Matrices AI HL

IB Mathematics: Applications and Interpretation · AI HL

Lesson 1.14 · Matrices

Organise data, perform matrix operations, solve systems and model practical problems.

Faith connection · God’s love

A matrix brings several pieces of information into one organised picture. The Bible tells us that God’s love has been shown clearly in Jesus: “For God so loved the world, that he gave his only begotten Son” (John 3:16, KJV). As you learn to recognise patterns and connections in mathematics, remember this invitation: come to Jesus, trust Him as your Savior, and follow Him as Lord.

Learning objectives

By the end of this lesson, you should be able to:

identify the order, rows, columns and elements of a matrix;
add, subtract and multiply matrices by scalars;
multiply compatible matrices and interpret the result;
recognise the identity and zero matrices and the main properties of matrix multiplication;
find determinants and inverses of $2\times2$ matrices by hand;
use technology for determinants and inverses of larger matrices;
write and solve a system of linear equations in the form $Ax=b$ ;
apply matrices to sales information and simple message coding.

1 · Matrix language and order

A matrix is a rectangular array of numbers. A matrix with $m$ rows and $n$ columns has order $m\times n$ . Each number is an element.

$A=\begin{pmatrix} 3&-2&5\\ 1&4&0 \end{pmatrix}$

Matrix $A$ has two rows and three columns, so its order is $2\times3$ . The element in row 2, column 3 is $a_{23}=0$ .

Worked example · Reading a matrix

For the matrix below, state its order and find $b_{31}$ .

$B=\begin{pmatrix} 2&8\\ -1&6\\ 7&0 \end{pmatrix}$

There are three rows and two columns, so the order is $\boxed{3\times2}$ . The element in row 3, column 1 is $\boxed{b_{31}=7}$ .

2 · Equality, addition, subtraction and scalar multiplication

Two matrices are equal when they have the same order and corresponding elements are equal. Matrices can be added or subtracted only when they have the same order.

$\begin{pmatrix}a&b\\c&d\end{pmatrix} +\begin{pmatrix}e&f\\g&h\end{pmatrix} =\begin{pmatrix}a+e&b+f\\c+g&d+h\end{pmatrix}.$

A scalar is an ordinary number. Multiplying a matrix by a scalar multiplies every element by that number.

Worked example · Combining operations

Let $A=\begin{pmatrix}2&-1\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}5&2\\-1&6\end{pmatrix}$ . Find $2A-B$ .

$2A-B =\begin{pmatrix}4&-2\\6&8\end{pmatrix} -\begin{pmatrix}5&2\\-1&6\end{pmatrix} =\boxed{\begin{pmatrix}-1&-4\\7&2\end{pmatrix}}.$

3 · Matrix multiplication

You can multiply an $m\times n$ matrix by an $n\times p$ matrix. The inner dimensions must match; the result has order $m\times p$ .

$(m\times n)(n\times p)\longrightarrow(m\times p).$

To calculate an element of the product, multiply the entries of one row by the corresponding entries of one column and add the results.

Worked example · A $2\times3$ product

Find $AB$ for the matrices below.

$A=\begin{pmatrix} 1&2&0\\ 3&-1&4 \end{pmatrix}, \qquad B=\begin{pmatrix} 2&1\\ 0&3\\ -1&2 \end{pmatrix}.$

The product has order $2\times2$ . For example, its top-right element comes from the first row of $A$ and second column of $B$ :

$1(1)+2(3)+0(2)=7.$

Calculating the other three elements gives:

$AB= \boxed{\begin{pmatrix} 2&7\\ 2&8 \end{pmatrix}}.$

Common mistake: Matrix multiplication is not performed by multiplying corresponding elements. Use row times column.

4 · Properties of matrix multiplication

Matrix multiplication is associative and distributive whenever the products have compatible dimensions.

$(AB)C=A(BC),\qquad A(B+C)=AB+AC.$

Matrix multiplication is generally not commutative: changing the order can change the answer.

$AB\ne BA\quad\text{in general.}$

Worked example · Why order matters

Let $A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}1&0\\3&1\end{pmatrix}$ .

$AB=\begin{pmatrix}7&2\\3&1\end{pmatrix}, \qquad BA=\begin{pmatrix}1&2\\3&7\end{pmatrix}.$

Both products exist, but $AB\ne BA$ .

5 · Identity and zero matrices

The identity matrix $I$ has 1s on its main diagonal and 0s elsewhere. Multiplying a matrix by a compatible identity matrix leaves it unchanged.

$I_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}, \qquad AI=IA=A \quad\text{for a square matrix }A.$

A zero matrix $O$ has 0 in every position. Adding it to a matrix of the same order leaves that matrix unchanged.

$A+O=A,\qquad AO=O \quad\text{when the product is defined.}$

Worked example · Identity and zero

For $A=\begin{pmatrix}2&-1\\3&4\end{pmatrix}$ :

$AI_2 =\begin{pmatrix}2&-1\\3&4\end{pmatrix}, \qquad A\begin{pmatrix}0&0\\0&0\end{pmatrix} =\begin{pmatrix}0&0\\0&0\end{pmatrix}.$

6 · Determinants and inverses

For a $2\times2$ matrix, its determinant is found by multiplying along the main diagonal and subtracting the other diagonal product.

$A=\begin{pmatrix}a&b\\c&d\end{pmatrix} \quad\Longrightarrow\quad \boxed{\det(A)=ad-bc}.$

If $\det(A)\ne0$ , the matrix has an inverse $A^{-1}$ , satisfying $A^{-1}A=AA^{-1}=I$ . For a $2\times2$ matrix:

$\boxed{ A^{-1}=\frac1{ad-bc} \begin{pmatrix}d&-b\\-c&a\end{pmatrix} },\qquad ad-bc\ne0.$

Worked example · Inverse by hand

Find the determinant and inverse of $A=\begin{pmatrix}2&1\\3&2\end{pmatrix}$ .

$\det(A)=2(2)-1(3)=\boxed{1}.$

$A^{-1} =\frac11\begin{pmatrix}2&-1\\-3&2\end{pmatrix} =\boxed{\begin{pmatrix}2&-1\\-3&2\end{pmatrix}}.$

Important: If $\det(A)=0$ , the matrix has no inverse. For larger $n\times n$ matrices, use technology to calculate determinants and inverses.

7 · Solving equations using $Ax=b$

A system of linear equations can be written as a matrix equation $Ax=b$ . Here $A$ contains the coefficients, $x$ is a column of unknowns, and $b$ is a column of constants. If $A$ is invertible, multiply both sides on the left by $A^{-1}$ .

$Ax=b \quad\Longrightarrow\quad \boxed{x=A^{-1}b}.$

Worked example · Two unknowns

Solve the system below using an inverse matrix.

$2x+y=7,\qquad 3x+2y=12.$

Write it as $Ax=b$ .

$\begin{pmatrix}2&1\\3&2\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}7\\12\end{pmatrix}.$

Use the inverse found in the previous section.

$\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2&-1\\-3&2\end{pmatrix} \begin{pmatrix}7\\12\end{pmatrix} = \begin{pmatrix}2\\3\end{pmatrix}.$

Therefore $\boxed{x=2,\ y=3}$ .

Order matters: From $Ax=b$ , the expression is $A^{-1}b$ . Matrix products cannot generally be reversed.

8 · Using the TI-84 Plus CE

For a numerical matrix calculation, press 2nd, then x⁻¹ to open the MATRIX menu. Under EDIT, choose a matrix, enter its dimensions, and fill in its elements. Return to the home screen to select the matrix under NAMES.

For a determinant, choose det( from MATRIX → MATH and enter the matrix name.
For an inverse, enter the matrix name followed by the calculator’s x⁻¹ key.
To solve $Ax=b$ , store $A$ and $b$ , then calculate $A^{-1}b$ .
Before multiplying, check the dimensions and the order of the matrices.

Worked example · Three unknowns with technology

Solve the following system by entering its coefficient matrix $A$ and constant column $b$ into a calculator, then calculating $A^{-1}b$ .

$x+y+z=6,\qquad x-y+z=2,\qquad 2x+y-z=1.$

$A=\begin{pmatrix} 1&1&1\\ 1&-1&1\\ 2&1&-1 \end{pmatrix}, \qquad b=\begin{pmatrix}6\\2\\1\end{pmatrix}.$

$A^{-1}b =\begin{pmatrix}1\\2\\3\end{pmatrix}.$

Hence $\boxed{x=1,\ y=2,\ z=3}$ . Substitution checks all three original equations.

IB-style Paper 1 · Skill practice

Question 1 · [3 marks]

State the order of $M$ and the values of $m_{13}$ and $m_{21}$ .

$M=\begin{pmatrix}4&0&-2\\7&5&1\end{pmatrix}$

Show solution

$M$ has two rows and three columns. Therefore its order is $\boxed{2\times3}$ , with $\boxed{m_{13}=-2}$ and $\boxed{m_{21}=7}$ .

Question 2 · [4 marks]

Let $A=\begin{pmatrix}1&3\\-2&4\end{pmatrix}$ and $B=\begin{pmatrix}5&-1\\0&2\end{pmatrix}$ . Find $3A-2B$ .

Show solution

$3A=\begin{pmatrix}3&9\\-6&12\end{pmatrix}, \qquad 2B=\begin{pmatrix}10&-2\\0&4\end{pmatrix}.$

$3A-2B =\boxed{\begin{pmatrix}-7&11\\-6&8\end{pmatrix}}.$

Question 3 · [5 marks]

Let $A=\begin{pmatrix}2&1\\-1&3\end{pmatrix}$ and $B=\begin{pmatrix}4&-2\\0&5\end{pmatrix}$ . Find $AB$ .

Show solution

$AB= \begin{pmatrix} 2(4)+1(0)&2(-2)+1(5)\\ (-1)(4)+3(0)&(-1)(-2)+3(5) \end{pmatrix} =\boxed{\begin{pmatrix}8&1\\-4&17\end{pmatrix}}.$

Question 4 · [5 marks]

Find the determinant and inverse of $A=\begin{pmatrix}4&1\\3&1\end{pmatrix}$ by hand.

Show solution

$\det(A)=4(1)-1(3)=\boxed{1}.$

$A^{-1} =\frac11\begin{pmatrix}1&-1\\-3&4\end{pmatrix} =\boxed{\begin{pmatrix}1&-1\\-3&4\end{pmatrix}}.$

Question 5 · [4 marks]

Explain why $C=\begin{pmatrix}2&4\\1&2\end{pmatrix}$ has no inverse.

Show solution

$\det(C)=2(2)-4(1)=0.$

A matrix with determinant zero is not invertible. Therefore $\boxed{C^{-1}\text{ does not exist}}$ .

Question 6 · [6 marks]

Use an inverse matrix to solve $3x+y=11$ and $x+y=5$ .

Show solution

$\begin{pmatrix}3&1\\1&1\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}11\\5\end{pmatrix}.$

The determinant is $3(1)-1(1)=2$ , so:

$A^{-1}=\frac12 \begin{pmatrix}1&-1\\-1&3\end{pmatrix}.$

$\begin{pmatrix}x\\y\end{pmatrix} =\frac12 \begin{pmatrix}1&-1\\-1&3\end{pmatrix} \begin{pmatrix}11\\5\end{pmatrix} =\begin{pmatrix}3\\2\end{pmatrix}.$

$\boxed{x=3,\ y=2}$ .

Question 7 · [6 marks]

Let $A=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ and $B=\begin{pmatrix}5&2\\3&1\end{pmatrix}$ . Solve the matrix equation $AX=B$ .

Show solution

Multiply on the left by $A^{-1}$ .

$X=A^{-1}B.$

Since $\det(A)=1$ , its inverse is $\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$ .

$X= \begin{pmatrix}1&-1\\-1&2\end{pmatrix} \begin{pmatrix}5&2\\3&1\end{pmatrix} =\boxed{\begin{pmatrix}2&1\\1&0\end{pmatrix}}.$

Applications · Advanced IB-style questions

Application 1 · Weekly sales [7 marks]

A shop sells three products. The rows of $S$ give the numbers sold in weeks 1 and 2; its columns correspond to products A, B and C. Matrix $p$ gives their prices in euros.

$S=\begin{pmatrix} 12&8&5\\ 15&6&9 \end{pmatrix}, \qquad p=\begin{pmatrix}4\\7\\10\end{pmatrix}.$

(a) Explain why $Sp$ can be calculated. (b) Find $Sp$ and interpret each element. (c) Find the difference between the weeks’ revenues.

Show solution

(a) The dimensions are $(2\times3)(3\times1)$ , so the inner dimensions match and the result is $2\times1$ .

(b)

$Sp= \begin{pmatrix} 12(4)+8(7)+5(10)\\ 15(4)+6(7)+9(10) \end{pmatrix} =\boxed{\begin{pmatrix}154\\192\end{pmatrix}}.$

The entries are the week 1 and week 2 revenues, respectively, in euros.

(c) The difference is $192-154=\boxed{€38}$ .

Application 2 · Coding and decoding [8 marks]

A message uses A = 1, B = 2, … , Z = 26. Each pair of letter numbers is written as a column $v$ . An encoder calculates $Kv$ , where:

$K=\begin{pmatrix}1&1\\1&2\end{pmatrix}.$

(a) Encode the pairs HE and LP. (b) Find $K^{-1}$ . (c) Use your inverse to decode the column $\begin{pmatrix}13\\18\end{pmatrix}$ .

Show solution

(a) HE corresponds to $(8,5)$ ; LP corresponds to $(12,16)$ .

$K\begin{pmatrix}8\\5\end{pmatrix} =\boxed{\begin{pmatrix}13\\18\end{pmatrix}}, \qquad K\begin{pmatrix}12\\16\end{pmatrix} =\boxed{\begin{pmatrix}28\\44\end{pmatrix}}.$

(b) Since $\det(K)=1(2)-1(1)=1$ :

$\boxed{K^{-1}=\begin{pmatrix}2&-1\\-1&1\end{pmatrix}}.$

(c)

$K^{-1}\begin{pmatrix}13\\18\end{pmatrix} =\begin{pmatrix}8\\5\end{pmatrix}.$

The decoded letters are $\boxed{\text{HE}}$ .

Application 3 · Production planning [8 marks]

A workshop produces three items. Let $x$ , $y$ and $z$ be the numbers of each item made in one day. The resource requirements give the system:

$x+y+z=30,\qquad 2x+y+3z=65,\qquad x+2y+z=38.$

(a) Write the system as $Ax=b$ . (b) Use an inverse matrix on your calculator to find $x$ , $y$ and $z$ . (c) Check the solution in one original equation.

Show solution

(a)

$\underbrace{\begin{pmatrix} 1&1&1\\ 2&1&3\\ 1&2&1 \end{pmatrix}}_{A} \underbrace{\begin{pmatrix}x\\y\\z\end{pmatrix}}_{\text{unknowns}} = \underbrace{\begin{pmatrix}30\\65\\38\end{pmatrix}}_{b}.$

(b) Calculate $A^{-1}b$ .

$\begin{pmatrix}x\\y\\z\end{pmatrix} =A^{-1}b =\boxed{\begin{pmatrix}10\\8\\12\end{pmatrix}}.$

(c) For example, $2(10)+8+3(12)=64$ . This does not equal 65, so the proposed result requires correction.

Solving the system carefully: subtract the first equation from the third to get $y=8$ . Then $x+z=22$ . The second equation gives $2x+3z=57$ ; subtract $2(x+z)=44$ to get $z=13$ and $x=9$ .

$\boxed{x=9,\qquad y=8,\qquad z=13}.$

Check: $2(9)+8+3(13)=65$ .

Application 4 · Two linked prices [7 marks]

A café sells a drink and a snack. Two drinks and three snacks cost €18; four drinks and one snack cost €22.

(a) Construct a matrix equation for the prices. (b) Solve it using the inverse matrix method. (c) Find the cost of five drinks and two snacks.

Show solution

(a) Let $d$ be the drink price and $s$ the snack price, in euros.

$\begin{pmatrix}2&3\\4&1\end{pmatrix} \begin{pmatrix}d\\s\end{pmatrix} = \begin{pmatrix}18\\22\end{pmatrix}.$

(b) The determinant is $2(1)-3(4)=-10$ .

$A^{-1} =-\frac1{10} \begin{pmatrix}1&-3\\-4&2\end{pmatrix}.$

$\begin{pmatrix}d\\s\end{pmatrix} =A^{-1}\begin{pmatrix}18\\22\end{pmatrix} =\boxed{\begin{pmatrix}4.8\\2.8\end{pmatrix}}.$

(c) The requested cost is $5(4.8)+2(2.8)=\boxed{€29.60}$ .

Quick recap

$(m\times n)(n\times p)\longrightarrow(m\times p)$

$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$

$Ax=b\quad\Longrightarrow\quad x=A^{-1}b \qquad\text{when }A^{-1}\text{ exists.}$

Exam habit: Check matrix dimensions before multiplying, keep factors in the correct order, and substitute a calculated solution back into the original context.

Looking ahead: Matrices also appear in transition models and Markov chains later in AI HL.

Lesson 1.14 · Matrices

Faith connection · God’s love

Learning objectives

1 · Matrix language and order

2 · Equality, addition, subtraction and scalar multiplication

3 · Matrix multiplication

4 · Properties of matrix multiplication

5 · Identity and zero matrices

6 · Determinants and inverses

7 · Solving equations using $Ax=b$

8 · Using the TI-84 Plus CE

IB-style Paper 1 · Skill practice

Applications · Advanced IB-style questions

Quick recap

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Matrices AI HL

Faith connection · God’s love

Learning objectives

1 · Matrix language and order

2 · Equality, addition, subtraction and scalar multiplication

3 · Matrix multiplication

4 · Properties of matrix multiplication

5 · Identity and zero matrices

6 · Determinants and inverses

7 · Solving equations using

8 · Using the TI-84 Plus CE

IB-style Paper 1 · Skill practice

Applications · Advanced IB-style questions

Quick recap

Related

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alphy school publishers

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7 · Solving equations using $Ax=b$