Hey everyone! Welcome to the ultimate deep dive into one of the trickiest, yet most powerful, tools in your IB Math toolkit: matrix multiplication. If you’ve ever looked at two grids of numbers and wondered how on earth they combine, you’re in the right place. It’s not like regular multiplication where $3 \times 5$ is the same as $5 \times 3$. Matrices have their own set of rules, a secret handshake they must perform before they can interact. Think of this post as your field guide to cracking that code. We’re going to move from the basic ‘can we even do this?’ check to uncovering the fundamental properties that make matrices so essential in fields from computer graphics to economics.
Table of Contents
The First Handshake: Are Your Matrices Compatible?
Before you can even think about multiplying matrices, you have to play matchmaker. There’s one non-negotiable rule they must follow. It’s the first thing you should check, every single time.
Rule: The Compatibility Check
To multiply two matrices, $A$ and $B$, to get the product $AB$, the number of columns in matrix $A$ must be exactly equal to the number of rows in matrix $B$.
If $A$ has dimensions $(m \times n)$ and $B$ has dimensions $(p \times q)$, the product $AB$ is only possible if $n = p$.
Example: Inventory Check
A bakery tracks its inventory of flour, sugar, and eggs at its two locations in a matrix $L$. They track the cost of each ingredient from their supplier in matrix $C$. Is the product $LC$ possible?
$L = \begin{pmatrix} 50 & 25 & 200 \\ 40 & 30 & 180 \end{pmatrix}$ (2 locations $\times$ 3 ingredients)
$C = \begin{pmatrix} 2 \\ 3 \\ 0.5 \end{pmatrix}$ (3 ingredients $\times$ 1 cost)
Analysis:
Dimensions of $L$: $(2 \times 3)$
Dimensions of $C$: $(3 \times 1)$
The ‘inner’ dimensions are the columns of $L$ (3) and the rows of $C$ (3). Since $3 = 3$, the multiplication is possible!
IB Exam Tip
This check is your best friend in an exam. It takes five seconds and can save you minutes of painful, pointless calculation. If a question asks you to find $AB$, always check the dimensions first. If they don’t match, you can just write ‘undefined’ and move on, scoring full marks for that insight.
Crystal Ball Math: Predicting Your Product’s Dimensions
Once you’ve confirmed that two matrices can be multiplied, you can actually predict the exact shape and size of the resulting matrix without calculating a single number. This is another super-fast check that helps you know what your final answer should look like.
Rule: Predicting Product Dimensions
If you multiply an $(m \times n)$ matrix by an $(n \times p)$ matrix, the resulting matrix will have dimensions determined by the ‘outer’ numbers: $(m \times p)$.
Example: Data Transformation
Let’s say we have a matrix $S$ representing the scores of 5 students on 2 assignments. This is a $(5 \times 2)$ matrix. We also have a weighting matrix $W$ that transforms these 2 scores into 4 different final grade components (e.g., homework, participation, quiz, final exam contribution). This would be a $(2 \times 4)$ matrix. If we calculate the final components matrix $F = SW$, what will its dimensions be?
Analysis:
Dimensions of $S$: $(5 \times 2)$
Dimensions of $W$: $(2 \times 4)$
The inner dimensions (2 and 2) match, so it’s possible. The outer dimensions are 5 and 4. Therefore, the resulting matrix $F$ will be a $(5 \times 4)$ matrix, giving us the 4 grade components for each of the 5 students.
The Detective Work: Calculating Individual Elements
Okay, we know *if* we can multiply and we know the *shape* of the answer. But how do we actually find the numbers that go inside? This is where you become a detective. Each element in the answer matrix is the result of a specific investigation involving one row from the first matrix and one column from the second.
Rule: The Dot Product
To find the element in row $i$ and column $j$ of the product matrix $C = AB$, you calculate the dot product of row $i$ from matrix $A$ and column $j$ from matrix $B$.
This means you multiply the corresponding elements of the row and column, and then sum up all those products.
Example: Mission: Find $c_{2,1}$
Let’s say $A = \begin{pmatrix} 1 & 5 \\ -2 & 0 \\ 4 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} 6 & 10 \\ 3 & -7 \end{pmatrix}$. Our mission is to find the single element $c_{2,1}$ in the product matrix $C = AB$.
Investigation:
The address $c_{2,1}$ tells us we need row 2 from matrix $A$ and column 1 from matrix $B$.
Row 2 of A is $(-2, 0)$.
Column 1 of B is $\begin{pmatrix} 6 \\ 3 \end{pmatrix}$.
Now, we calculate the dot product:
$c_{2,1} = (-2 \times 6) + (0 \times 3)$
$c_{2,1} = -12 + 0$
$c_{2,1} = -12$
Mission accomplished. The element in the second row, first column of the answer is -12.
Paper 1 Pro Tip
When you’re doing this by hand in a no-calculator exam, physically trace the row with your left index finger and the column with your right index finger. This simple physical action helps your brain focus and dramatically cuts down on silly mistakes made under pressure. Trust me, it works!
The Rules of Engagement: Key Properties
Now that we have the mechanics down, let’s explore the fundamental laws that govern matrix multiplication. Some of these might surprise you, especially if you’re used to the comfortable, predictable world of regular numbers!
Commutativity: Does Order Matter? (Spoiler: YES.)
In regular algebra, $a \times b = b \times a$. This is called the commutative property. With matrices, this is almost never true. The order in which you multiply matrices is critically important.
Example: The Order Question
Let $A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 4 \\ 5 & -2 \end{pmatrix}$. Let’s calculate $AB$ and $BA$ to see if they match.
Calculating AB:
$AB = \begin{pmatrix} (2)(0)+(-1)(5) & (2)(4)+(-1)(-2) \\ (1)(0)+(3)(5) & (1)(4)+(3)(-2) \end{pmatrix} = \begin{pmatrix} -5 & 10 \\ 15 & -2 \end{pmatrix}$
Calculating BA:
$BA = \begin{pmatrix} (0)(2)+(4)(1) & (0)(-1)+(4)(3) \\ (5)(2)+(-2)(1) & (5)(-1)+(-2)(3) \end{pmatrix} = \begin{pmatrix} 4 & 12 \\ 8 & -11 \end{pmatrix}$
Conclusion: As you can see, $AB \neq BA$. Matrix multiplication is not commutative.
IA Idea Spark
The fact that matrix multiplication isn’t commutative is a fantastic starting point for a Math IA. You could explore its impact on geometric transformations (a rotation then a translation is different from a translation then a rotation) or investigate the special conditions under which $AB$ *does* equal $BA$.
Associativity: Can You Regroup? (Yes, you can!)
While order matters, the way you group multiplications in a chain doesn’t. This is the associative property, and thankfully, it works just like it does with regular numbers.
Rule: The Associative Property
For any three compatible matrices $A$, $B$, and $C$, it is always true that $(AB)C = A(BC)$. You can multiply $A$ and $B$ first, or you can multiply $B$ and $C$ first—the final answer will be the same.
Distributivity: Can You ‘Share’ the Multiplication? (Yes, again!)
Just like in algebra where $a(b+c) = ab+ac$, you can distribute matrix multiplication over matrix addition.
Rule: The Distributive Property
For any three compatible matrices $A$, $B$, and $C$, it is always true that $A(B+C) = AB + AC$. This is extremely useful for simplifying complex matrix expressions.
The Special Agents: Identity and Zero Matrices
In the world of matrices, there are two special agents that have unique and predictable effects on any matrix they interact with: the Identity matrix and the Zero matrix.
The Identity Matrix: The ‘Do Nothing’ Operator
The Identity Matrix, denoted as $I$, is the matrix equivalent of the number 1. Multiplying any matrix by a compatible identity matrix leaves the original matrix completely unchanged.
Definition: The Identity Matrix
An identity matrix $I$ is a square matrix (it has the same number of rows and columns) with 1s on the main diagonal (from top-left to bottom-right) and 0s everywhere else. For any matrix $A$, $AI = A$ and $IA = A$, provided the dimensions are compatible.
Examples: $I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, $I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
IB HL Connection
The identity matrix is the foundation for matrix inverses. The inverse of a matrix $A$, written as $A^{-1}$, is defined as the matrix that gives you the identity when multiplied by $A$: $AA^{-1} = A^{-1}A = I$. This is a huge concept for solving systems of linear equations!
The Zero Matrix: The Great Annihilator
The Zero Matrix, denoted as $O$, is the matrix equivalent of the number 0. Multiplying by a compatible zero matrix wipes out the original matrix, resulting in another zero matrix.
Example: The Zero Effect
Let $A = \begin{pmatrix} 10 & 20 \\ 30 & 40 \\ 50 & 60 \end{pmatrix}$ and the zero matrix $O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. What is $AO$?
Analysis:
The dimensions are $(3 \times 2)$ and $(2 \times 2)$, so the product is possible and will be a $(3 \times 2)$ matrix.
Let’s find the element in the first row, first column: $(10 \times 0) + (20 \times 0) = 0$.
You’ll quickly see that every dot product will result in 0 because you are always multiplying by and adding zeros. Therefore:
$AO = \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} = O_{3,2}$
Conclusion
And that’s the code, cracked! Matrix multiplication might seem intimidating with its peculiar rules, but it’s completely logical. Always start with the compatibility check, know how to predict your result’s dimensions, and master the dot product. Understanding the core properties like non-commutativity and the roles of the Identity and Zero matrices will elevate your skills from just calculating to truly understanding how matrices work. Keep practicing, and these operations will become second nature!
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 $P$, a $(4 \times 2)$ matrix, and $Q$, a $(2 \times 7)$ matrix. Is the product $QP$ possible? If so, what are the dimensions of the resulting matrix?
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Solution: The dimensions of $Q$ are $(2 \times 7)$ and the dimensions of $P$ are $(4 \times 2)$. To find $QP$, we check the inner dimensions: columns of $Q$ (7) and rows of $P$ (4). Since $7 \neq 4$, the product $QP$ is not possible (undefined).
Question 2: Matrix $A$ has dimensions $(1 \times 5)$ and matrix $B$ has dimensions $(5 \times 3)$. What are the dimensions of the product matrix $C = AB$?
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Solution: The inner dimensions are 5 and 5, so multiplication is possible. The dimensions of the product matrix are determined by the outer dimensions. The result will be a $(1 \times 3)$ matrix.
Question 3: Let $X = \begin{pmatrix} 3 & -1 & 0 \\ 2 & 5 & 1 \end{pmatrix}$ and $Y = \begin{pmatrix} 4 & 7 \\ -2 & 0 \\ 1 & 6 \end{pmatrix}$. If $Z = XY$, find the element $z_{1,2}$.
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Solution: To find $z_{1,2}$, we need the dot product of row 1 of $X$ and column 2 of $Y$.
Row 1 of $X$: $(3, -1, 0)$
Column 2 of $Y$: $\begin{pmatrix} 7 \\ 0 \\ 6 \end{pmatrix}$
$z_{1,2} = (3 \times 7) + (-1 \times 0) + (0 \times 6) = 21 + 0 + 0 = 21$.
Question 4: Given the matrices from Question 3, find the element $z_{2,1}$.
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Solution: To find $z_{2,1}$, we need the dot product of row 2 of $X$ and column 1 of $Y$.
Row 2 of $X$: $(2, 5, 1)$
Column 1 of $Y$: $\begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$
$z_{2,1} = (2 \times 4) + (5 \times -2) + (1 \times 1) = 8 – 10 + 1 = -1$.
Question 5: Calculate the full product $XY$ for the matrices in Question 3.
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Solution: We already know $z_{1,2}=21$ and $z_{2,1}=-1$. We need $z_{1,1}$ and $z_{2,2}$.
$z_{1,1}$ (Row 1 of X $\cdot$ Col 1 of Y): $(3)(4)+(-1)(-2)+(0)(1) = 12+2+0 = 14$.
$z_{2,2}$ (Row 2 of X $\cdot$ Col 2 of Y): $(2)(7)+(5)(0)+(1)(6) = 14+0+6 = 20$.
So, $Z = XY = \begin{pmatrix} 14 & 21 \\ -1 & 20 \end{pmatrix}$.
Question 6: Let $M = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$ and $N = \begin{pmatrix} 5 & 0 & -1 \end{pmatrix}$. Calculate the product $MN$.
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Solution: $M$ is $(3 \times 1)$ and $N$ is $(1 \times 3)$. The product $MN$ will be a $(3 \times 3)$ matrix.
$MN = \begin{pmatrix} (1)(5) & (1)(0) & (1)(-1) \\ (-3)(5) & (-3)(0) & (-3)(-1) \\ (2)(5) & (2)(0) & (2)(-1) \end{pmatrix} = \begin{pmatrix} 5 & 0 & -1 \\ -15 & 0 & 3 \\ 10 & 0 & -2 \end{pmatrix}$.
Question 7: Let $F = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $G = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$. Calculate both $FG$ and $GF$. Are they equal?
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Solution:
$FG = \begin{pmatrix} (1)(1)+(1)(1) & (1)(0)+(1)(1) \\ (0)(1)+(1)(1) & (0)(0)+(1)(1) \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$.
$GF = \begin{pmatrix} (1)(1)+(0)(0) & (1)(1)+(0)(1) \\ (1)(1)+(1)(0) & (1)(1)+(1)(1) \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$.
No, $FG \neq GF$.
Question 8: Let $A = \begin{pmatrix} 2 & 0 \\ 1 & -1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $C = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$. Verify that $A(B+C) = AB + AC$.
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Solution:
Left Side: First, $B+C = \begin{pmatrix} 1+3 & 1+0 \\ 1+0 & 1+2 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 1 & 3 \end{pmatrix}$.
Then, $A(B+C) = \begin{pmatrix} 2 & 0 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 4 & 1 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 8 & 2 \\ 3 & -2 \end{pmatrix}$.
Right Side: First, $AB = \begin{pmatrix} 2 & 2 \\ 0 & 0 \end{pmatrix}$ and $AC = \begin{pmatrix} 6 & 0 \\ 3 & -2 \end{pmatrix}$.
Then, $AB+AC = \begin{pmatrix} 2+6 & 2+0 \\ 0+3 & 0-2 \end{pmatrix} = \begin{pmatrix} 8 & 2 \\ 3 & -2 \end{pmatrix}$.
The left side equals the right side, verifying the distributive property.
Question 9: Let $M = \begin{pmatrix} 3 & -2 & 1 \\ 0 & 5 & 4 \end{pmatrix}$. Multiply $M$ by the appropriate identity matrix $I$ such that $MI=M$. What is $I$?
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Solution: $M$ is a $(2 \times 3)$ matrix. For $MI$ to be defined, $I$ must have 3 rows. For the result to be a $(2 \times 3)$ matrix, $I$ must have 3 columns. Therefore, $I$ must be the $(3 \times 3)$ identity matrix, $I_3$.
$I = I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.
IB Exam Style Question
Let the matrices be defined as $A = \begin{pmatrix} k & 2 \\ 1 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & -2 \\ 0 & 1 \end{pmatrix}$.
- (a) Find the product $AB$ in terms of $k$.
(b) Given that $(AB) = \begin{pmatrix} 9 & -4 \\ 3 & -3 \end{pmatrix}$, find the value of $k$.
Reveal Solution
Solution:
(a) Find the product AB.
We perform the dot product for each element:
$AB = \begin{pmatrix} (k)(3)+(2)(0) & (k)(-2)+(2)(1) \\ (1)(3)+(-1)(0) & (1)(-2)+(-1)(1) \end{pmatrix}$
$AB = \begin{pmatrix} 3k & -2k+2 \\ 3 & -3 \end{pmatrix}$
(b) Find the value of k.
We are given that $AB = \begin{pmatrix} 9 & -4 \\ 3 & -3 \end{pmatrix}$. We can equate the corresponding elements of our calculated matrix with the given matrix.
From the top-left element (position 1,1):
$3k = 9$
$k = 3$
We can verify this using the top-right element (position 1,2):
$-2k+2 = -4$
$-2k = -6$
$k = 3$
Both elements give the same result. Therefore, the value of $k$ is 3.