Sec 4 · G3 (Express)

Matrices

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1. Order and elements

Term	Meaning	Example
order	rows $\times$ columns	$\begin{pmatrix} 5 & 9 & 6 \end{pmatrix}$ is $1 \times 3$
		$\begin{pmatrix} -1 & 6 & -1 \\ 2 & 3 & -5 \end{pmatrix}$ is $2 \times 3$
element	the number in a given row and column	in $\begin{pmatrix} -1 & 6 & -1 \\ 2 & 3 & -5 \end{pmatrix}$ , row 2, column 1: $2$
row matrix	one row	$\begin{pmatrix} 5 & 9 & 6 \end{pmatrix}$
column matrix	one column	$\begin{pmatrix} 5 \\ 9 \end{pmatrix}$
square matrix	rows $=$ columns	$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

📖 Worked example

Shop P sold 4 laptops, 4 tablets and 11 phones. Shop Q sold 3 laptops, 6 tablets and 3 phones. Write down a matrix showing this information, with one row for each shop and one column for each product, in the order given.

Step 1 of 3

Row 1: Shop P

4 \quad 4 \quad 11

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2. Adding, subtracting and scalar multiples

Add or subtract: same order, element by element

\displaystyle A + B

e.g.

\displaystyle \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}

\displaystyle = \begin{pmatrix} 3 & 2 \\ 1 & 6 \end{pmatrix}

Scalar multiple: multiply every element

\displaystyle kA

e.g.

\displaystyle 3\begin{pmatrix} 1 & -2 \\ -4 & 0 \end{pmatrix}

\displaystyle = \begin{pmatrix} 3 & -6 \\ -12 & 0 \end{pmatrix}

Combining

\displaystyle 3A + 2B

e.g.

\displaystyle 3\begin{pmatrix} 1 & -2 \\ -4 & -2 \end{pmatrix} + 2\begin{pmatrix} 2 & 4 \\ -1 & 4 \end{pmatrix}

\displaystyle = \begin{pmatrix} 7 & 2 \\ -14 & 2 \end{pmatrix}

📖 Worked example

$A = \begin{pmatrix} 3 & -4 \\ -5 & -6 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & -4 \\ 3 & -1 \end{pmatrix}$ . Find $A - B$ .

Step 1 of 3

Row 1: subtract the elements in the same position

Column 1:

3 - 3 = 0

Column 2:

-4 - (-4) = 0

📖 Worked example

$A = \begin{pmatrix} 12 & 8 \\ 16 & 16 \end{pmatrix}$ . Find $\dfrac{1}{4}A$ .

Step 1 of 3

Row 1: multiply every element by the fraction

Column 1:

\dfrac{1}{4} \times 12 = 3

Column 2:

\dfrac{1}{4} \times 8 = 2

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3. Multiplying matrices

Rule	Example
$(m \times n)(n \times p)$ gives $m \times p$	$(2 \times 3)(3 \times 1)$ gives $2 \times 1$
middle numbers must match	$(2 \times 1)(2 \times 3)$ : not possible
row of the first $\times$ column of the second	$\begin{pmatrix} 2 & 3 \end{pmatrix}\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ gives $2 \times 4 + 3 \times 1$
order matters	$AB \neq BA$ in general

2 × 2 times 2 × 2

\displaystyle AB

e.g.

\displaystyle \begin{pmatrix} 5 & -3 \\ -2 & -1 \end{pmatrix}\begin{pmatrix} -3 & 4 \\ -3 & -3 \end{pmatrix}

\displaystyle = \begin{pmatrix} -6 & 29 \\ 9 & -5 \end{pmatrix}

The other order gives a different answer

\displaystyle BA

e.g.

\displaystyle \begin{pmatrix} -3 & 4 \\ -3 & -3 \end{pmatrix}\begin{pmatrix} 5 & -3 \\ -2 & -1 \end{pmatrix}

\displaystyle = \begin{pmatrix} -23 & 5 \\ -9 & 12 \end{pmatrix}

📖 Worked example

$A$ is a $2 \times 1$ matrix and $B$ is a $2 \times 3$ matrix. State the order of $AB$ , or write "not possible" if $AB$ cannot be found.

Step 1 of 3

Write the orders side by side

(2 \mathbin{\times} 1)(2 \mathbin{\times} 3)

📖 Worked example

$A = \begin{pmatrix} 4 & -1 \\ -1 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} -3 \\ 6 \end{pmatrix}$ . Find $AB$ .

Step 1 of 4

The order of AB

(2 \times 2)(2 \times 1) \rightarrow 2 \times 1

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4. Matrices in context

Cups sold (rows: stalls) times prices (column): money taken by each stall

\displaystyle PQ

e.g.

\displaystyle \begin{pmatrix} 8 & 15 & 9 \\ 10 & 5 & 14 \end{pmatrix}\begin{pmatrix} 1.5 \\ 2 \\ 2.5 \end{pmatrix}

\displaystyle = \begin{pmatrix} 64.5 \\ 60 \end{pmatrix}

📖 Worked example

The matrix $P = \begin{pmatrix} 8 & 15 & 9 \\ 10 & 5 & 14 \end{pmatrix}$ shows the numbers of small, medium and large cups of juice sold at Stall A and Stall B (one row for each stall). The matrix $Q = \begin{pmatrix} 1.5 \\ 2 \\ 2.5 \end{pmatrix}$ shows the prices, in dollars, of a small, medium and large cup. Find $PQ$ .

Step 1 of 4

The order of PQ

(2 \times 3)(3 \times 1) \rightarrow 2 \times 1

PQ = \begin{pmatrix} 8 & 15 & 9 \\ 10 & 5 & 14 \end{pmatrix}\begin{pmatrix} 1.5 \\ 2 \\ 2.5 \end{pmatrix}

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Sec 4 · G3 (Express)

Matrices

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1. Order and elements

Term	Meaning	Example
order	rows $\times$ columns	$\begin{pmatrix} 5 & 9 & 6 \end{pmatrix}$ is $1 \times 3$
		$\begin{pmatrix} -1 & 6 & -1 \\ 2 & 3 & -5 \end{pmatrix}$ is $2 \times 3$
element	the number in a given row and column	in $\begin{pmatrix} -1 & 6 & -1 \\ 2 & 3 & -5 \end{pmatrix}$ , row 2, column 1: $2$
row matrix	one row	$\begin{pmatrix} 5 & 9 & 6 \end{pmatrix}$
column matrix	one column	$\begin{pmatrix} 5 \\ 9 \end{pmatrix}$
square matrix	rows $=$ columns	$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

📖 Worked example

Step 1 of 3

Row 1: Shop P

4 \quad 4 \quad 11

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2. Adding, subtracting and scalar multiples

Add or subtract: same order, element by element

\displaystyle A + B

e.g.

\displaystyle \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}

\displaystyle = \begin{pmatrix} 3 & 2 \\ 1 & 6 \end{pmatrix}

Scalar multiple: multiply every element

\displaystyle kA

e.g.

\displaystyle 3\begin{pmatrix} 1 & -2 \\ -4 & 0 \end{pmatrix}

\displaystyle = \begin{pmatrix} 3 & -6 \\ -12 & 0 \end{pmatrix}

Combining

\displaystyle 3A + 2B

e.g.

\displaystyle 3\begin{pmatrix} 1 & -2 \\ -4 & -2 \end{pmatrix} + 2\begin{pmatrix} 2 & 4 \\ -1 & 4 \end{pmatrix}

\displaystyle = \begin{pmatrix} 7 & 2 \\ -14 & 2 \end{pmatrix}

📖 Worked example

$A = \begin{pmatrix} 3 & -4 \\ -5 & -6 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & -4 \\ 3 & -1 \end{pmatrix}$ . Find $A - B$ .

Step 1 of 3

Row 1: subtract the elements in the same position

Column 1:

3 - 3 = 0

Column 2:

-4 - (-4) = 0

📖 Worked example

$A = \begin{pmatrix} 12 & 8 \\ 16 & 16 \end{pmatrix}$ . Find $\dfrac{1}{4}A$ .

Step 1 of 3

Row 1: multiply every element by the fraction

Column 1:

\dfrac{1}{4} \times 12 = 3

Column 2:

\dfrac{1}{4} \times 8 = 2

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3. Multiplying matrices

Rule	Example
$(m \times n)(n \times p)$ gives $m \times p$	$(2 \times 3)(3 \times 1)$ gives $2 \times 1$
middle numbers must match	$(2 \times 1)(2 \times 3)$ : not possible
row of the first $\times$ column of the second	$\begin{pmatrix} 2 & 3 \end{pmatrix}\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ gives $2 \times 4 + 3 \times 1$
order matters	$AB \neq BA$ in general

2 × 2 times 2 × 2

\displaystyle AB

e.g.

\displaystyle \begin{pmatrix} 5 & -3 \\ -2 & -1 \end{pmatrix}\begin{pmatrix} -3 & 4 \\ -3 & -3 \end{pmatrix}

\displaystyle = \begin{pmatrix} -6 & 29 \\ 9 & -5 \end{pmatrix}

The other order gives a different answer

\displaystyle BA

e.g.

\displaystyle \begin{pmatrix} -3 & 4 \\ -3 & -3 \end{pmatrix}\begin{pmatrix} 5 & -3 \\ -2 & -1 \end{pmatrix}

\displaystyle = \begin{pmatrix} -23 & 5 \\ -9 & 12 \end{pmatrix}

📖 Worked example

$A$ is a $2 \times 1$ matrix and $B$ is a $2 \times 3$ matrix. State the order of $AB$ , or write "not possible" if $AB$ cannot be found.

Step 1 of 3

Write the orders side by side

(2 \mathbin{\times} 1)(2 \mathbin{\times} 3)

📖 Worked example

$A = \begin{pmatrix} 4 & -1 \\ -1 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} -3 \\ 6 \end{pmatrix}$ . Find $AB$ .

Step 1 of 4

The order of AB

(2 \times 2)(2 \times 1) \rightarrow 2 \times 1

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4. Matrices in context

Cups sold (rows: stalls) times prices (column): money taken by each stall

\displaystyle PQ

e.g.

\displaystyle \begin{pmatrix} 8 & 15 & 9 \\ 10 & 5 & 14 \end{pmatrix}\begin{pmatrix} 1.5 \\ 2 \\ 2.5 \end{pmatrix}

\displaystyle = \begin{pmatrix} 64.5 \\ 60 \end{pmatrix}

📖 Worked example

Step 1 of 4

The order of PQ

(2 \times 3)(3 \times 1) \rightarrow 2 \times 1

PQ = \begin{pmatrix} 8 & 15 & 9 \\ 10 & 5 & 14 \end{pmatrix}\begin{pmatrix} 1.5 \\ 2 \\ 2.5 \end{pmatrix}

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