matrices. Write down what you entered.
✓
Standard form first: arrange every equation as $ax+by=c$ (or $ax+by+cz=d$ ) with the variables in the same order before reading off coefficients.
✓
Interpret the outcome: a unique solution, no solution (inconsistent) or infinitely many solutions (dependent).

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a Paper 1 system marked: set up $A\mathbf{x}=\mathbf{b}$, find the inverse and solve with full working

Get a Paper 1 system marked: set up $A\mathbf{x}=\mathbf{b}$ , find the inverse and solve with full working

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

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Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a Paper 1 system marked: set up $A\mathbf{x}=\mathbf{b}$, find the inverse and solve with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS1.4 Exponents and logarithms NEXT →1.6 Matrices: determinants, inverses and applications

Systems of equations and matrix operations

BY THE END, YOU CAN…

1Solve a system of two linear equations in two unknowns algebraically, and a system of up to three unknowns using the GDC.
2Translate a worded context into a system of linear equations, defining variables clearly.
3Add, subtract and scalar-multiply matrices, and multiply matrices, applying the conformability rule to decide when a product exists.
4State the role of the identity matrix, and find the determinant and inverse of a 2x2 matrix.
5Write a system as a matrix equation $A\mathbf{x}=\mathbf{b}$ and solve it using the inverse $\mathbf{x}=A^{-1}\mathbf{b}$ or the GDC, presenting working so method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Many clues, one tidy grid

THINK OF IT LIKE…

1
Translate the context into equations. Define each variable in words, then write every equation in standard form, for example $ax+by=c$ , with the variables in the same order.
2
Choose your route. Two unknowns can be solved algebraically; three unknowns are solved on the GDC. Either way you can package the coefficients into a matrix $A$ , the unknowns into a column $\mathbf{x}$ , and the constants into a column $\mathbf{b}$ .
3
Solve. For a 2x2 system find $\det A$ and the inverse $A^{-1}$ , then compute $\mathbf{x}=A^{-1}\mathbf{b}$ ; or type the system straight into the GDC.
4
Interpret and check. Read the solution back into the context, and substitute into an original equation to confirm it works.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$k\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}ka&kb\\kc&kd\end{pmatrix}$

$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$

Full topic notes

Formal explanation with the rigour you need for the exam.

Solving systems of linear equations

Two unknowns: solve by elimination or substitution by hand, or with the GDC's simultaneous-equation solver. Show at least one algebraic step for the method mark.
Up to three unknowns: use the GDC — the system solver, or entering the coefficients and constants as matrices. Write down what you entered.
Standard form first: arrange every equation as $ax+by=c$ (or $ax+by+cz=d$ ) with the variables in the same order before reading off coefficients.
Interpret the outcome: a unique solution, no solution (inconsistent) or infinitely many solutions (dependent).

Setting up a system from a context

Matrix operations at SL

Addition and subtraction: only for matrices of the SAME order; work element by element.
Scalar multiplication: multiply EVERY element by the scalar.
Matrix multiplication: row-by-column. The product $AB$ exists only when the columns of $A$ match the rows of $B$ .
Order of a product: an $(m\times n)$ times an $(n\times p)$ gives an $(m\times p)$ matrix — inner numbers must match, outer numbers give the size.

The identity, determinant and inverse of a 2x2 matrix

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ : determinant $\det A=ad-bc$ , and inverse $A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ , defined only when $\det A\neq0$ .

The matrix equation $A\mathbf{x}=\mathbf{b}$

Package it: $A$ = coefficients, $\mathbf{x}$ = unknowns (a column), $\mathbf{b}$ = constants (a column).
Left-multiply by $A^{-1}$ : $A^{-1}A\mathbf{x}=A^{-1}\mathbf{b}$ gives $\mathbf{x}=A^{-1}\mathbf{b}$ .
Order matters: it is $A^{-1}\mathbf{b}$ , never $\mathbf{b}A^{-1}$ .
Or use the GDC: enter $A$ and $\mathbf{b}$ and compute $A^{-1}\mathbf{b}$ , or use the system solver directly. Both are accepted; show the set-up.

EXAM TIP

Common mistakes examiners penalise

Adding or subtracting matrices of different orders — these operations only exist for matrices of the SAME order; check the orders before you start.
Forcing a product that is not conformable — $AB$ exists only if the columns of $A$ equal the rows of $B$ . If the inner numbers disagree, write 'not defined' rather than inventing entries.
Swapping the order of a product — $AB\neq BA$ in general. Copy the factors down in the order the question gives them and never reverse them.
Mis-signing the determinant — $\det A=ad-bc$ , so watch the subtraction and the signs of $b$ and $c$ ; a sign slip here corrupts the whole inverse.
Forgetting to divide by the determinant — the inverse is $\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ ; writing only the adjusted matrix without the $\frac{1}{\det A}$ factor is a classic lost accuracy mark.
Solving as $\mathbf{b}A^{-1}$ instead of $A^{-1}\mathbf{b}$ — you must left-multiply; the reversed product is wrong and usually not conformable.
Trying to invert a singular matrix — if $\det A=0$ there is no inverse and no unique solution; say so instead of dividing by zero.
Setting up equations with misaligned variables — line up $x$ , $y$ , $z$ in the same order in every equation and use a coefficient of 0 for any missing variable before reading them into $A$ .

Model answer — marked the way our engine marks it

Where this leads

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

1
Define the variables (given): $a=$ adult tickets, $c=$ child tickets.\n\nSet up the system. 'Total tickets' and 'total takings' give one equation each:\n $a+c=200$ \n $12a+8c=2040$ [M1: two correct equations from the context]\n\nSolve. From the first equation $a=200-c$ . Substitute into the second:\n $12(200-c)+8c=2040$ \n $2400-4c=2040$ \n $-4c=-360$ , so $c=90$ . [M1: a valid elimination/substitution step] [A1: $c=90$ ]\n\nThen $a=200-90=110$ . [A1: $a=110$ ]\n\nInterpret. 110 adult tickets and 90 child tickets were sold. Check: $12(110)+8(90)=1320+720=2040$ . Correct. [A1: both values stated in context, with a check]

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EXAMPLE 2

Let $A=\begin{pmatrix}1&2\\3&0\end{pmatrix}$ and $B=\begin{pmatrix}4&-1\\2&5\end{pmatrix}$ . Find (a) $2A-B$ , and (b) the product $AB$ , and (c) state whether $BA$ equals $AB$ . [5]

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(a) $2A-B$ . Scalar-multiply first, then subtract element by element:\n $2A=\begin{pmatrix}2&4\\6&0\end{pmatrix}$ , so $2A-B=\begin{pmatrix}2-4&4-(-1)\\6-2&0-5\end{pmatrix}=\begin{pmatrix}-2&5\\4&-5\end{pmatrix}$ . [M1: scalar multiple and subtraction of like elements] [A1]\n\n**(b) $AB$ .** Both are $2\times2$ , so the product is $2\times2$ . Multiply row by column:\n $AB=\begin{pmatrix}1(4)+2(2)&1(-1)+2(5)\\3(4)+0(2)&3(-1)+0(5)\end{pmatrix}=\begin{pmatrix}8&9\\12&-3\end{pmatrix}$ . [M1: row-by-column method] [A1]\n\n**(c) Does $BA=AB$ ?** Computing $BA=\begin{pmatrix}4(1)+(-1)(3)&4(2)+(-1)(0)\\2(1)+5(3)&2(2)+5(0)\end{pmatrix}=\begin{pmatrix}1&8\\17&4\end{pmatrix}$ , which is different from $AB$ . So $BA\neq AB$ : matrix multiplication is not commutative. [A1: correct comparison and conclusion]

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EXAMPLE 3

Let $A=\begin{pmatrix}4&3\\2&1\end{pmatrix}$ . (a) Find $\det A$ . (b) Hence find $A^{-1}$ . (c) Verify that $AA^{-1}=I$ . [5]

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(a) Determinant. $\det A=ad-bc=(4)(1)-(3)(2)=4-6=-2$ . [A1]\n\n**(b) Inverse.** Since $\det A=-2\neq0$ , the inverse exists. Swap the leading diagonal, negate the other diagonal, and divide by $-2$ :\n $A^{-1}=\dfrac{1}{-2}\begin{pmatrix}1&-3\\-2&4\end{pmatrix}=\begin{pmatrix}-0.5&1.5\\1&-2\end{pmatrix}$ . [M1: correct inverse structure] [A1]\n\n**(c) Verify.** $AA^{-1}=\begin{pmatrix}4&3\\2&1\end{pmatrix}\begin{pmatrix}-0.5&1.5\\1&-2\end{pmatrix}=\begin{pmatrix}4(-0.5)+3(1)&4(1.5)+3(-2)\\2(-0.5)+1(1)&2(1.5)+1(-2)\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ . [M1: product attempted] [A1: shown equal to $I$ ]

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EXAMPLE 4

Solve the system $2x+3y=13$ , $x-y=1$ using matrices. [5]

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Model answer — full working.\n\nWrite the system as a matrix equation. Collect the coefficients, unknowns and constants:\n $A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad \mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}13\\1\end{pmatrix},\qquad A\mathbf{x}=\mathbf{b}.$ \n\nDeterminant and inverse. $\det A=(2)(-1)-(3)(1)=-2-3=-5\neq0$ , so $A^{-1}$ exists:\n $A^{-1}=\dfrac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.$ \n\nSolve. Left-multiply by $A^{-1}$ : $\mathbf{x}=A^{-1}\mathbf{b}=\dfrac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}\begin{pmatrix}13\\1\end{pmatrix}=\dfrac{1}{-5}\begin{pmatrix}-16\\-11\end{pmatrix}=\begin{pmatrix}3.2\\2.2\end{pmatrix}.$ \n\nConclusion: $x=3.2,\ y=2.2$ . (Equivalently, entering the two equations in the GDC's simultaneous-equation solver returns $x=3.2$ , $y=2.2$ directly.) Check: $2(3.2)+3(2.2)=6.4+6.6=13$ . Correct.\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — set up the matrix equation. Awarded for correctly packaging the system as $A\mathbf{x}=\mathbf{b}$ with $A=\begin{pmatrix}2&3\\1&-1\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}13\\1\end{pmatrix}$ . It is the correct set-up that is rewarded, so it survives an arithmetic slip further down.\n- M1 — determinant/inverse or GDC solve. A method mark for finding $\det A=-5$ and forming $A^{-1}$ , or equivalently for entering the system into the GDC's solver. A valid technology route earns this mark just as the by-hand inverse does.\n- A1 — value of $x$ . Awarded for $x=3.2$ . This accuracy mark depends on the two method marks above and is protected by follow-through: a candidate whose inverse differed but who multiplied correctly still earns it on their own figures.\n- A1 — value of $y$ . Awarded for $y=2.2$ , again dependent and FT-protected.\n- A1 — method/answer presentation. Awarded for showing the solution reached through $\mathbf{x}=A^{-1}\mathbf{b}$ (or a clearly evidenced GDC solve) and stating both values; the engine checks the inverse was applied on the LEFT, not as $\mathbf{b}A^{-1}$ .\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $x=3.2$ written as $\frac{16}{5}$ and $y=2.2$ as $\frac{11}{5}$ , and accepts the answer whether reached by the inverse or by the GDC solver. Once the correct pair appears, subsequent working is ignored (ISW).\n\nBottom line: two of the five marks are method marks that survive an arithmetic slip, and the accuracy marks for $x$ and $y$ are shielded by follow-through — but only because the matrix equation and the inverse (or GDC set-up) are on the page. A student who writes only ' $x=3.2$ , $y=2.2$ ' with no set-up risks losing the three method-bearing marks.

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How it all connects

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MAIN IDEASystems of equations and matrix operations

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QUESTION · 1 / 14

System of linear equations

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ANSWER

Two or more linear equations sharing the same unknowns, for example $2x+3y=13$ and $x-y=1$ . A solution is a set of values that satisfies every equation at once.

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Key takeaways

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Two unknowns: solve by elimination or substitution by hand, or with the GDC's simultaneous-equation solver. Show at least one algebraic step for the method mark.
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Up to three unknowns: use the GDC — the system solver, or entering the coefficients and constants as matrices. Write down what you entered.
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Standard form first: arrange every equation as $ax+by=c$ (or $ax+by+cz=d$ ) with the variables in the same order before reading off coefficients.
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Interpret the outcome: a unique solution, no solution (inconsistent) or infinitely many solutions (dependent).

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Get a Paper 1 system marked: set up $A\mathbf{x}=\mathbf{b}$, find the inverse and solve with full working

Get a Paper 1 system marked: set up $A\mathbf{x}=\mathbf{b}$ , find the inverse and solve with full working

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Revise Systems of equations and matrix operations for IB Mathematics: Applications and Interpretation (1.5)

In simple terms

Explore the concept

Key formulas

Full topic notes

Solving systems of linear equations

Setting up a system from a context

Matrix operations at SL

The identity, determinant and inverse of a 2x2 matrix

The matrix equation $A\mathbf{x}=\mathbf{b}$

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 1 system marked: set up $A\mathbf{x}=\mathbf{b}$, find the inverse and solve with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

Solving systems of linear equations

Setting up a system from a context

Matrix operations at SL

The identity, determinant and inverse of a 2x2 matrix

The matrix equation $A\mathbf{x}=\mathbf{b}$

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 1 system marked: set up $A\mathbf{x}=\mathbf{b}$, find the inverse and solve with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Systems of equations and matrix operations