Matrix multiplication and non-commutativity | Year 11

01 / LEARN

The idea, step by step

Check dimensions before calculating

If $A$ has order $m \times n$ and $B$ has order $n \times p$ , then $A B$ is defined and has order $m \times p$ . The inner dimensions must match; the outer dimensions give the answer size.

The $(i, j)$ entry of $A B$ is the dot product of row $i$ of $A$ and column $j$ of $B$ . Multiply matching terms and add. Ordinary matrix multiplication is not entry-by-entry multiplication.

Order is part of the operation

Usually $A B \neq B A$ . Sometimes one order is defined and the other is not. Even for square matrices both products can exist but be different.

Use a small numerical example to test a claimed rule. To calculate $A B C$ , associativity allows $(A B) C = A (B C)$ when the dimensions match, but it does not permit swapping $A$ and $B$ .

A common trap

Multiplying only entries in the same positions.

Instead: A product entry uses an entire row and an entire column, not one pair of matching positions.

02 / WORKED EXAMPLES

See the thinking

Example 1

A rectangular product

Calculate $(\begin{array}{cccccccccc} 1 & 2 & - 1 \\ 0 & 3 & 4 \end{array}) (\begin{array}{cccccccccc} 2 & 1 \\ 1 & - 2 \\ 3 & 0 \end{array})$ .

Top-left: $1 (2) + 2 (1) - 1 (3) = 1$ .
Top-right: $1 (1) + 2 (- 2) - 1 (0) = - 3$ .
Bottom-left is $0 + 3 + 12 = 15$ ; bottom-right is $0 - 6 + 0 = - 6$ .

(\begin{array}{cccccccccc} 1 & - 3 \\ 15 & - 6 \end{array})

Example 2

Demonstrate non-commutativity

Let $A = (\begin{array}{cccccccccc} 1 & 1 \\ 0 & 1 \end{array})$ , $B = (\begin{array}{cccccccccc} 2 & 0 \\ 0 & 3 \end{array})$ . Compare $A B$ and $B A$ .

Row-column multiplication gives $A B = (\begin{array}{cccccccccc} 2 & 3 \\ 0 & 3 \end{array})$ .
Reversing the order gives $B A = (\begin{array}{cccccccccc} 2 & 2 \\ 0 & 3 \end{array})$ .
The top-right entries differ.

A B \neq B A

Check your understanding

If $A$ is $2 \times 3$ and $B$ is $3 \times 4$ , what is the order of $A B$ ?

Show answer

$2 \times 4$ .

The inner dimensions 3 agree and the outer dimensions are 2 and 4.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find $(\begin{array}{cccccccccc} 2 & 1 \end{array}) (\begin{array}{cccccccccc} 3 \\ - 4 \end{array})$ .
2 marks
Find $(\begin{array}{cccccccccc} 1 & 2 \\ 3 & 0 \end{array}) (\begin{array}{cccccccccc} 2 & 1 \\ 1 & 4 \end{array})$ .
2 marks
A is $3 \times 2$ and B is $2 \times 5$ . State the order of AB and explain whether BA exists.
2 marks
Find $(\begin{array}{cccccccccc} 1 \\ 2 \end{array}) (\begin{array}{cccccccccc} 3 & 4 \end{array})$ .
2 marks
For $A = (\begin{array}{cccccccccc} 0 & 1 \\ 1 & 0 \end{array})$ , calculate $A^{2}$ .
2 marks
A shop sells 6 notebooks at 4 dollars each and 3 folders at 5 dollars each. Express the total as a row-column matrix product and evaluate it.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $(\begin{array}{cccccccccc} 2 & 1 \end{array}) (\begin{array}{cccccccccc} 3 \\ - 4 \end{array})$ .
2 marks
The product is $2 (3) + 1 (- 4) = 2$ .
$2$ .
Find $(\begin{array}{cccccccccc} 1 & 2 \\ 3 & 0 \end{array}) (\begin{array}{cccccccccc} 2 & 1 \\ 1 & 4 \end{array})$ .
2 marks
Rows dotted with columns give $(2 + 2, 1 + 8)$ and $(6 + 0, 3 + 0)$ .
$(\begin{array}{cccccccccc} 4 & 9 \\ 6 & 3 \end{array})$ .
A is $3 \times 2$ and B is $2 \times 5$ . State the order of AB and explain whether BA exists.
2 marks
The AB inner dimensions are both 2.
For BA the inner dimensions would be 5 and 3, which do not match.
AB is $3 \times 5$ ; BA is not defined.
Find $(\begin{array}{cccccccccc} 1 \\ 2 \end{array}) (\begin{array}{cccccccccc} 3 & 4 \end{array})$ .
2 marks
A $2 \times 1$ by $1 \times 2$ product has order $2 \times 2$ .
Multiply each column-vector entry by the row.
$(\begin{array}{cccccccccc} 3 & 4 \\ 6 & 8 \end{array})$ .
For $A = (\begin{array}{cccccccccc} 0 & 1 \\ 1 & 0 \end{array})$ , calculate $A^{2}$ .
2 marks
Each output column is found by applying A twice.
Direct multiplication gives diagonal entries 1 and off-diagonal entries 0.
$I = (\begin{array}{cccccccccc} 1 & 0 \\ 0 & 1 \end{array})$ .
A shop sells 6 notebooks at 4 dollars each and 3 folders at 5 dollars each. Express the total as a row-column matrix product and evaluate it.
2 marks
Quantities form a row and prices form a column.
$6 (4) + 3 (5) = 39$ .
$(\begin{array}{cccccccccc} 6 & 3 \end{array}) (\begin{array}{cccccccccc} 4 \\ 5 \end{array}) = 39$ dollars.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: matrices and transformations; original lessons follow the approved school-sized chapter structure.

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