Consider the matrices $A = \begin{pmatrix} 4 & -2 \\ 1 & 5 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 3 \\ -1 & 2 \end{pmatrix}$. Calculate exactly the matrix product $AB$.

1. Verify the dimensions: both are $2 \times 2$ matrices, so they can be multiplied to produce a $2 \times 2$ result. 2. Calculate the top-left element (Row 1 $\times$ Col 1): $(4)(0) + (-2)(-1) = 0 + 2 = 2$. 3. Calculate the top-right element (Row 1 $\times$ Col 2): $(4)(3) + (-2)(2) = 12 - 4 = 8$. 4. Calculate the bottom-left element (Row 2 $\times$ Col 1): $(1)(0) + (5)(-1) = 0 - 5 = -5$. 5. Calculate the bottom-right element (Row 2 $\times$ Col 2): $(1)(3) + (5)(2) = 3 + 10 = 13$. The final matrix is $\begin{pmatrix} 2 & 8 \\ -5 & 13 \end{pmatrix}$ .

A system of equations is written in matrix form as $MX = C$, where $M = \begin{pmatrix} 2 & 1 & -1 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, and $C = \begin{pmatrix} 1 \\ 2 \\ 7 \end{pmatrix}$. Using your Graphic Display Calculator, find the exact values of $x$, $y$, and $z$.

1. Rearrange the matrix equation to solve for $X$ by pre-multiplying both sides by the inverse of $M$: $X = M^{-1}C$. 2. Enter matrix $M$ and matrix $C$ into the matrix workspace of your GDC. 3. Compute the inverse matrix $M^{-1}$ using the $x^{-1}$ function on your calculator. 4. Multiply $M^{-1}$ by $C$ directly on your GDC to prevent intermediate rounding errors. 5. Verify the exact solutions by direct substitution: $2(1) + 2 - 3 = 1$; $1 + 2(2) - 3 = 2$; $3(1) - 2 + 2(3) = 7$. All three equations balance, so $x = 1$, $y = 2$, and $z = 3$ .

Find the exact values of $k$ for which the matrix $A = \begin{pmatrix} k & 4 \\ 3 & k-1 \end{pmatrix}$ is singular.

1. Recall that a matrix is singular if and only if its determinant is exactly zero. 2. Set up the determinant formula for a $2 \times 2$ matrix ($\det(A) = ad - bc$): $k(k-1) - (4)(3) = 0$. 3. Expand the brackets: $k^2 - k - 12 = 0$. 4. Factorise the resulting quadratic equation: $(k - 4)(k + 3) = 0$. 5. Solve to find the exact values that make the matrix singular: $k = 4$ and $k = -3$ .

Matrices, Determinants and Inverses Questions

What you need to know

2×2 and 3×3 matrices: products, determinants and inverses, and solving a system of linear equations with the inverse matrix. Know your GDC's matrix menu — Paper 2 relies on it. Matrices — multiplication, inverse, and determinant overview →

What's examined in AI HL matrices, determinants and inverses

The question bank covers these matrices, determinants and inverses question types (number of questions in brackets):

Inverses and Equations (22)
Matrix Algebra Basics (13)
Determinants and Singularity (7)

Matrices, Determinants and Inverses worked examples

Worked example 1: Executing Matrix Multiplication · easy

Consider the matrices $A = (\begin{matrix} 4 & - 2 \\ 1 & 5 \end{matrix})$ and $B = (\begin{matrix} 0 & 3 \\ - 1 & 2 \end{matrix})$ . Calculate exactly the matrix product $A B$ .

Solution

5 steps · 1 to fill in

Examiner tip: Matrix multiplication operates by multiplying the rows of the first matrix by the columns of the second matrix; it is generally not commutative (

A B \neq B A

).

Worked example 2: Using Matrix Inverses to Solve Systems · medium

A system of equations is written in matrix form as $M X = C$ , where $M = (\begin{matrix} 2 & 1 & - 1 \\ 1 & 2 & - 1 \\ 3 & - 1 & 2 \end{matrix})$ , $X = (\begin{matrix} x \\ y \\ z \end{matrix})$ , and $C = (\begin{matrix} 1 \\ 2 \\ 7 \end{matrix})$ . Using your Graphic Display Calculator, find the exact values of $x$ , $y$ , and $z$ .

Solution

5 steps · 1 to fill in

Examiner tip: When isolating

X

in the equation

M X = C

, you must strictly pre-multiply by the inverse to get

X = M^{- 1} C

. Post-multiplying (

C M^{- 1}

) is dimensionally impossible here.

Worked example 3: Identifying Singular Matrices · hard

Find the exact values of $k$ for which the matrix $A = (\begin{matrix} k & 4 \\ 3 & k - 1 \end{matrix})$ is singular.

Solution

5 steps · 1 to fill in

Examiner tip: A singular matrix cannot be inverted because dividing by its determinant (which is zero) is mathematically undefined. This means the associated system of equations has no unique solution.

Try these IB Maths AI HL matrices, determinants and inverses questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area. One of them has its full worked solution here, open to everyone.

Question 1 · easy · 4 marks · Paper 1

A matrix $M$ is defined as $M = (\begin{matrix} 6 & 2 \\ 5 & 3 \end{matrix})$ .
(a) Calculate the determinant of $M$ , denoted as $det (M)$ or $| M |$ . [2 marks]
(b) Hence, write down the exact inverse matrix, $M^{- 1}$ . [2 marks]

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Question 2 · medium · 5 marks · Paper 1

A company sells two types of bundles: Bundle A contains 3 phones and 2 cases, and costs €1850. Bundle B contains 4 phones and 5 cases, and costs €2700.
Let $p$ be the price of one phone and $c$ be the price of one case.
(a) Write this information as a system of linear equations in the form $A X = B$ , where $X = (\begin{matrix} p \\ c \end{matrix})$ . [2 marks]
(b) Find $A^{- 1}$ and use it to calculate the exact price of one phone and one case. [3 marks]

Full worked solution Free, no sign-in

(a) Write the equations: $3 p + 2 c = 1850$ and $4 p + 5 c = 2700$ [M1]
Matrix form: $(\begin{matrix} 3 & 2 \\ 4 & 5 \end{matrix}) (\begin{matrix} p \\ c \end{matrix}) = (\begin{matrix} 1850 \\ 2700 \end{matrix})$ [A1]
(b) Find the determinant of $A$ : $(3) (5) - (2) (4) = 15 - 8 = 7$
Form the inverse: $A^{- 1} = \frac{1}{7} (\begin{matrix} 5 & - 2 \\ - 4 & 3 \end{matrix})$ [A1]
Multiply $X = A^{- 1} B$ :
$X = \frac{1}{7} (\begin{matrix} 5 & - 2 \\ - 4 & 3 \end{matrix}) (\begin{matrix} 1850 \\ 2700 \end{matrix}) = \frac{1}{7} (\begin{matrix} 9250 - 5400 \\ - 7400 + 8100 \end{matrix})$ [M1]
$X = \frac{1}{7} (\begin{matrix} 3850 \\ 700 \end{matrix}) = (\begin{matrix} 550 \\ 100 \end{matrix})$
Price of a phone $= € 550$ , price of a case $= € 100$ [A1]

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Question 3 · hard · 8 marks · Paper 1

Consider the matrix $A = (\begin{matrix} x & 2 \\ - 1 & y \end{matrix})$ , where $x, y \in R$ .

Given that $A^{2} = I$ , where $I$ is the $2 \times 2$ identity matrix:

(a) Find the possible values for $x$ and $y$ . [5 marks]

(b) Using your answers from part (a), determine the value of $det (2 A)$ . [3 marks]

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All 42 matrices, determinants and inverses questions with mark schemes →

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FAQ

How many IB Maths AI HL matrices, determinants and inverses questions are there?

There are 42 exam-style matrices, determinants and inverses questions in the AI HL question bank (Paper 1: 27 · Paper 2: 15), graded 11 easy, 14 medium, 16 hard, 1 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is matrices, determinants and inverses on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 27 · Paper 2: 15. Practise with your GDC — AI papers expect calculator methods throughout.

Where can I get the mark schemes?

Open the AI HL Unit 1 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

IB Maths AI HL Matrices, Determinants and Inverses Questions

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What you need to know

What's examined in AI HL matrices, determinants and inverses

Matrices, Determinants and Inverses worked examples

Worked example 1: Executing Matrix Multiplication · easy

Solution

Worked example 2: Using Matrix Inverses to Solve Systems · medium

Solution

Worked example 3: Identifying Singular Matrices · hard

Solution

Try these IB Maths AI HL matrices, determinants and inverses questions

Question 1 · easy · 4 marks · Paper 1

Question 2 · medium · 5 marks · Paper 1

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Question 3 · hard · 8 marks · Paper 1

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