SACE Board SACE 2022-style practice: Calculator-assumed. Given S_1 = bmatrix 10 & 4 \\ 6 & 8 bmatrix and S_2 = bmatrix 3 & 5 \\ 2 & 1 bmatrix, find (a) S_1 + S_2 and (b) 2S_1.

(a) Add matching entries: S_1 + S_2 = bmatrix 13 & 9 \\ 8 & 9 bmatrix. (2 marks) (b) Multiply every entry by 2: 2S_1 = bmatrix 20 & 8 \\ 12 & 16 bmatrix. (1 mark) Marks: two for the element-by-element sum, one for the scalar multiple.

§-Syllabus dot point

SAGeneral MathematicsSyllabus dot point

How do we add, scale and multiply matrices, and when is each operation defined?

Perform matrix addition, subtraction, scalar multiplication and matrix multiplication, applying the order and conformability rules correctly.

How to read a matrix order, add and subtract matrices, multiply by a scalar, and multiply two matrices using the row-by-column rule, including when each operation is defined.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Order and notation
Addition, subtraction and scalar multiplication
Matrix multiplication
The identity and zero matrices
Why order matters
A larger multiplication

What this dot point is asking

You must know the order of a matrix, perform the four operations correctly, and state when an operation is or is not defined.

Order and notation

A matrix is a rectangular array of numbers. Its order is rows $\times$ columns. The matrix

A = \begin{bmatrix} 2 & 5 & 1 \\ 0 & 3 & 4 \end{bmatrix}

has order $2 \times 3$ (2 rows, 3 columns). The entry $a_{ij}$ sits in row $i$ , column $j$ , so $a_{23} = 4$ .

Addition, subtraction and scalar multiplication

These three operations act on each element separately. Same-position entries are combined; for scalar multiplication every entry is multiplied by the scalar.

\begin{bmatrix} 2 & 5 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 1 & -2 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 4 & 9 \end{bmatrix}, \qquad 3\begin{bmatrix} 2 & 5 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 6 & 15 \\ 0 & 9 \end{bmatrix}

Worked example

Two stores hold stock as $S_1 = \begin{bmatrix} 10 & 4 \\ 6 & 8 \end{bmatrix}$ and $S_2 = \begin{bmatrix} 3 & 5 \\ 2 & 1 \end{bmatrix}$ , where rows represent different products and columns represent sizes. Find the total combined stock, then find $2S_1$ to represent a planned doubled order for store 1.

Step 1: Check the matrices are conformable for addition

Before adding, we confirm both matrices have the same order. Both $S_1$ and $S_2$ are $2 \times 2$ matrices, so addition is defined. The result will also be $2 \times 2$ .

Step 2: Add the matrices by combining matching entries

Matrix addition works element by element: each entry in the result comes from adding the entries in the same position of both matrices. We add the entry in row 1, column 1 of $S_1$ to the entry in row 1, column 1 of $S_2$ , and so on for every position:

S_1 + S_2 = \begin{bmatrix} 10 + 3 & 4 + 5 \\ 6 + 2 & 8 + 1 \end{bmatrix} = \begin{bmatrix} 13 & 9 \\ 8 & 9 \end{bmatrix}.

This result gives the total number of each product in each size across both stores.

Step 3: Apply scalar multiplication to find the doubled order

To double $S_1$ , we multiply every entry in $S_1$ by the scalar 2. This scales the entire matrix uniformly, which makes sense here because we want twice as many of every item in the order:

2S_1 = \begin{bmatrix} 2 \times 10 & 2 \times 4 \\ 2 \times 6 & 2 \times 8 \end{bmatrix} = \begin{bmatrix} 20 & 8 \\ 12 & 16 \end{bmatrix}.

Final answers: Total stock is $\begin{bmatrix} 13 & 9 \\ 8 & 9 \end{bmatrix}$ ; the doubled order for store 1 is $2S_1 = \begin{bmatrix} 20 & 8 \\ 12 & 16 \end{bmatrix}$ .

Matrix multiplication

To find the entry in row $i$ , column $j$ of the product $AB$ , multiply the elements of row $i$ of $A$ by the matching elements of column $j$ of $B$ and add the results.

If $A$ is $m \times n$ and $B$ is $n \times p$ , the product $AB$ has order $m \times p$ . The two inner numbers must match; the two outer numbers give the size of the answer.

Worked example

A cafe order is $Q = \begin{bmatrix} 3 & 2 \end{bmatrix}$ (3 coffees, 2 muffins). Prices are $P = \begin{bmatrix} 4.50 \\ 3.00 \end{bmatrix}$ dollars. Find the total cost of the order using matrix multiplication, and state the order of the resulting matrix.

Step 1: Check that the product $QP$ is defined

Before multiplying, we confirm conformability. $Q$ has order $1 \times 2$ (one row, two columns) and $P$ has order $2 \times 1$ (two rows, one column). The inner dimensions are both 2, so the product is defined. The outer dimensions tell us the result will have order $1 \times 1$ , that is, a single number, which makes sense for a total cost.

Step 2: Apply the row-by-column rule to compute $QP$

To find the single entry in the $1 \times 1$ product, we pair each element of row 1 of $Q$ with the matching element of column 1 of $P$ and sum the products. This is why matrix multiplication is suited to "quantity times price" problems: it automatically multiplies each quantity by its price and adds them:

QP = \begin{bmatrix} 3 & 2 \end{bmatrix}\begin{bmatrix} 4.50 \\ 3.00 \end{bmatrix} = 3(4.50) + 2(3.00) = 13.50 + 6.00 = 19.50.

Step 3: Interpret the result in context

The resulting $1 \times 1$ matrix holds the single value $19.50$ . This represents the total cost of the order: 3 coffees at $\$4.50$ each plus 2 muffins at $\$3.00$ each comes to $\$19.50$ .

Final answer: $QP = [19.50]$ , so the total cost of the order is $\$19.50$ .

The identity and zero matrices

Two special matrices behave like the numbers $1$ and $0$ . The zero matrix has every entry zero and leaves a matrix unchanged under addition. The identity matrix $I$ has $1$ s down the leading diagonal and $0$ s elsewhere; multiplying any conformable matrix by $I$ returns that matrix, so $AI = IA = A$ . The identity plays the role of "one" in matrix multiplication and is essential when you later meet inverse matrices, where the goal is to find a matrix that multiplies with $A$ to give $I$ .

Why order matters

Because multiplication combines rows of the first matrix with columns of the second, swapping the order changes both whether the product is defined and what it equals. In the costing example, $PQ$ is also defined ( $2\times1$ times $1\times2$ ) but gives a $2 \times 2$ matrix, which means something entirely different.

A larger multiplication

When both matrices are larger, work systematically through every row-column pairing. For $A$ of order $2\times 2$ times $B$ of order $2\times 2$ , the result is $2\times 2$ , and each of its four entries is a separate row-by-column sum. For example, the top-left entry of the product uses row 1 of $A$ with column 1 of $B$ , the top-right uses row 1 of $A$ with column 2 of $B$ , and so on. Laying the working out entry by entry, rather than trying to do it all at once, is the reliable way to avoid pairing the wrong row with the wrong column - the most common source of multiplication errors.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20234 marksCalculator-assumed. A cafe sells coffees and muffins. The order matrix is

Q = \begin{bmatrix} 3 & 2 \end{bmatrix}

(3 coffees, 2 muffins) and the price matrix is

P = \begin{bmatrix} 4.50 \\ 3.00 \end{bmatrix}

dollars. (a) State the order of the product

QP

and explain why

QP

is defined. (b) Calculate

QP

and interpret the result.

Show worked answer →

(a) $Q$ is $1 \times 2$ and $P$ is $2 \times 1$ . The inner numbers (2 and 2) match, so $QP$ is defined and has order $1 \times 1$ . (2 marks)

(b) $QP = \begin{bmatrix} 3 & 2 \end{bmatrix}\begin{bmatrix} 4.50 \\ 3.00 \end{bmatrix} = 3(4.50) + 2(3.00) = 13.50 + 6.00 = 19.50$ . This is the total cost of the order, 19.50 dollars. (2 marks)

Marks: two for the conformability reasoning and order, two for the product and its interpretation.

SACE 20223 marksCalculator-assumed. Given

S_1 = \begin{bmatrix} 10 & 4 \\ 6 & 8 \end{bmatrix}

and

S_2 = \begin{bmatrix} 3 & 5 \\ 2 & 1 \end{bmatrix}

, find (a)

S_1 + S_2

and (b)

2S_1

Show worked answer →

(a) Add matching entries: $S_1 + S_2 = \begin{bmatrix} 13 & 9 \\ 8 & 9 \end{bmatrix}$ . (2 marks)

(b) Multiply every entry by 2: $2S_1 = \begin{bmatrix} 20 & 8 \\ 12 & 16 \end{bmatrix}$ . (1 mark)

Marks: two for the element-by-element sum, one for the scalar multiple.