When a matrix product is defined and what its dimensions are, the row-by-column recipe for each element, computing a full product, why matrix multiplication is not commutative, and modelling a cost calculation as a matrix product.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 3 — Linear Systems and Matrices
Multiply Matrices
Objectives
Five outcomes. The first decides whether the rest is even possible.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 195-201 — the lesson these objectives are drawn from
Warm-up
Lesson 3.5 added matrices entry by entry, and every operation preserved the dimensions. Multiplication does neither.
Discussion prompt
You have a table of quantities — three products bought by two shops — and a list of three prices. How would you work out each shop's total bill, and how many numbers does the answer have?
Hint: Do one shop first and count the multiplications.
Answer:
For each shop you multiply each quantity by its price and add the three results. That is three multiplications and two additions per shop, giving two totals from six quantities and three prices.
Notice what happened to the shape: a 2 by 3 table and a 3 by 1 list produced a 2 by 1 answer. Matrix multiplication is that calculation, given a name and a rule.
Concept
The element in row i, column j of a product comes from row i of the first matrix and column j of the second: multiply corresponding entries and add. Everything else in this lesson follows from that one sentence.
matrix product — The matrix AB, defined when the number of columns of A equals the number of rows of B. Each element pairs a row of A with a column of B.
\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix} \]
Because a row must pair off against a column, the row and the column have to be the same length — which is the dimension condition, arrived at rather than memorised.
Figure (svg): One row of the first matrix paired with one column of the second, multiplied term by term and summed
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 195-195 — Multiplying Matrices
Section
Section 1
Concept
Write the two sets of dimensions side by side. The product exists only if the inner two numbers are equal, and when it does exist the outer two numbers are its dimensions.
\[ (m \times n)(n \times p) = m \times p \]
This is not an arbitrary rule. A row of A has n entries and a column of B has as many entries as B has rows, so they can only pair off if those counts agree.
Figure (svg): The dimension rule: the inner numbers must match and the outer numbers give the size of the product
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 195-195 — Describe matrix products
Picture it
Write the dimensions next to each other and read the inner and outer pairs.
Figure (svg): The dimension rule: the inner numbers must match and the outer numbers give the size of the product
A 4 by 3 times a 3 by 2 works and gives a 4 by 2. A 3 by 4 times a 3 by 2 does not, because 4 and 3 disagree — and no amount of care in the arithmetic could rescue it.
Worked example
Example 1, both parts.
\[ \text{Is } AB \text{ defined, and what are its dimensions? (a) } A: 4 \times 3, \, B: 3 \times 2. \text{ (b) } A: 3 \times 4, \, B: 3 \times 2. \]
Write the first pair of dimensions side by side
Why: Four by three, then three by two.
\[ 4 x 3\text{ and } 3 x 2 \]
Compare the inner numbers
Why: Three and three — equal, so the product is defined.
\[ 3 = 3,\text{ defined} \]
Read the outer numbers for the answer's shape
Why: Four and two.
\[ AB\text{ is } 4 x 2 \]
Check the second pair
Why: Three by four, then three by two: the inner numbers are 4 and 3, which differ.
Figure (svg): The solution to Worked example two dimension checks shown as a ladder of expressions, one row per algebraic move
\[ \text{(a) } 4 \times 2 \qquad \text{(b) undefined} \]
Verify: say why the second one fails in terms of rows and columns
Why: In the second case a row of A has four entries but a column of B has only three, so there is nothing for the fourth entry of the row to pair with. The dimension rule is that mismatch, stated in shorthand.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 195-195
Sorting
Compare the inner numbers.
Sort into buckets
Sort each product AB by whether it exists.
Notice that the two undefined cases would become defined if either matrix were transposed. Order and shape both matter.
Worked example
Guided Practice 1 and 2.
\[ \text{(1) } A: 5 \times 2, \, B: 2 \times 2. \text{ (2) } A: 3 \times 2, \, B: 3 \times 2. \]
Check the first pair's inner numbers
Why: Two and two, so the product exists.
Read its dimensions
Why: The outer numbers are 5 and 2.
\[ AB\text{ is } 5 x 2 \]
Check the second pair
Why: The inner numbers are 2 and 3, which differ.
Notice what the second pair CAN do
Why: Two matrices of identical dimensions can be added, even when they cannot be multiplied. The two conditions are different.
Figure (svg): The solution to Worked example the guided practice pair shown as a ladder of expressions, one row per algebraic move
\[ \text{(1) } 5 \times 2 \qquad \text{(2) undefined} \]
Verify: contrast the two conditions
Why: Addition needs both dimensions to match; multiplication needs only the columns of the first to match the rows of the second. So a 3 by 2 and a 3 by 2 can be added but not multiplied, while a 4 by 3 and a 3 by 2 can be multiplied but not added. The two operations have almost opposite requirements.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 195-195
Trap
\[ A: 3 \times 2, \quad B: 3 \times 2 \]
Say AB is defined because the dimensions are identical
Why: The rule for addition is carried over to multiplication.
Identical dimensions is exactly what ADDITION needs. Multiplication needs the columns of A to match the rows of B, which here is 2 against 3.
\[ A: 3 \times 2, \quad B: 3 \times 2 \;\Longrightarrow\; AB \text{ undefined} \]
Compare the columns of A with the rows of B
Why: Two columns against three rows: they do not match, so no row of A can pair with a column of B.
These two matrices can be added, and they can be multiplied in neither order. Same dimensions is a hindrance to multiplication, not a help.
Fill the middle
A is 2 by 3 and B is 3 by 5.
Fill in the blanks
(2 \times 3)(3 \times 5) = 2 \times 5
Why: The inner numbers are both 3, so the product is defined, and the outer numbers 2 and 5 give its dimensions. Note that the product has 10 entries while A has 6 and B has 15 — the shape of the answer matches neither input, which never happens with addition.
Prediction
Commit before reasoning.
Predict first
A is 4 by 3 and B is 3 by 2. Which products are defined?
Correct: AB only.
\[ AB: (4 \times 3)(3 \times 2) = 4 \times 2 \qquad BA: (3 \times 2)(4 \times 3) \text{ undefined} \]
Why: For AB the inner numbers are 3 and 3, which match. For BA the dimensions read 3 by 2 then 4 by 3, so the inner numbers are 2 and 4, which do not. So one order works and the other does not even exist — a situation with no parallel in ordinary arithmetic, where ab and ba are both always defined.
Comparison
Fill the blanks. The two operations demand almost opposite things.
Comparison matrix
| Feature | A + B | AB |
|---|---|---|
| Condition | both dimensions match | columns of A equal rows of B |
| Dimensions of the result | same as the inputs | rows of A by columns of B |
| How each entry is built | one sum of two numbers | a sum of several products |
| Order matters? | no | yes |
Every row differs. Addition and multiplication of matrices share a notation and almost nothing else.
Section
Section 2
Concept
To find the element in row i and column j of AB, take row i of A and column j of B, multiply corresponding entries, and add the products. The position of the answer is named by which row and which column you used.
\[ (AB)_{11} = 1(5) + 4(9) = 41 \]
Row of the FIRST matrix, column of the SECOND. Getting that round the wrong way is what produces BA instead of AB.
Figure (svg): One row of the first matrix paired with one column of the second, multiplied term by term and summed
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 195-196 — Multiplying Matrices
Picture it
Example 2, step one: the top-left entry of the product.
Figure (svg): One row of the first matrix paired with one column of the second, multiplied term by term and summed
The highlighted row and the highlighted column are both of length two, which is exactly what the dimension condition guaranteed. The answer sits where they cross.
Worked example
Example 2, step one, in full.
\[ \text{For } A = \begin{bmatrix} 1 & 4 \\ 3 & -2 \end{bmatrix}, \; B = \begin{bmatrix} 5 & -7 \\ 9 & 6 \end{bmatrix}, \text{ find the entry in row 1, column 1 of } AB. \]
Take row 1 of A
Why: The entries 1 and 4.
\[ \text{row } 1\text{ of } A: 1, 4 \]
Take column 1 of B
Why: The entries 5 and 9, reading downward.
\[ \text{col } 1\text{ of } B: 5, 9 \]
Multiply corresponding entries
Why: One times five, and four times nine.
\[ 5\text{ and } 36 \]
Add the products
Why: Five plus thirty-six.
\[ 41 \]
Figure (svg): The solution to Worked example compute one entry shown as a ladder of expressions, one row per algebraic move
\[ (AB)_{11} = 1(5) + 4(9) = 41 \]
Verify: check the lengths matched
Why: Row 1 of A had two entries and column 1 of B had two entries, so every term found a partner and none was left over. Had they been different lengths the sum would have been impossible, which is the dimension condition doing its work at the level of a single entry.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 196-196
Fill the middle
Example 2, row 1 column 2.
Fill in the blanks
1(-7) + 4(6) = 17
Why: One times negative seven is negative seven, and four times six is twenty-four; their sum is seventeen. This entry used row 1 of A with column 2 of B, so it belongs in row 1, column 2 of the product — the same row as the previous entry and one column across.
Worked example
Example 2, step three. A different row, so a different position.
\[ \text{Find the entry in row 2, column 1 of } AB \text{ for the same } A \text{ and } B. \]
Take row 2 of A
Why: The entries 3 and negative 2.
\[ \text{row } 2\text{ of } A: 3, -2 \]
Take column 1 of B again
Why: The same column as before: 5 and 9.
\[ \text{col } 1\text{ of } B: 5, 9 \]
Multiply corresponding entries
Why: Three times five is fifteen; negative two times nine is negative eighteen.
\[ 15\text{ and } -18 \]
Add
Why: Fifteen plus negative eighteen.
\[ -3 \]
Figure (svg): The solution to Worked example a different entry shown as a ladder of expressions, one row per algebraic move
\[ (AB)_{21} = 3(5) + (-2)(9) = -3 \]
Verify: confirm the position of the answer
Why: Row 2 of A and column 1 of B were used, so the result belongs in row 2, column 1 of the product. Using the same column with a different row moves the answer down a row and never across, which is a useful check when you are working systematically.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 196-196
Trap
\[ \begin{bmatrix} 1 & 4 \\ 3 & -2 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ 9 & 6 \end{bmatrix} \]
Multiply the numbers in matching positions, as for addition
Why: The element-by-element rule from Lesson 3.5 is carried over.
\[ \begin{bmatrix} 5 & -28 \\ 27 & -12 \end{bmatrix} \quad \text{(wrong)} \]
Nothing was summed and no row met any column, so this is not the matrix product at all.
\[ \begin{bmatrix} 1 & 4 \\ 3 & -2 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ 9 & 6 \end{bmatrix} = \begin{bmatrix} 41 & 17 \\ -3 & -33 \end{bmatrix} \]
Pair each row of the first with each column of the second, and SUM the products
Why: Every entry of the answer is a sum, not a single product.
The element-wise product exists and has a use in other subjects, but it is not what matrix multiplication means. The row-and-column rule is what makes matrices able to represent systems of equations.
Matching
For the product AB of two 2 by 2 matrices.
Match the pairs
Why: The position of an entry names its ingredients exactly: the row number picks the row of A and the column number picks the column of B. That is why the rule is easy to apply systematically — you never have to decide which pieces to use, only to work through the positions in order.
Prediction
Commit before counting.
Predict first
Computing the product of a 2 by 3 matrix and a 3 by 4 matrix requires how many individual multiplications?
Correct: 24 — eight entries, each needing three multiplications.
\[ (2 \times 4) \text{ entries} \times 3 \text{ products each} = 24 \]
Why: The product is 2 by 4, so it has eight entries. Each pairs a row of length three with a column of length three, needing three multiplications and two additions. Eight times three is 24. Counting this before starting tells you how long the calculation will take, and it explains why matrix multiplication by hand is reserved for small matrices.
Explain it to yourself
The dimension condition, explained from the recipe.
\[ (AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots \]
Discussion prompt
Explain why the number of columns of A must equal the number of rows of B, using the way a single entry is computed rather than quoting the rule.
Hint: Ask what pairs up with what.
Answer:
An entry pairs the first item of the row with the first item of the column, the second with the second, and so on. If the row has more items than the column, the extra ones have nothing to multiply.
A row of A has as many entries as A has COLUMNS, and a column of B has as many as B has ROWS. So those two counts must agree, which is exactly the condition. The rule is a consequence of the recipe rather than an extra requirement.
Section
Section 3
Concept
Compute each entry of the product in turn, moving along the first row and then the second. Being systematic is what prevents the commonest error, which is using the wrong row or column somewhere in the middle.
\[ \begin{bmatrix} 1 & 4 \\ 3 & -2 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ 9 & 6 \end{bmatrix} = \begin{bmatrix} 41 & 17 \\ -3 & -33 \end{bmatrix} \]
Write the position next to each calculation as you go. It costs nothing and it makes a misplaced entry obvious.
Figure (svg): All four entries of a two by two product, each labelled with the row and column that produced it
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 196-196 — Find the product of two matrices
Picture it
Example 2 in full.
Figure (svg): All four entries of a two by two product, each labelled with the row and column that produced it
The first two entries share a row of A and the last two share the other row. Working across a row at a time means you only change one of the two ingredients at each step.
Worked example
Example 2, all five steps.
\[ \text{Find } AB \text{ for } A = \begin{bmatrix} 1 & 4 \\ 3 & -2 \end{bmatrix}, \; B = \begin{bmatrix} 5 & -7 \\ 9 & 6 \end{bmatrix}. \]
Check the dimensions and note the answer's shape
Why: Two by two times two by two, so the product exists and is two by two.
\[ AB\text{ is } 2 x 2 \]
Row 1 with column 1
Why: One times five plus four times nine.
\[ 5 + 36 = 41 \]
Row 1 with column 2
Why: One times negative seven plus four times six.
\[ -7 + 24 = 17 \]
Row 2 with column 1
Why: Three times five plus negative two times nine.
\[ 15 - 18 = -3 \]
Row 2 with column 2
Why: Three times negative seven plus negative two times six.
\[ -21 - 12 = -33 \]
Figure (svg): The solution to Worked example the full product shown as a ladder of expressions, one row per algebraic move
\[ AB = \begin{bmatrix} 41 & 17 \\ -3 & -33 \end{bmatrix} \]
Verify: recompute one entry from scratch
Why: The bottom-right entry: row 2 of A is 3 and negative 2; column 2 of B is negative 7 and 6. Three times negative seven is negative twenty-one, and negative two times six is negative twelve, giving negative thirty-three. Choosing a different entry to re-derive, rather than re-reading the working, is what makes this a real check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 196-196
Ranking
Computing a matrix product.
Put in order
Why: The dimension check comes first because a product that does not exist is not worth starting. Knowing the answer's shape before computing means you know how many entries to produce, which catches a missed one. And recomputing a single entry independently is a far better check than re-reading, because it uses different arithmetic to reach the same number.
Worked example
A 2 by 3 times a 3 by 2, where the answer's shape differs from both inputs.
\[ \begin{bmatrix} 2 & 0 & 1 \\ -1 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & -1 \\ 3 & 1 \end{bmatrix} \]
Check the dimensions
Why: The inner numbers are both 3, so the product exists and is 2 by 2.
\[ AB\text{ is } 2 x 2 \]
Row 1 with column 1
Why: Two times one, plus zero times zero, plus one times three.
\[ 2 + 0 + 3 = 5 \]
Row 1 with column 2
Why: Two times two, plus zero times negative one, plus one times one.
\[ 4 + 0 + 1 = 5 \]
Row 2 with column 1
Why: Negative one times one, plus three times zero, plus four times three.
\[ -1 + 0 + 12 = 11 \]
Row 2 with column 2
Why: Negative one times two, plus three times negative one, plus four times one.
\[ -2 - 3 + 4 = -1 \]
Figure (svg): The solution to Worked example a non-square product shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 5 & 5 \\ 11 & -1 \end{bmatrix} \]
Verify: check the shape against the dimension rule
Why: A 2 by 3 times a 3 by 2 should give a 2 by 2, and the answer has two rows and two columns. Note that neither input was 2 by 2 — the product's shape came from the outer numbers, not from either matrix, which is unlike anything in Lesson 3.5.
Error analysis
A student computes the top-left entry of a product using the wrong pieces.
Annotate
On: \( (AB)_{11} = 1(5) + 3(9) = 32 \quad \text{for } A = \begin{bmatrix} 1 & 4 \\ 3 & -2 \end{bmatrix}, \; B = \begin{bmatrix} 5 & -7 \\ 9 & 6 \end{bmatrix} \)
Row of the first, column of the second. Saying it aloud each time is slower for a week and reliable thereafter.
Fill the middle
The 2 by 3 times 3 by 2 product from the worked example.
Fill in the blanks
(-1)(1) + 3(0) + 4(3) = 11
Why: Negative one plus zero plus twelve is eleven. The zero term is worth noticing: an entry of zero in either the row or the column contributes nothing, which is why matrices with many zeros are so much faster to multiply — and why the identity matrix of Lesson 3.8 behaves the way it does.
Discrimination
For a product AB where A is 3 by 2 and B is 2 by 4.
Sort into buckets
Sort each description by whether it correctly names the ingredients of the entry in row 3, column 2 of AB.
Prediction
Commit before computing.
Predict first
A is 3 by 2 and B is 2 by 4. What are the dimensions of AB?
Correct: 3 by 4 — the outer numbers.
\[ (3 \times 2)(2 \times 4) = 3 \times 4 \]
Why: The inner numbers are both 2, so the product exists, and the outer numbers 3 and 4 give the dimensions. The answer has twelve entries where A had six and B had eight, which is a good illustration of how differently multiplication treats shape compared with addition — a sum would have had to have exactly the same shape as both inputs.
Section
Section 4
Concept
For matrices, AB and BA are generally different matrices — and sometimes one is defined while the other is not. Matrix multiplication is not commutative, which is the sharpest break from ordinary arithmetic in this chapter.
\[ AB = \begin{bmatrix} 41 & 17 \\ -3 & -33 \end{bmatrix} \quad BA = \begin{bmatrix} -16 & 34 \\ 27 & 24 \end{bmatrix} \]
The reason is visible in the recipe: AB pairs rows of A with columns of B, and BA pairs rows of B with columns of A. Those are different pairings, so different numbers.
Figure (svg): Two columns showing that AB and BA give different matrices for the same A and B
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 196-196 — the note that BA is not the same as AB
Picture it
The A and B of Example 2, multiplied in both orders.
Figure (svg): Two columns showing that AB and BA give different matrices for the same A and B
Not one entry agrees. This is not a near miss caused by rounding — the two products are entirely different matrices.
Worked example
The same matrices as Example 2, in the other order.
\[ \text{Find } BA \text{ for } A = \begin{bmatrix} 1 & 4 \\ 3 & -2 \end{bmatrix}, \; B = \begin{bmatrix} 5 & -7 \\ 9 & 6 \end{bmatrix}. \]
Row 1 of B with column 1 of A
Why: Five times one plus negative seven times three.
\[ 5 - 21 = -16 \]
Row 1 of B with column 2 of A
Why: Five times four plus negative seven times negative two.
\[ 20 + 14 = 34 \]
Row 2 of B with column 1 of A
Why: Nine times one plus six times three.
\[ 9 + 18 = 27 \]
Row 2 of B with column 2 of A
Why: Nine times four plus six times negative two.
\[ 36 - 12 = 24 \]
Figure (svg): The solution to Worked example compute BA shown as a ladder of expressions, one row per algebraic move
\[ BA = \begin{bmatrix} -16 & 34 \\ 27 & 24 \end{bmatrix} \]
Verify: compare with AB entry by entry
Why: AB was 41, 17, negative 3, negative 33; BA is negative 16, 34, 27, 24. Not one of the four positions agrees, and even the signs differ in three of them. The two products are different matrices, which is the point of computing both.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 196-196
Two truths and a lie
All three are about matrix multiplication.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: C
Why: The survivor is the false one. Example 2's matrices are both 2 by 2, so both products are defined and both are 2 by 2 — and they still differ in every entry. Same size is not enough; matrix multiplication is simply not commutative, with a few special exceptions such as multiplying by the identity.
Worked example
A stronger failure of commutativity than merely giving different answers.
\[ \text{For } A: 2 \times 3 \text{ and } B: 3 \times 1, \text{ which of } AB \text{ and } BA \text{ exist?} \]
Check AB
Why: Two by three then three by one: inner numbers 3 and 3, which match.
\[ AB\text{ is } 2 x 1 \]
Check BA
Why: Three by one then two by three: inner numbers 1 and 2, which do not match.
State what this shows
Why: Not only can the two orders give different answers — one of them may not be a legal expression at all.
Compare with numbers
Why: For real numbers ab and ba are both always defined and always equal. Neither holds for matrices.
Figure (svg): The solution to Worked example one order defined, the other not shown as a ladder of expressions, one row per algebraic move
\[ AB: 2 \times 1 \qquad BA: \text{undefined} \]
Verify: say when both orders ARE defined
Why: Both exist only when A is m by n and B is n by m, so that the inner numbers match in each order. Even then AB is m by m and BA is n by n, so unless m equals n the two products are not even the same size — let alone equal.
Error analysis
A student simplifies a matrix expression by reordering a product.
Annotate
On: \( ABC = ACB \quad \text{since multiplication is commutative} \)
Check which properties survive rather than assuming them all do. Associativity transfers to matrices; commutativity does not.
Counterexample
A classmate insists that squaring works the same way for matrices.
\[ (A + B)^2 = A^2 + 2AB + B^2 \]
Discussion prompt
Expand the left side using the distributive property but WITHOUT assuming commutativity, and say where the familiar formula goes wrong.
Hint: Expand carefully and look at the two middle terms.
Answer:
\[ (A+B)(A+B) = A^2 + AB + BA + B^2 \]
The two middle terms are AB and BA, and they cannot be combined into 2AB unless AB happens to equal BA — which for matrices it generally does not.
So the correct expansion keeps both terms separately. Every identity from ordinary algebra that relies on reordering a product has to be re-derived for matrices, and many of them fail.
Sorting
Some of Lesson 1.1's properties transfer to matrix multiplication and some do not.
Sort into buckets
Sort each property.
One property out of five. That is why matrix algebra looks so familiar until the moment you try to reorder a product.
Commit first
Answer, then rate your confidence honestly.
Predict first
Are there any pairs of matrices for which AB does equal BA?
Correct: Yes, in special cases — multiplying by the identity is the standard one.
\[ AI = IA = A \quad \text{for the identity } I \]
Lesson 3.8 makes the identity matrix central, and the fact that it commutes with everything is exactly what makes it behave like the number 1.
Why: Not commutative means it does not hold in general, not that it never holds. The identity matrix commutes with every square matrix of the right size, and so does any scalar multiple of the identity. A matrix also commutes with itself and with its own powers. What is false is the general rule, and relying on it in a particular case requires checking that case.
Section
Section 5
Concept
A matrix of quantities multiplied by a column of prices gives one total per row. The row-times-column rule is exactly the arithmetic a cost calculation already performs, which is what makes matrix multiplication useful rather than merely defined.
\[ \begin{bmatrix} 12 & 45 & 15 \\ 15 & 38 & 17 \end{bmatrix}\begin{bmatrix} 35 \\ 4 \\ 30 \end{bmatrix} = \begin{bmatrix} 1050 \\ 1112 \end{bmatrix} \]
The dimensions carry the meaning: teams by items, times items by one, gives teams by one — one total per team. Checking the dimensions is checking that the model makes sense.
Figure (svg): A quantity matrix multiplied by a cost column, giving one total per team
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 195-195 — in the shape of the textbook's Example 4; the numbers here are built for this deck
Picture it
Two teams, three items, one price list.
Figure (svg): A quantity matrix multiplied by a cost column, giving one total per team
Each row of quantities meets the single column of prices, and the answer has one entry per team. Scalar multiplication could not do this, because the three items have different prices.
Worked example
A quantity matrix times a price column. The numbers here are built for this deck.
\[ \text{Team A buys 12 bats, 45 balls, 15 gloves; team B buys 15, 38, 17. Prices are } \$35, \$4, \$30. \]
Arrange the quantities as a matrix
Why: Teams down the side, items across the top: a 2 by 3 matrix.
\[ 2 x 3\text{ quantities} \]
Arrange the prices as a column
Why: One price per item, in the SAME order as the columns of the quantity matrix.
\[ 3 x 1\text{ prices} \]
Check the dimensions
Why: Two by three times three by one: inner numbers match, and the product is 2 by 1 — one total per team.
\[ \text{product is } 2 x 1 \]
Compute team A's total
Why: Twelve times 35, plus 45 times 4, plus 15 times 30.
\[ 420 + 180 + 450 = 1050 \]
Compute team B's total
Why: Fifteen times 35, plus 38 times 4, plus 17 times 30.
\[ 525 + 152 + 510 = 1187 \]
Figure (svg): The solution to Worked example total equipment cost shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 1050 \\ 1187 \end{bmatrix} \]
Verify: check the units and one total by hand
Why: Items times dollars per item gives dollars, so the answer's entries really are costs. Team A: 12 bats at 35 is 420, 45 balls at 4 is 180, 15 gloves at 30 is 450, and 420 plus 180 plus 450 is 1050. The matrix product is doing exactly the arithmetic anyone would do by hand, all at once.
Real world
Three cafes each sell coffee, tea and cake. Yesterday cafe 1 sold 80, 40 and 25; cafe 2 sold 65, 55 and 30; cafe 3 sold 90, 30 and 20. Prices are 3, 2 and 4 dollars.
Discussion prompt
Set this up as a matrix product, state the dimensions at each stage, and compute the revenue for cafe 2. What would you change to compare two different price lists?
Hint: Cafes down the side, items across.
Answer:
\[ (3 \times 3)(3 \times 1) = 3 \times 1 \]
\[ \text{cafe 2}: \; 65(3) + 55(2) + 30(4) = 195 + 110 + 120 = 425 \]
To compare two price lists, widen the price matrix to 3 by 2 — items down, price lists across. The product becomes 3 by 3 times 3 by 2, which is 3 by 2: one revenue figure per cafe per price list, six answers from one multiplication.
Worked example
Widening the price matrix to compare two suppliers.
\[ \text{The same quantities, with a second supplier charging } \$32, \$5, \$28. \]
Put both price lists in one matrix
Why: Three items down the side, two suppliers across: a 3 by 2 matrix.
\[ 3 x 2\text{ prices} \]
Check the dimensions
Why: Two by three times three by two gives a 2 by 2 product: two teams by two suppliers.
\[ \text{product is } 2 x 2 \]
Compute team A with supplier 2
Why: Twelve times 32, plus 45 times 5, plus 15 times 28.
\[ 384 + 225 + 420 = 1029 \]
Read what the 2 by 2 answer means
Why: Each entry is one team's total from one supplier, so the matrix answers four questions at once.
Figure (svg): The solution to Worked example two price lists at once shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 1050 & 1029 \\ 1187 & 1146 \end{bmatrix} \]
Verify: check that supplier 2 is cheaper for both teams
Why: Team A: 1029 against 1050, so supplier 2 is 21 dollars cheaper. Team B: 1146 against 1187, cheaper by 41 dollars. Supplier 2 charges less for bats and gloves and more for balls, so which is cheaper could in principle have gone either way — the matrix answers it for both teams in one calculation.
Error analysis
A student multiplies quantities by prices but arranges the price column carelessly.
Annotate
On: \( \begin{bmatrix} 12 & 45 & 15 \\ 15 & 38 & 17 \end{bmatrix}\begin{bmatrix} 4 \\ 30 \\ 35 \end{bmatrix} \)
A matrix carries no labels, so nothing warns you when the orders disagree. Write the labels beside the matrices while setting the problem up, then take them away.
Matching
In a quantity-times-price model.
Match the pairs
Why: The inner dimension — the items — is what gets summed over and therefore disappears from the answer. The outer dimensions, buyers and suppliers, survive into the product's shape. That is a general pattern: the thing you multiply and add over vanishes, and the two things you are cross-tabulating remain.
Prediction
Commit before reasoning.
Predict first
Four shops, six products, and two different price lists. What are the dimensions of the revenue matrix?
Correct: 4 by 2 — shops by price lists.
\[ (4 \times 6)(6 \times 2) = 4 \times 2 \]
Why: The quantity matrix is 4 by 6, shops by products, and the price matrix is 6 by 2, products by price lists. The inner sixes match and vanish, leaving 4 by 2: one revenue figure for each shop under each price list. Predicting the shape before computing tells you how many answers to expect, which is the fastest check that the model was set up the right way round.
Socratic
One question, and nothing else on this slide.
\[ \begin{bmatrix} 12 & 45 & 15 \\ 15 & 38 & 17 \end{bmatrix}\begin{bmatrix} 35 \\ 4 \\ 30 \end{bmatrix} \]
Discussion prompt
Lesson 3.5's scalar multiplication also converts quantities into money. Why can it not do this particular calculation, and what does that tell you about what matrix multiplication adds?
Hint: Ask what a scalar would have to assume about the three prices.
Answer:
A scalar multiplies every entry by the SAME number, so it can only handle the case where all three items cost the same. Here bats cost 35 and balls cost 4, so no single multiplier works.
Matrix multiplication allows a different multiplier for each column and then SUMS across the row, which is what turns a table of quantities into a total rather than into another table of the same shape.
That summing is the real addition. Scalar multiplication rescales; matrix multiplication combines and collapses, which is why the answer's shape differs from the input's.
Comparison
Fill the blanks. This operation breaks every pattern from Lesson 3.5.
Comparison matrix
| Feature | Addition (3.5) | Multiplication (3.6) |
|---|---|---|
| Condition | identical dimensions | columns of A equal rows of B |
| Shape of the answer | same as the inputs | rows of A by columns of B |
| Each entry is | one sum | a sum of several products |
| Commutative | yes | no |
| Associative | yes | yes |
Four rows differ and one agrees. The surviving associativity is what makes long matrix expressions manageable at all.
Pattern
One routine computes any matrix product.
Step three's empty grid is worth drawing. It tells you how many entries to produce and stops a missed one from going unnoticed.
OpenStax Algebra and Trigonometry 2e, §11.5 Matrices and Matrix Operations §11.5
Check
The dimension rule.
Check your understanding
A is 3 by 4 and B is 4 by 2. Is AB defined, and if so what are its dimensions?
Answer: A
Why: The inner numbers are 4 and 4, which match, so the product exists. The outer numbers 3 and 2 give its dimensions.
Check
One entry of a product. Row of the first, column of the second.
Check your understanding
For A = [[1,4],[3,-2]] and B = [[5,-7],[9,6]], what is the entry in row 2, column 2 of AB?
Answer: A
Why: Row 2 of A is 3 and -2; column 2 of B is -7 and 6. So the entry is 3(-7) + (-2)(6), which is -21 - 12, or -33.
Check
The property that fails.
Check your understanding
Which statement about matrix multiplication is FALSE?
Answer: A
Why: Matrix multiplication is not commutative. For A = [[1,4],[3,-2]] and B = [[5,-7],[9,6]], AB is [[41,17],[-3,-33]] while BA is [[-16,34],[27,24]] — no entry agrees.
Real world
A search engine ranks pages by repeatedly multiplying a vector of scores by a matrix describing which pages link to which. A social network finds friends-of-friends by multiplying a connection matrix by itself.
Discussion prompt
In the friends-of-friends case, the connection matrix has a 1 in row i column j when person i knows person j. Explain what the entry in row i, column j of the SQUARE of that matrix counts, using the row-times-column rule.
Hint: Write out what a single term of the sum contributes.
Answer:
The entry pairs row i with column j, so its terms are products of the form: does i know k, times does k know j. Each product is 1 exactly when i knows k AND k knows j, and 0 otherwise.
\[ (A^2)_{ij} = \sum_k a_{ik}a_{kj} \]
Summing over every k therefore COUNTS the number of two-step paths from i to j — the number of mutual friends. The row-times-column rule is not an arbitrary convention; it is the operation that counts paths, which is why the same multiplication turns up in networks, in probability and in geometry.
Commit first
Answer, then rate your confidence honestly.
Predict first
If AB is the zero matrix, must A or B be the zero matrix?
Correct: No — two nonzero matrices can multiply to give the zero matrix.
\[ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]
So from AB equal to AC you may NOT conclude that B equals C. Lesson 3.8's inverse matrices are what make cancelling possible, and only for some matrices.
Why: Take A with rows 1, 0 and 0, 0, and B with rows 0, 0 and 0, 1. Neither is the zero matrix, but every entry of AB is a sum of products in which one factor is always zero, so AB is entirely zero. This has no parallel in ordinary arithmetic, where a product of two nonzero numbers is never zero — and it is the reason you cannot cancel matrices the way you cancel numbers.
Explain it
They have just learned to add matrices and expect multiplication to work the same way.
Discussion prompt
In four sentences or fewer, warn them what is different, give them the rule for one entry, and tell them the check to do before starting any product.
Hint: The check is about dimensions.
Answer:
Multiplication is nothing like addition here: you do not multiply matching positions. Each entry of the answer comes from a whole row of the first matrix and a whole column of the second — multiply them term by term and add up.
Before starting, write the two sets of dimensions side by side and check the inner numbers match; if they do not, the product does not exist. And never swap the order, because AB and BA are usually different matrices.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For existence, write the dimensions side by side and look at the inner pair. For the rule, say it aloud each time until it is automatic. For positions, draw the empty answer grid first and fill it in order. For reordering, remember Example 2, where AB and BA share not one entry. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top of a page write two sets of dimensions side by side, circle the inner pair and box the outer pair, and write beside them what each pair decides. Below, choose two 2 by 2 matrices of your own and compute AB in full, writing the row and column used beside each of the four entries. Then compute BA and write one sentence about how the two answers compare. In the lower half, invent a small quantity-times-price situation with at least two buyers and three items, set it up as a matrix product, label what the rows and columns mean, and compute one total by hand to check. Finally, in a margin, list the four properties from Lesson 1.1 that still hold for matrices and the one that does not.
The margin should show exactly one failure, and it should be the commutativity of multiplication. If you listed more, check each against the worked examples — associativity and distributivity both survive.
Recap
Five things, and the fourth is the one that changes how you write everything afterwards.
| If you see | Then |
|---|---|
| Two matrices to multiply | Compare the inner dimensions first |
| Inner numbers unequal | The product is undefined; stop |
| An entry to compute | Row of the first, column of the second |
| A product to reorder | Do not — check whether it even exists |
| Quantities and prices | Match the item orders before multiplying |
Lesson 3.7 attaches a single number to a square matrix — its determinant — which will turn out to decide whether a system has a unique solution at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.6 Multiply Matrices §3.6, pp. 195-201 — everything on these slides traces back here
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