Addition/subtraction/scalar multiplication | Year 11

01 / LEARN

Understand the idea.

Combine matching positions

Addition and subtraction combine entries in the same positions. Check that the orders are identical before calculating; the same number of entries alone is insufficient, because the row and column positions must match.

For subtraction, keep the sign of each entry in the second matrix inside brackets. For example, $4 - (- 3) = 7$ . Writing the unsimplified entry first makes double negatives easier to handle correctly.

Scale and rearrange

A scalar multiplies every entry of a matrix, including negative and zero entries. You may distribute a scalar across a matrix sum, or factor out a common scalar, just as with ordinary addition.

Solve an equation such as $2 X + A = B$ by subtracting $A$ on both sides, then multiplying every entry by $\frac{1}{2}$ . These are scalar and addition operations; they do not involve dividing by a matrix.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Subtract negative entries

Find $(\begin{array}{cccccccccc} 3 & - 2 \\ 1 & 4 \end{array}) - (\begin{array}{cccccccccc} - 1 & 5 \\ 2 & - 3 \end{array})$ .

Worked solution

1
Subtract entries in their matching positions.
$(\begin{array}{cccccccccc} 3 - (- 1) & - 2 - 5 \\ 1 - 2 & 4 - (- 3) \end{array})$
Both matrices are of order $2 \times 2$ , so each position has a partner.
2
Simplify the first row.
$3 - (- 1) = 4, - 2 - 5 = - 7$
Subtracting a negative is addition.
3
Simplify the second row and assemble.
$(\begin{array}{cccccccccc} 4 & - 7 \\ - 1 & 7 \end{array})$
The lower entries are $1 - 2 = - 1$ and $4 - (- 3) = 7$ .

Answer

$(\begin{array}{cccccccccc} 4 & - 7 \\ - 1 & 7 \end{array})$ .

EXAMPLE 02

Solve for an unknown matrix

Solve $2 X + (\begin{array}{cccccccccc} 1 & 3 \\ - 2 & 0 \end{array}) = (\begin{array}{cccccccccc} 5 & 1 \\ 4 & 6 \end{array})$ .

Worked solution

1
Subtract the known matrix on both sides.
$2 X = (\begin{array}{cccccccccc} 5 - 1 & 1 - 3 \\ 4 - (- 2) & 6 - 0 \end{array})$
Subtracting the same matrix preserves equality and leaves the scalar multiple of $X$ .
2
Simplify the difference.
$2 X = (\begin{array}{cccccccccc} 4 & - 2 \\ 6 & 6 \end{array})$
Each entry is calculated independently in its original position.
3
Halve every entry.
$X = \frac{1}{2} (\begin{array}{cccccccccc} 4 & - 2 \\ 6 & 6 \end{array}) = (\begin{array}{cccccccccc} 2 & - 1 \\ 3 & 3 \end{array})$
The scalar inverse of 2 is $\frac{1}{2}$ .
4
Substitute to verify.
$\begin{aligned} 2 X + (\begin{array}{cccccccccc} 1 & 3 \\ - 2 & 0 \end{array}) & = 2 (\begin{array}{cccccccccc} 2 & - 1 \\ 3 & 3 \end{array}) + (\begin{array}{cccccccccc} 1 & 3 \\ - 2 & 0 \end{array}) \\ = (\begin{array}{cccccccccc} 5 & 1 \\ 4 & 6 \end{array}) \end{aligned}$
Rebuilding the left side recovers the stated right side.

Answer

$X = (\begin{array}{cccccccccc} 2 & - 1 \\ 3 & 3 \end{array})$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

What order condition is needed to calculate $A - B$ ?

Show answer & reasoning

The orders must be equal.

$A - B$ requires equal orders. Each position in $A$ must match the same row and column position in $B$ ; equal total entry counts do not guarantee this.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Calculate $(\begin{array}{cccccccccc} 1 & 4 \end{array}) + (\begin{array}{cccccccccc} 3 & - 2 \end{array})$ .
2 marks
Calculate $- 3 (\begin{array}{cccccccccc} 2 & - 1 \\ 0 & 4 \end{array})$ .
2 marks
Calculate $(\begin{array}{cccccccccc} 5 & 2 \\ - 1 & 7 \end{array}) - (\begin{array}{cccccccccc} 2 & - 3 \\ 4 & 1 \end{array})$ .
2 marks
Find $X$ if $X + (\begin{array}{cccccccccc} 2 & 0 \\ 1 & 3 \end{array}) = (\begin{array}{cccccccccc} 5 & 4 \\ - 2 & 8 \end{array})$ .
2 marks
Find $\frac{1}{2} A + 2 B$ for $A = (\begin{array}{cccccccccc} 4 & 2 \end{array})$ , $B = (\begin{array}{cccccccccc} - 1 & 3 \end{array})$ .
2 marks
Explain why a $2 \times 2$ matrix and a $2 \times 3$ matrix cannot be added.
2 marks

Read the worked solutions

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

Calculate $(\begin{array}{cccccccccc} 1 & 4 \end{array}) + (\begin{array}{cccccccccc} 3 & - 2 \end{array})$ .
2 marks
Add corresponding entries: $1 + 3 = 4$ , $4 - 2 = 2$ .
$(\begin{array}{cccccccccc} 4 & 2 \end{array})$ .
Calculate $- 3 (\begin{array}{cccccccccc} 2 & - 1 \\ 0 & 4 \end{array})$ .
2 marks
Multiply all four entries by $- 3$ .
$(\begin{array}{cccccccccc} - 6 & 3 \\ 0 & - 12 \end{array})$ .
Calculate $(\begin{array}{cccccccccc} 5 & 2 \\ - 1 & 7 \end{array}) - (\begin{array}{cccccccccc} 2 & - 3 \\ 4 & 1 \end{array})$ .
2 marks
Subtract entries in matching positions.
In particular, $2 - (- 3) = 5$ and $- 1 - 4 = - 5$ .
$(\begin{array}{cccccccccc} 3 & 5 \\ - 5 & 6 \end{array})$ .
Find $X$ if $X + (\begin{array}{cccccccccc} 2 & 0 \\ 1 & 3 \end{array}) = (\begin{array}{cccccccccc} 5 & 4 \\ - 2 & 8 \end{array})$ .
2 marks
Subtract the known matrix from the right side.
$X = (\begin{array}{cccccccccc} 3 & 4 \\ - 3 & 5 \end{array})$ .
Find $\frac{1}{2} A + 2 B$ for $A = (\begin{array}{cccccccccc} 4 & 2 \end{array})$ , $B = (\begin{array}{cccccccccc} - 1 & 3 \end{array})$ .
2 marks
Half of $A$ is $(2, 1)$ .
Twice $B$ is $(- 2, 6)$ ; add to obtain $(0, 7)$ .
$(\begin{array}{cccccccccc} 0 & 7 \end{array})$ .
Explain why a $2 \times 2$ matrix and a $2 \times 3$ matrix cannot be added.
2 marks
The first matrix has no third-column entries to match the second matrix.
The orders differ.

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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.

Official curriculum source ↗

In this lesson

Understand the idea.

Combine matching positions

Scale and rearrange

See the method in action.

Subtract negative entries

Subtract entries in their matching positions.

Simplify the first row.

Simplify the second row and assemble.

Solve for an unknown matrix

Subtract the known matrix on both sides.

Simplify the difference.

Halve every entry.

Substitute to verify.

Try it yourself.

Check your understanding

Keep practising