What a matrix is, its dimensions and elements, when two matrices are equal, adding and subtracting element by element, scalar multiplication, combining the operations, and organising real data in matrix form.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 3 — Linear Systems and Matrices
Perform Basic Matrix Operations
Objectives
Five outcomes. The first is vocabulary the rest of the chapter assumes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-189 — the lesson these objectives are drawn from
Warm-up
Lesson 3.4 solved a system by carrying x, y and z through every line. Most of that writing was bookkeeping.
Discussion prompt
Write out the system 2x plus 3y equals 7 and 4x minus y equals 1. Which symbols actually changed from line to line while you solved it, and which were just repeated?
Hint: Underline everything that carried information.
Answer:
\[ \begin{cases} 2x + 3y = 7 \\ 4x - y = 1 \end{cases} \;\longrightarrow\; \begin{bmatrix} 2 & 3 & 7 \\ 4 & -1 & 1 \end{bmatrix} \]
Only the numbers ever changed. The letters, the plus signs and the equals signs were copied unchanged down every line. A matrix keeps the numbers and drops the rest, which is why the next four lessons are about matrices.
Concept
A matrix is a rectangular arrangement of numbers in rows and columns. Its size is described by its dimensions, rows first, and the numbers inside are its elements. Once data is in this form, whole tables can be added, subtracted and scaled in one operation.
dimensions — The size of a matrix, written as the number of rows by the number of columns, in that order. A matrix with 2 rows and 3 columns is 2 by 3.
Rows before columns is a convention worth over-learning, because everything in the next three lessons depends on getting the order right.
Figure (svg): A two by three matrix with its rows, columns, dimensions and one element labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-187
Section
Section 1
Concept
Count the rows first, then the columns, and write the dimensions in that order. An element is located by giving its row and then its column, again in that order.
equal matrices — Two matrices are equal when their dimensions are the same and the elements in corresponding positions are equal.
\[ A = \begin{bmatrix} 4 & -1 & 5 \\ 0 & 6 & 3 \end{bmatrix} \quad \text{is } 2 \times 3 \]
Two matrices with the same numbers arranged differently are not equal. Shape is part of the identity of a matrix, not just a detail of how it is written.
Figure (svg): A two by three matrix with its rows, columns, dimensions and one element labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-187
Picture it
A two by three matrix, fully labelled.
Figure (svg): A two by three matrix with its rows, columns, dimensions and one element labelled
The element in row 1, column 3 is 5. Reversing the order would name the element in row 3, column 1, which does not exist in a matrix with only two rows.
Worked example
Naming the parts before doing anything with them.
\[ \text{For } A = \begin{bmatrix} 4 & -1 & 5 \\ 0 & 6 & 3 \end{bmatrix}, \text{ give the dimensions and the element in row 2, column 2.} \]
Count the rows
Why: There are two horizontal lines of numbers.
\[ 2\text{ rows} \]
Count the columns
Why: There are three vertical lines of numbers.
\[ 3\text{ columns} \]
Write the dimensions rows first
Why: Two by three, never three by two.
\[ 2 x 3 \]
Locate row 2, column 2
Why: Go down to the second row, then across to the second column.
\[ \text{the element is } 6 \]
Figure (svg): The solution to Worked example read a matrix shown as a ladder of expressions, one row per algebraic move
\[ 2 \times 3, \quad a_{2,2} = 6 \]
Verify: count the elements two ways
Why: Two rows of three gives six elements, and counting the numbers directly also gives six. A matrix always has exactly rows times columns elements, which is a quick check that nothing was miscounted.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-187
Sorting
Rows first, then columns.
Sort into buckets
Sort each matrix by its dimensions.
The last bucket is worth remembering: only square matrices reach the two most powerful ideas of this chapter.
Worked example
Equality needs matching shape and matching entries.
\[ \text{Are } \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \text{ and } \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \text{ equal?} \]
Compare the dimensions
Why: Both are 2 by 2, so the first condition is met.
Compare row 1, column 2
Why: The first has 2 there; the second has 3.
\[ 2\text{ is not } 3 \]
Conclude
Why: One corresponding pair differs, which is enough.
Note what they do share
Why: The same four numbers appear in both, arranged differently. That is not enough for equality.
Figure (svg): The solution to Worked example decide whether two matrices are equal shown as a ladder of expressions, one row per algebraic move
\[ \text{not equal: position matters} \]
Verify: check every corresponding pair
Why: Row 1 column 1 matches at 1 and row 2 column 2 matches at 4, but both off-diagonal entries differ. Equality requires every position to agree, so two agreements out of four is a failure. Compare with numbers: 12 and 21 use the same digits and are not the same number.
Trap
\[ A = \begin{bmatrix} 4 & -1 & 5 \\ 0 & 6 & 3 \end{bmatrix} \]
Report the dimensions as 3 by 2, counting across first
Why: The eye reads left to right, so columns are counted before rows.
But then the element in row 3, column 1 would have to exist, and there is no third row.
\[ A = \begin{bmatrix} 4 & -1 & 5 \\ 0 & 6 & 3 \end{bmatrix} \text{ is } 2 \times 3 \]
Count rows first, then columns
Why: The convention is rows by columns, in that order, and it is universal.
A quick check: the first number of the dimensions must equal the number of horizontal lines of numbers. Here that is two.
Fill the middle
In the matrix with rows 4, -1, 5 and 0, 6, 3.
Fill in the blanks
\text-1 1, \text___ 2 \;\longrightarrow\; ___
Why: The first row is 4, negative 1, 5, so its second entry is negative 1. Reading row then column, in that order, is what keeps this unambiguous — the element in row 2, column 1 is 0, a completely different number.
Prediction
Commit before counting.
Predict first
How many elements does a 4 by 7 matrix have?
Correct: 28 — four rows of seven.
\[ 4 \times 7 = 28 \text{ elements} \]
Why: A matrix is a full rectangle, so every row has the same number of entries and the total is rows times columns. Adding the dimensions gives 11, which counts nothing meaningful, and 47 simply writes the two numbers next to each other. This product is also why a 4 by 7 and a 7 by 4 matrix have the same number of elements while being entirely different objects.
Explain it to yourself
Two matrices can contain the same numbers and still be different.
\[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \quad \text{versus} \quad \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \]
Discussion prompt
Both contain the numbers 1 through 6. Explain why they are not the same matrix, using an example of real data to make the difference concrete.
Hint: Think about what the rows and columns would label.
Answer:
The first is 2 by 3 and the second is 3 by 2, so no element of one corresponds to an element of the other — row 3 does not exist in the first at all.
Concretely: the first might be two teams and three statistics, the second three teams and two statistics. The same six numbers mean entirely different things, and only the shape records which.
Section
Section 2
Concept
To add or subtract two matrices, add or subtract the numbers in corresponding positions. This is only possible when the two matrices have the same dimensions, because otherwise some positions have no partner.
\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} + \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} a+e & b+f \\ c+g & d+h \end{bmatrix} \]
The textbook flags this with an Avoid Errors note: check the dimensions before starting, not after.
Figure (svg): Two matrices being added element by element, with each corresponding pair highlighted
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-187 — Adding and Subtracting Matrices
Picture it
Example 1a: two 2 by 2 matrices added.
Figure (svg): Two matrices being added element by element, with each corresponding pair highlighted
Each of the four sums is a completely separate calculation. Nothing in one position affects any other, which is what makes matrix addition so easy compared with the multiplication of Lesson 3.6.
Worked example
Example 1a. Four independent additions.
\[ \begin{bmatrix} 3 & 0 \\ -5 & -1 \end{bmatrix} + \begin{bmatrix} -1 & 4 \\ 2 & 0 \end{bmatrix} \]
Check the dimensions match
Why: Both are 2 by 2, so the sum exists.
\[ \text{both } 2 x 2 \]
Add the top-left pair
Why: Three plus negative one.
\[ 2 \]
Add the top-right pair
Why: Zero plus four.
\[ 4 \]
Add the bottom row
Why: Negative five plus two is negative three; negative one plus zero is negative one.
\[ -3\text{ and } -1 \]
Figure (svg): The solution to Worked example add two matrices shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 2 & 4 \\ -3 & -1 \end{bmatrix} \]
Verify: subtract one of the originals from the answer
Why: Taking the second matrix away from the result should recover the first: 2 minus negative 1 is 3, 4 minus 4 is 0, negative 3 minus 2 is negative 5, and negative 1 minus 0 is negative 1. That is exactly the first matrix, so the addition was done correctly in every position.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-187
Picture it
The dimension condition, drawn.
Figure (svg): Two columns contrasting pairs of matrices that can be added with pairs that cannot
Both numbers of the dimensions must match. A 2 by 3 and a 3 by 2 have the same number of elements and still cannot be added, because no element of one corresponds to an element of the other.
Worked example
Example 1b. Six subtractions, several involving negatives.
\[ \begin{bmatrix} 7 & 4 & 0 \\ -2 & -1 & 6 \end{bmatrix} - \begin{bmatrix} -2 & 5 & 3 \\ -10 & -3 & 1 \end{bmatrix} \]
Check the dimensions
Why: Both are 2 by 3, so the difference exists.
\[ \text{both } 2 x 3 \]
Subtract along the first row
Why: Seven minus negative two is nine; four minus five is negative one; zero minus three is negative three.
\[ 9, -1, -3 \]
Subtract along the second row
Why: Negative two minus negative ten is eight; negative one minus negative three is two; six minus one is five.
\[ 8, 2, 5 \]
Assemble the answer
Why: The result has the same dimensions as the two originals.
\[ a 2 x 3\text{ matrix} \]
Figure (svg): The solution to Worked example subtract two matrices shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 9 & -1 & -3 \\ 8 & 2 & 5 \end{bmatrix} \]
Verify: add the second matrix back to the answer
Why: Nine plus negative two is 7, negative one plus five is 4, negative three plus three is 0, eight plus negative ten is negative 2, two plus negative three is negative 1, and five plus one is 6. That reconstructs the first matrix exactly, confirming all six subtractions — including the four involving a negative being subtracted.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-187
Trap
\[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \]
Add the entries that do have partners and copy the rest
Why: The mismatch is treated as a minor obstacle rather than a fatal one.
But the elements in the third column have nothing to be added to, so the answer would be inventing information.
\[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad \text{is undefined} \]
Check the dimensions first and stop if they differ
Why: A 2 by 3 and a 2 by 2 cannot be added: the sum simply does not exist.
Undefined is a legitimate answer here, exactly as it was for the slope of a vertical line in Lesson 2.2. Some operations have conditions.
Sorting
Judge from the dimensions alone.
Sort into buckets
Sort each sum or difference by whether it exists.
The two undefined cases each involve a matrix and its transpose, which is the commonest near-miss and the one worth recognising instantly.
Fill the middle
One entry of Example 1b.
Fill in the blanks
-2 - (-10) = 8
Why: Subtracting negative ten is the same as adding ten, so negative two becomes positive eight. This is the definition of subtraction from Lesson 1.1 — subtract by adding the opposite — applied inside a matrix. Four of the six entries in that example involve a negative being subtracted, which is why the whole computation is really a sign exercise.
Prediction
Commit before computing.
Predict first
What are the dimensions of the sum of two 3 by 4 matrices?
Correct: 3 by 4 — the same as both originals.
Keep this contrast in mind: addition preserves shape, and multiplication generally does not.
Why: Addition is done position by position, so the answer has exactly one entry for each position of the originals and therefore the same shape. Nothing about matrix addition changes the dimensions. This will be strikingly different in Lesson 3.6, where multiplying a 2 by 3 by a 3 by 4 gives a 2 by 4 — a shape that matches neither input.
Section
Section 3
Concept
In matrix algebra an ordinary number is called a scalar. Multiplying a matrix by a scalar means multiplying every element by it, which scales the whole table at once without changing its shape.
scalar — A real number, in a context where matrices are also present. Multiplying a matrix by a scalar multiplies each of its elements by that number.
\[ -2\begin{bmatrix} 4 & -1 \\ 1 & 0 \\ 2 & 7 \end{bmatrix} = \begin{bmatrix} -8 & 2 \\ -2 & 0 \\ -4 & -14 \end{bmatrix} \]
Every element, without exception — including any zeros, which stay zero, and including negatives, which change sign when the scalar is negative.
Figure (svg): A matrix multiplied by a scalar, with the multiplier reaching every element
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 188-188 — Scalar multiplication
Picture it
Example 2a: negative two times a 3 by 2 matrix.
Figure (svg): A matrix multiplied by a scalar, with the multiplier reaching every element
The result is still 3 by 2. A scalar changes the values inside a matrix and never its shape, which distinguishes it from every operation in the next lesson.
Worked example
Example 2a. Six multiplications, each with a sign to watch.
\[ -2\begin{bmatrix} 4 & -1 \\ 1 & 0 \\ 2 & 7 \end{bmatrix} \]
Multiply the first row
Why: Negative two times four is negative eight; negative two times negative one is positive two.
\[ -8\text{ and } 2 \]
Multiply the second row
Why: Negative two times one is negative two; negative two times zero is zero.
\[ -2\text{ and } 0 \]
Multiply the third row
Why: Negative two times two is negative four; negative two times seven is negative fourteen.
\[ -4\text{ and } -14 \]
Note the shape is unchanged
Why: Still three rows and two columns.
\[ \text{still } 3 x 2 \]
Figure (svg): The solution to Worked example multiply by a negative scalar shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} -8 & 2 \\ -2 & 0 \\ -4 & -14 \end{bmatrix} \]
Verify: divide the answer by the scalar
Why: Dividing every element by negative two gives 4, negative 1, 1, 0, 2, 7 — exactly the original matrix. Note the one element that stayed the same: zero times any scalar is zero, so a zero entry survives every scalar multiplication unchanged.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 188-188
Fill the middle
One entry of Example 2a.
Fill in the blanks
-2 \times (-1) = 2
Why: Negative two times negative one is positive two. In a matrix with mixed signs, a negative scalar flips every sign, so entries that were negative become positive and vice versa. Scanning the answer for the right pattern of signs is a quick partial check: with a negative scalar, no sign should be the same as it was.
Worked example
The same procedure with a fractional multiplier.
\[ \tfrac{1}{2}\begin{bmatrix} 6 & -4 \\ 0 & 10 \end{bmatrix} \]
Halve the first row
Why: Six halved is three; negative four halved is negative two.
\[ 3\text{ and } -2 \]
Halve the second row
Why: Zero halved is zero; ten halved is five.
\[ 0\text{ and } 5 \]
Assemble
Why: Same shape, every value halved.
\[ a 2 x 2\text{ matrix} \]
Note what a fractional scalar does
Why: It shrinks the entries rather than growing them, exactly as a fractional coefficient shrank a graph in Lesson 2.7.
Figure (svg): The solution to Worked example a scalar with a fraction shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 3 & -2 \\ 0 & 5 \end{bmatrix} \]
Verify: multiply the answer by 2
Why: Doubling gives 6, negative 4, 0 and 10 — the original. Multiplying by one half and then by two returns the matrix unchanged, because the two scalars are reciprocals. That is a genuine check, not a restatement.
Trap
\[ -2\begin{bmatrix} 4 & -1 \\ 1 & 0 \end{bmatrix} \]
Multiply the first row and copy the second
Why: The multiplier is applied where the eye starts and then attention moves on.
\[ \begin{bmatrix} -8 & 2 \\ 1 & 0 \end{bmatrix} \quad \text{(wrong)} \]
The second row was untouched, so the answer is a mixture of the scaled and unscaled matrix.
\[ -2\begin{bmatrix} 4 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -8 & 2 \\ -2 & 0 \end{bmatrix} \]
Multiply EVERY element, working systematically across each row in turn
Why: A scalar reaches the whole matrix, and skipping any element produces something that is not a scalar multiple of anything.
The check is to divide the answer back by the scalar. A row that does not return to its original value was missed.
Discrimination
Compare the size and sign of the multiplier.
Sort into buckets
Sort each scalar by its effect on a matrix of positive entries.
Prediction
Commit before computing.
Predict first
You multiply a 3 by 5 matrix by the scalar 7. What are the dimensions of the result?
Correct: 3 by 5 — unchanged.
This is why scalar multiplication and addition combine freely: both preserve dimensions, so any expression built from them stays the same shape throughout.
Why: A scalar multiplies the values stored in each position; it does not create or remove positions. So the result has exactly the same fifteen positions as the original. Nothing a scalar can do will change a matrix's shape, which makes scalar multiplication the safest of all the matrix operations to combine with addition.
Explain it to yourself
The word is doing descriptive work.
\[ 3\begin{bmatrix} 2 & 1 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 0 & 12 \end{bmatrix} \]
Discussion prompt
Explain why an ordinary number is called a scalar in this context. What does it do to the matrix, and why would calling it a matrix instead be confusing?
Hint: The word shares a root with scale.
Answer:
It scales the matrix: every entry grows or shrinks by the same factor, so the whole table changes size while keeping its proportions and its shape.
Calling it a matrix would be confusing because a 1 by 1 matrix is a different object with different rules — you could not add it to a 2 by 2, whereas a scalar multiplies any matrix at all. The separate name records that a scalar interacts with matrices in a way matrices do not interact with each other.
Section
Section 4
Concept
When an expression mixes scalar multiplication with addition or subtraction, do the scalar multiplications first and then combine. It is the order of operations from Lesson 1.2, applied to whole matrices.
\[ 4\begin{bmatrix} -2 & -8 \\ 5 & 0 \end{bmatrix} + \begin{bmatrix} -3 & 8 \\ 6 & -5 \end{bmatrix} \]
Because addition and scalar multiplication both preserve dimensions, any such expression stays the same shape from beginning to end.
Figure (svg): A matrix multiplied by a scalar, with the multiplier reaching every element
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 188-188
Picture it
Example 2b: a scalar multiple added to another matrix.
Figure (svg): A matrix multiplied by a scalar, with the multiplier reaching every element
Adding before scaling would give a different answer, exactly as it would with ordinary numbers. Multiplication outranks addition here for the same reason it does everywhere.
Worked example
Example 2b, worked in the required order.
\[ 4\begin{bmatrix} -2 & -8 \\ 5 & 0 \end{bmatrix} + \begin{bmatrix} -3 & 8 \\ 6 & -5 \end{bmatrix} \]
Do the scalar multiplication first
Why: Four times each element: negative eight, negative thirty-two, twenty, zero.
\[ [[-8, -32], [20, 0]] \]
Check the shapes before adding
Why: Both matrices are now 2 by 2, so the sum exists.
\[ \text{both } 2 x 2 \]
Add the first row
Why: Negative eight plus negative three is negative eleven; negative thirty-two plus eight is negative twenty-four.
\[ -11\text{ and } -24 \]
Add the second row
Why: Twenty plus six is twenty-six; zero plus negative five is negative five.
\[ 26\text{ and } -5 \]
Figure (svg): The solution to Worked example scalar multiplication then addition shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} -11 & -24 \\ 26 & -5 \end{bmatrix} \]
Verify: undo both operations
Why: Subtracting the second matrix from the answer gives negative 8, negative 32, 20 and 0, and dividing that by 4 gives negative 2, negative 8, 5 and 0 — the original first matrix. Reversing both steps in the opposite order recovers the input, which checks every element.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 188-188
Ranking
Evaluating 4A plus B, where A and B are matrices.
Put in order
Why: Checking dimensions first avoids doing a scalar multiplication that turns out to be useless because the addition is undefined. Scaling comes before adding by the order of operations. The final dimension check is quick and catches any slip that dropped or invented a row.
Worked example
Two scalars, then a subtraction.
\[ 3\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix} - 2\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \]
Scale the first matrix by 3
Why: Six, negative three, zero, twelve.
\[ [[6, -3], [0, 12]] \]
Scale the second matrix by 2
Why: Two, six, negative four, ten.
\[ [[2, 6], [-4, 10]] \]
Subtract element by element
Why: Six minus two is four; negative three minus six is negative nine; zero minus negative four is four; twelve minus ten is two.
\[ 4, -9, 4, 2 \]
Assemble
Why: Both scalings preserved the 2 by 2 shape, so the difference does too.
\[ a 2 x 2\text{ matrix} \]
Figure (svg): The solution to Worked example a subtraction of scalar multiples shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 4 & -9 \\ 4 & 2 \end{bmatrix} \]
Verify: check one entry by an independent route
Why: The bottom-left entry: three times zero is 0, and two times negative two is negative 4, so the entry is 0 minus negative 4, which is 4. Computing a single entry from scratch, rather than re-reading the working, is a genuine spot check.
Error analysis
A student evaluates a mixed expression in the wrong order.
Annotate
On: \( 4\begin{bmatrix} -2 & -8 \\ 5 & 0 \end{bmatrix} + \begin{bmatrix} -3 & 8 \\ 6 & -5 \end{bmatrix} \;\Longrightarrow\; 4\begin{bmatrix} -5 & 0 \\ 11 & -5 \end{bmatrix} = \begin{bmatrix} -20 & 0 \\ 44 & -20 \end{bmatrix} \)
If the 4 had been meant to multiply the whole sum, the expression would have been written with brackets around it. Read the expression before computing.
Comparison
Fill the blanks. The rules transfer almost unchanged.
Comparison matrix
| Rule | For numbers | For matrices |
|---|---|---|
| Order of operations | multiply before adding | scale before adding |
| Adding requires | nothing special | matching dimensions |
| Multiplying by 1 | changes nothing | changes nothing |
| Multiplying by 0 | gives 0 | gives a matrix of zeros |
| Result of adding | a number | a matrix of the same shape |
The one genuinely new rule is in the second row. Everything else carries over from arithmetic, which is why these operations feel familiar.
Fill the middle
Example 2b, after the scaling.
Fill in the blanks
\begin-11 -8 & -32 \\ 20 & 0 \end___ + \begin___ -3 & 8 \\ 6 & -5 \end___ = \begin___ ___ & -24 \\ 26 & -5 \end___
Why: Negative eight plus negative three is negative eleven. Notice that this entry was negative 2 before scaling, so the 4 turned it into negative 8 before the addition ever happened — computing it as negative 2 plus negative 3 and then scaling would give negative 20 instead, which is the wrong-order error.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is matrix addition commutative — does A plus B always equal B plus A?
Correct: Yes — each position is an ordinary sum of numbers, and number addition is commutative.
\[ (A + B)_{ij} = a_{ij} + b_{ij} = b_{ij} + a_{ij} = (B + A)_{ij} \]
Why: The entry in any position of A plus B is the sum of two numbers, and swapping them changes nothing. So every position agrees and the two matrices are equal. This is worth stating explicitly because matrix MULTIPLICATION in Lesson 3.6 is emphatically not commutative — the properties do not all transfer, and knowing which ones do is what stops you assuming the rest.
Section
Section 5
Concept
Real data usually arrives as a two-way table: rows for one category, columns for another. Stripping the labels leaves a matrix, and then whole tables can be combined in a single operation.
The labels have to be remembered, because the matrix itself does not record them. Two matrices can only be added meaningfully if their rows and columns mean the same things.
Figure (svg): A table of sports data written as a matrix, with rows and columns labelled by what they mean
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-189
Picture it
Two teams down the side, three statistics across the top.
Figure (svg): A table of sports data written as a matrix, with rows and columns labelled by what they mean
Adding a second season's matrix gives the combined record, and multiplying by 3 would convert wins into points. Both operations act on the whole table at once.
Worked example
Two matrices of the same shape and the same meaning, added.
\[ \text{Season 1: } \begin{bmatrix} 12 & 4 & 6 \\ 9 & 7 & 6 \end{bmatrix}, \text{ Season 2: } \begin{bmatrix} 10 & 6 & 6 \\ 14 & 3 & 5 \end{bmatrix}. \text{ Find the two-season totals.} \]
Check the rows and columns mean the same things
Why: Both have teams down the side in the same order and wins, draws, losses across the top. Without that, the sum would be meaningless even though it is defined.
Add element by element
Why: Twelve plus ten is 22; four plus six is 10; six plus six is 12.
\[ r o w 1: 22, 10, 12 \]
Do the second row
Why: Nine plus fourteen is 23; seven plus three is 10; six plus five is 11.
\[ r o w 2: 23, 10, 11 \]
Interpret the result
Why: Each entry is that team's two-season total for that statistic.
Figure (svg): The solution to Worked example combine two seasons shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 22 & 10 & 12 \\ 23 & 10 & 11 \end{bmatrix} \]
Verify: check each row against the number of games
Why: Team A played 22 games and 22 more, so its totals should sum to 44: 22 plus 10 plus 12 is 44 — correct. Team B: 23 plus 10 plus 11 is 44 as well, and both seasons had 22 games each. Row totals that match the games played confirm nothing was lost.
Real world
Two shops each sell three products. In week 1, shop A sold 20, 15 and 8; shop B sold 12, 22 and 9.
Discussion prompt
Write this as a matrix, state its dimensions, and say what the rows and columns mean. Then say what adding week 2's matrix would give, and what multiplying by a price would and would not achieve.
Hint: Prices differ by product, which matters for the last part.
Answer:
\[ \begin{bmatrix} 20 & 15 & 8 \\ 12 & 22 & 9 \end{bmatrix} \quad \text{2 by 3: shops by products} \]
Adding week 2's matrix gives the two-week totals for every shop and product. A single scalar would multiply everything by one price, which is only right if all three products cost the same.
Applying different prices to different products is not scalar multiplication at all — it needs matrix multiplication, which is exactly what Lesson 3.6 is for.
Worked example
Scalar multiplication used to change the unit of a whole table.
\[ \text{At 3 points a win, convert } \begin{bmatrix} 12 \\ 9 \end{bmatrix} \text{ wins into points.} \]
Identify what the scalar means
Why: Three points per win, so it converts a count of wins into a count of points.
\[ 3\text{ points per win} \]
Multiply every element
Why: Three times twelve is thirty-six; three times nine is twenty-seven.
\[ 36\text{ and } 27 \]
State the units of the result
Why: The entries are now points, not wins — the numbers changed meaning as well as value.
Note what stayed the same
Why: Still a 2 by 1 matrix, with the same teams in the same order.
Figure (svg): The solution to Worked example convert wins to points shown as a ladder of expressions, one row per algebraic move
\[ 3\begin{bmatrix} 12 \\ 9 \end{bmatrix} = \begin{bmatrix} 36 \\ 27 \end{bmatrix} \]
Verify: check against a direct count
Why: Team A won 12 games at 3 points each, which is 36 points — matching. And the ordering is preserved: the team with more wins still has more points, which any positive scalar guarantees.
Error analysis
A student adds two 2 by 3 matrices whose columns mean different things.
Annotate
On: \( \begin{bmatrix} 12 & 4 & 6 \\ 9 & 7 & 6 \end{bmatrix} + \begin{bmatrix} 30 & 24 & 18 \\ 27 & 21 & 18 \end{bmatrix} \quad \text{(the second is goals for, against, difference)} \)
Defined and meaningful are different tests. The dimensions decide the first; only knowing what the rows and columns stand for decides the second.
Matching
Each matrix operation does something recognisable to real data.
Match the pairs
Why: Subtraction is the one worth noticing: it gives the change in every cell at once, which is exactly what a year-on-year comparison table shows. Doing that cell by cell for a large table is tedious; doing it as one matrix subtraction is a single operation.
Definition probe
Two tests: the dimensions, and what the entries stand for.
Sort into buckets
Sort each proposed addition.
Socratic
One question, and nothing else on this slide.
\[ \begin{bmatrix} 12 & 4 & 6 \\ 9 & 7 & 6 \end{bmatrix} \]
Discussion prompt
This matrix came from a labelled table. What information was lost when the labels were dropped, and why is losing it worth the trouble? What would go wrong if you came back to this matrix in a year?
Hint: Ask what you would need in order to read it.
Answer:
Everything about what the numbers MEAN was lost: which teams, which statistics, which season, what units. The matrix records only the values and their arrangement.
It is worth it because arithmetic on the whole table becomes possible — you cannot add two labelled tables, but you can add two matrices, and then reattach the labels afterwards.
Coming back in a year, the matrix alone would be unreadable. That is why real data work keeps the labels alongside, and it is why the middle bucket of the last sort exists: the mathematics will happily add things that should never be added.
Comparison
Fill the blanks. All three preserve the shape of the matrix.
Comparison matrix
| Operation | Condition | Effect on dimensions |
|---|---|---|
| Addition | dimensions must match | unchanged |
| Subtraction | dimensions must match | unchanged |
| Scalar multiplication | none - any matrix, any scalar | unchanged |
| Mixed expression | scale first, then add | unchanged |
Every row says unchanged, which is why these three operations combine so freely. Lesson 3.6 breaks that pattern completely.
Pattern
One routine handles any expression built from these three operations.
Step five's second half is the one the mathematics cannot do for you. A defined operation on mismatched data produces a number that means nothing.
OpenStax Algebra and Trigonometry 2e, §11.5 Matrices and Matrix Operations §11.5
Check
Dimensions, rows first.
Check your understanding
What are the dimensions of a matrix with 3 rows and 5 columns, and how many elements does it have?
Answer: A
Why: Dimensions are written rows by columns, so 3 by 5, and the number of elements is the product, which is 15.
Check
Subtraction with negatives. Take care with the signs.
Check your understanding
Compute the entry in row 2, column 1 of [[7,4,0],[-2,-1,6]] minus [[-2,5,3],[-10,-3,1]].
Answer: A
Why: The entry is -2 minus -10, and subtracting a negative is adding, so -2 plus 10 is 8.
Check
A mixed expression. Watch the order.
Check your understanding
Compute 4[[-2,-8],[5,0]] + [[-3,8],[6,-5]].
Answer: A
Why: Scaling first gives [[-8,-32],[20,0]], and adding the second matrix gives [[-11,-24],[26,-5]].
Real world
A shop records monthly sales of three products across two branches as a 2 by 3 matrix, one matrix per month.
Discussion prompt
Describe what each of these gives: adding twelve monthly matrices; subtracting January from December; multiplying a monthly matrix by 1.2. Then say which of them a spreadsheet does for you, and why the matrix notation is still worth having.
Hint: Think about what each operation says about the business.
Answer:
Adding twelve months gives the annual totals for every branch and product at once. December minus January gives the change over the year in every cell — where growth happened and where it did not.
Multiplying by 1.2 models a uniform twenty percent increase, which is a forecast rather than a record.
A spreadsheet does all three cell by cell. The matrix notation is worth having because it names the operation as a single object — you can then say things like the annual total is the sum of the monthly matrices, and reason about it, which is the step that leads to matrix multiplication in the next lesson.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is there a matrix that behaves like zero, so that adding it changes nothing?
Correct: Yes — the matrix whose every entry is zero, with the same dimensions.
\[ A + O = A \quad \text{where } O \text{ is the zero matrix of the same size} \]
Why: Adding zero to each element leaves each element unchanged, so the whole matrix is unchanged. This is the additive identity, exactly parallel to the number 0 from Lesson 1.1. Note the dimension proviso: a 2 by 3 matrix needs a 2 by 3 zero matrix, so there is one zero matrix per shape rather than a single universal one. The matrix with ones down the diagonal is a different object entirely, and it will turn out to be the identity for MULTIPLICATION in Lesson 3.8.
Explain it
They have never seen a matrix and think it looks alarming.
Discussion prompt
In four sentences or fewer, explain what a matrix is, how you add two of them, and the one condition that has to hold. Give them a familiar example of data that is already in this shape.
Hint: They have seen tables.
Answer:
A matrix is just a table of numbers with the labels taken off — a sports league table or a shop's sales by branch and product is already one. To add two matrices you add the numbers that sit in the same position, and nothing else happens.
The one condition is that the two matrices must be the same size, both in rows and in columns, because otherwise some numbers have nothing to be added to.
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 dimensions, count the horizontal lines of numbers first, always. For negative subtractions, rewrite each as adding the opposite before computing. For scalars, work systematically across each row rather than jumping about. For order, look for brackets — if the scalar is outside them it multiplies only what it touches. 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 draw a 2 by 3 matrix of your own and label its rows, its columns, its dimensions and one named element. Below it, write two matrices of the same shape and add them, showing every individual sum, then subtract them, again showing every difference. Beside that, multiply one of them by a negative scalar and circle every element whose sign changed. In the lower half, write a mixed expression combining a scalar multiple and an addition, and evaluate it twice — once in the correct order and once in the wrong order — writing one sentence about why the answers differ. Finally, in a margin, list the three operations from this lesson and write next to each what it does to the dimensions.
All three margin entries should say the dimensions are unchanged. That is what makes the next lesson a genuine surprise.
Recap
Five things, and all of them are element-by-element.
| If you see | Then |
|---|---|
| Two matrices to add | Check the dimensions first |
| Different dimensions | The sum is undefined; stop |
| A number in front of a matrix | Multiply every element by it |
| A scalar and an addition together | Scale first, then add |
| Real data in two matrices | Check the labels mean the same things |
Lesson 3.6 introduces the one matrix operation that is not element-by-element, changes the dimensions, and refuses to be commutative.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.5 Perform Basic Matrix Operations §3.5, pp. 187-189 — everything on these slides traces back here
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