Hello everyone! Ever felt like you’re on a secret mission when you’re deep into an IB Math problem? That feeling of cracking a code or unlocking a puzzle? That’s exactly what working with matrices can feel like. They aren’t just random grids of numbers; they’re powerful tools for organizing and manipulating data. Think of them as classified dossiers holding vital information—sales figures, inventory levels, or even coordinates for a 3D model.
Today, we’re accepting three fundamental missions that form the basis of matrix algebra: fusing data sets with addition, auditing changes with subtraction, and uniformly scaling information with scalar multiplication. By the end of this briefing, you’ll be navigating these operations with confidence. Let’s gear up and dive in!
Table of Contents
Mission 1: The Data Fusion (Addition)
Our first mission involves combining information. Imagine you’re managing two different streaming platforms and you want to see your total weekly new subscribers for different genres. This is a perfect job for matrix addition, where we essentially merge two datasets into one comprehensive picture.
The Golden Rule of Matrix Addition
To add two matrices, they must have the exact same dimensions. This means they need to have the same number of rows and the same number of columns. If one matrix is a $2 \times 3$ and the other is a $3 \times 2$, you can’t add them. It’s a non-negotiable rule!
Example: Streaming Service Subscribers
Let’s say Matrix $N$ represents new subscribers for Netflix and Matrix $H$ for HBO Max over a weekend. The rows represent genres (Sci-Fi, Comedy) and the columns represent days (Saturday, Sunday).
$N = \begin{pmatrix} 500 & 750 \\ 1200 & 1400 \end{pmatrix}$ and $H = \begin{pmatrix} 450 & 600 \\ 900 & 1150 \end{pmatrix}$
To find the total new subscribers, we calculate $T = N + H$.
First, we check the dimensions. Both $N$ and $H$ are $2 \times 2$ matrices. The rule is satisfied, so we’re cleared for data fusion.
The process is simple: add the elements in the corresponding positions.
$T = \begin{pmatrix} 500+450 & 750+600 \\ 1200+900 & 1400+1150 \end{pmatrix} = \begin{pmatrix} 950 & 1350 \\ 2100 & 2550 \end{pmatrix}$
And there you have it! Our combined subscriber matrix, $T$.
But what happens if the dimensions don’t align? Imagine trying to add data from a third service, Disney+, which also tracks a ‘Documentary’ genre, making its matrix a $3 \times 2$.
$D = \begin{pmatrix} 300 & 400 \\ 800 & 850 \\ 500 & 620 \end{pmatrix}$
If we try to calculate $N + D$, the operation breaks down. The third row in $D$ has nothing to add to in matrix $N$. The dimensions are incompatible, so the operation is undefined.
IB Exam Tip
On an exam, if you’re asked to add or subtract matrices with different dimensions, don’t just leave it blank! Explicitly state that the operation is undefined and, crucially, explain why. A sentence like, “The operation is not possible because the matrices have incompatible dimensions ($2 \times 2$ and $3 \times 2$)” demonstrates clear understanding and will earn you marks.
Mission 2: The Inventory Audit (Subtraction)
Next up, an audit. You need to determine net change, like figuring out how many products were sold from an initial inventory. Matrix subtraction works almost exactly like addition, including its most important rule.
Rule for Matrix Subtraction
Just like addition, matrix subtraction is only possible if the two matrices have the exact same dimensions. No exceptions!
Example: Monthly Budgeting
Let’s track a personal budget. Matrix $I$ shows your income sources and Matrix $E$ shows your expenses. The rows are for Week 1 and Week 2; the columns are for ‘Job’/’Side Hustle’ income and ‘Rent’/’Groceries’ expenses.
$I = \begin{pmatrix} 800 & 150 \\ 800 & 210 \end{pmatrix}$ and $E = \begin{pmatrix} 600 & 90 \\ 600 & 110 \end{pmatrix}$
To find your net savings, we’ll calculate $S = I – E$.
Both are $2 \times 2$ matrices, so we can proceed. We subtract each corresponding element.
$S = \begin{pmatrix} 800-600 & 150-90 \\ 800-600 & 210-110 \end{pmatrix} = \begin{pmatrix} 200 & 60 \\ 200 & 100 \end{pmatrix}$
This matrix $S$ clearly shows your savings for each category over the two weeks. Mission successful!
IB Paper 2 GDC Hint
On Paper 2, your Graphic Display Calculator (GDC) is your best friend for matrix calculations. It saves a ton of time. But be warned: a tiny typo during data entry can lead to a completely wrong answer. Always, always, always do a quick visual scan of the matrix on your calculator’s screen to ensure you’ve typed it in correctly before hitting that calculate button. Accuracy is just as important as speed!
Mission 3: The Price Adjustment (Scalar Multiplication)
Our final mission is about scaling. Imagine a company needs to apply a 10% holiday discount to all its prices. Instead of changing each price one by one, we can use a single operation: scalar multiplication.
Rule for Scalar Multiplication
A ‘scalar’ is just a fancy word for a regular number (not a matrix). In scalar multiplication, you multiply an entire matrix by a single scalar value. Here’s the great news: there are no dimension restrictions for this operation! You can multiply any matrix, of any size, by a scalar.
The process feels a lot like the distributive property. You take the scalar and multiply it by every single element inside the matrix.
Example: Currency Conversion
You have a matrix, $P$, representing the prices of some tech gadgets in US Dollars ($). The rows represent different models, and the columns represent the cost for the basic and premium versions.
$P = \begin{pmatrix} 999 & 1199 \\ 499 & 649 \\ 1299 & 1599 \end{pmatrix}$
You need to convert these prices to Euros (€). Let’s say the conversion rate is $0.92$ EUR for every $1$ USD. Our scalar is $0.92$. We need to calculate the new price matrix, $E = 0.92 \times P$.
We just multiply every entry in $P$ by $0.92$.
$E = 0.92 \times \begin{pmatrix} 999 & 1199 \\ 499 & 649 \\ 1299 & 1599 \end{pmatrix} = \begin{pmatrix} 0.92 \times 999 & 0.92 \times 1199 \\ 0.92 \times 499 & 0.92 \times 649 \\ 0.92 \times 1299 & 0.92 \times 1599 \end{pmatrix}$
$E = \begin{pmatrix} 919.08 & 1103.08 \\ 459.08 & 597.08 \\ 1195.08 & 1471.08 \end{pmatrix}$
And just like that, all prices are converted. Mission complete!
Math IA Idea
Scalar multiplication is an amazing tool for your Math Internal Assessment (IA). Are you modeling something that scales uniformly? Maybe you’re applying a percentage-based tax to a list of prices, calculating compound interest across multiple accounts, or even scaling the vertices of a 2D shape represented by a coordinate matrix. Using scalar multiplication shows that you can apply mathematical concepts to real-world modeling scenarios elegantly—a key to a high-scoring IA.
Key Takeaways & IB Insights
Congratulations, agent! You’ve successfully completed your initial training. Let’s debrief the key intel:
- Addition & Subtraction: These operations are twins. They both require matrices to have the exact same dimensions. The calculation is done element by corresponding element.
- Scalar Multiplication: This is the lone wolf. You multiply a single number (the scalar) by every element in the matrix. It has no dimension restrictions.
- Communication is Key: In your IB exams, always state your reasoning. If an operation is impossible, say so and explain that the dimensions do not match.
Mastering these basics is your launchpad into the more complex and fascinating world of matrices, like multiplication and finding determinants and inverses. Keep practicing, and you’ll be an expert in no time.
Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.
Additional Practice
Question 1: Given $A = \begin{pmatrix} 5 & -2 \\ 3 & 7 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 10 \\ -4 & 0 \end{pmatrix}$, calculate $A + B$.
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Solution: $A + B = \begin{pmatrix} 5+1 & -2+10 \\ 3+(-4) & 7+0 \end{pmatrix} = \begin{pmatrix} 6 & 8 \\ -1 & 7 \end{pmatrix}$
Question 2: Given $C = \begin{pmatrix} 20 & 30 \\ 15 & 25 \end{pmatrix}$ and $D = \begin{pmatrix} 8 & 12 \\ 10 & 5 \end{pmatrix}$, calculate $C – D$.
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Solution: $C – D = \begin{pmatrix} 20-8 & 30-12 \\ 15-10 & 25-5 \end{pmatrix} = \begin{pmatrix} 12 & 18 \\ 5 & 20 \end{pmatrix}$
Question 3: Let $E = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$. Calculate $5E$.
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Solution: $5E = 5 \times \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} = \begin{pmatrix} 5 & 10 & 15 \\ 20 & 25 & 30 \end{pmatrix}$
Question 4: Given $F = \begin{pmatrix} 1 & 9 \\ 8 & 2 \end{pmatrix}$ and $G = \begin{pmatrix} 3 & 7 & 1 \\ 6 & 5 & 4 \end{pmatrix}$. Can you calculate $F + G$? Explain your answer.
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Solution: No, the operation is undefined. Matrix $F$ has dimensions $2 \times 2$, while matrix $G$ has dimensions $2 \times 3$. They must be identical for addition to be possible.
Question 5: Calculate $2A – B$ using the matrices from Question 1: $A = \begin{pmatrix} 5 & -2 \\ 3 & 7 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 10 \\ -4 & 0 \end{pmatrix}$.
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Solution: First, find $2A$: $2A = 2 \times \begin{pmatrix} 5 & -2 \\ 3 & 7 \end{pmatrix} = \begin{pmatrix} 10 & -4 \\ 6 & 14 \end{pmatrix}$.
Now subtract B: $\begin{pmatrix} 10 & -4 \\ 6 & 14 \end{pmatrix} – \begin{pmatrix} 1 & 10 \\ -4 & 0 \end{pmatrix} = \begin{pmatrix} 10-1 & -4-10 \\ 6-(-4) & 14-0 \end{pmatrix} = \begin{pmatrix} 9 & -14 \\ 10 & 14 \end{pmatrix}$.
Question 6: Given $H = \begin{pmatrix} 1.5 & 2.1 \\ -0.5 & 3.0 \end{pmatrix}$, calculate $10H$.
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Solution: $10H = 10 \times \begin{pmatrix} 1.5 & 2.1 \\ -0.5 & 3.0 \end{pmatrix} = \begin{pmatrix} 15 & 21 \\ -5 & 30 \end{pmatrix}$.
Question 7: Let $J = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $K = \begin{pmatrix} 5 & 5 \\ 5 & 5 \end{pmatrix}$. Find a scalar $x$ such that $xK – J$ results in a matrix with an element of $19$ in the top-left position.
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Solution: The expression is $x \begin{pmatrix} 5 & 5 \\ 5 & 5 \end{pmatrix} – \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5x-1 & 5x \\ 5x & 5x-1 \end{pmatrix}$. We need the top-left element, $5x-1$, to be equal to $19$.
$5x – 1 = 19 \implies 5x = 20 \implies x = 4$.
Question 8: Given $M = \begin{pmatrix} 4 \\ 8 \\ 12 \end{pmatrix}$ and $N = \begin{pmatrix} -3 \\ 6 \\ -9 \end{pmatrix}$. Calculate $\frac{1}{2}M + \frac{1}{3}N$.
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Solution: First, calculate the scalar products:
$\frac{1}{2}M = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$ and $\frac{1}{3}N = \begin{pmatrix} -1 \\ 2 \\ -3 \end{pmatrix}$.
Now add them: $\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} + \begin{pmatrix} -1 \\ 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 2-1 \\ 4+2 \\ 6-3 \end{pmatrix} = \begin{pmatrix} 1 \\ 6 \\ 3 \end{pmatrix}$.
Question 9: If $P + Q = P$, what can you conclude about matrix $Q$?
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Solution: For $P+Q=P$ to be true, every element in $Q$ must be zero. This means $Q$ is the zero matrix. It must also have the same dimensions as $P$.
IB Exam Style Question
A company sells two models of electric scooters, Model X and Model Y, in two cities, Amsterdam and Berlin. The sales for 2023 are given by matrix $S$. The rows represent the cities and the columns represent the scooter models.
- $S = \begin{pmatrix} 500 & 350 \\ 420 & 480 \end{pmatrix} \quad \begin{matrix} \text{Amsterdam} \\ \text{Berlin} \end{matrix}$
The company sets sales targets for 2024, given by matrix $T$.
$T = \begin{pmatrix} 600 & 450 \\ 500 & 550 \end{pmatrix}$
(a) The company projects that sales in 2024 will increase by 20% from 2023 levels across all models and cities. Calculate the projected sales matrix for 2024, let’s call it $P$.
(b) Calculate the matrix $D = T – P$.
(c) Explain what the elements of matrix $D$ represent.
Reveal Solution
(a) Solution:
A 20% increase means we need to multiply the 2023 sales by a scalar of 1.2.
$P = 1.2 \times S = 1.2 \times \begin{pmatrix} 500 & 350 \\ 420 & 480 \end{pmatrix}$
$P = \begin{pmatrix} 1.2 \times 500 & 1.2 \times 350 \\ 1.2 \times 420 & 1.2 \times 480 \end{pmatrix} = \begin{pmatrix} 600 & 420 \\ 504 & 576 \end{pmatrix}$
(b) Solution:
We need to subtract the projected sales ($P$) from the target sales ($T$).
$D = T – P = \begin{pmatrix} 600 & 450 \\ 500 & 550 \end{pmatrix} – \begin{pmatrix} 600 & 420 \\ 504 & 576 \end{pmatrix}$
$D = \begin{pmatrix} 600-600 & 450-420 \\ 500-504 & 550-576 \end{pmatrix} = \begin{pmatrix} 0 & 30 \\ -4 & -26 \end{pmatrix}$
(c) Solution:
The elements of matrix $D$ represent the difference between the sales targets and the projected sales. A positive number means the projection is below target, while a negative number means the projection exceeds the target.
- The element $d_{11}=0$ means the projected sales for Model X in Amsterdam exactly meet the target.
- The element $d_{12}=30$ means the projected sales for Model Y in Amsterdam are 30 units short of the target.
- The element $d_{21}=-4$ means the projected sales for Model X in Berlin are 4 units above the target.
- The element $d_{22}=-26$ means the projected sales for Model Y in Berlin are 26 units above the target.