How can you check a claimed equivalence between transformations?

Write each transformation as a matrix and multiply them in the correct order. If the product equals the matrix for the single transformation claimed, the two really are the same.

GCSEFurther MathsAQAFlashcards5. Matrix TransformationsMatrix Transformations

Matrix Transformations (AQA GCSE Further Maths): Flashcards

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FrontTransformations with Matrices
How do you find the image of the point $(x, y)$ under a matrix $M$ ?

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How do you find the image of the point $(x, y)$ under a matrix $M$ ?
Write the point as a column vector and multiply, so $M (\begin{matrix} x \\ y \end{matrix})$ gives $(\begin{matrix} x' \\ y' \end{matrix})$ .
The starting point is the object and the result is the image.
Find the image of $(2, 3)$ under $M = (\begin{matrix} 4 & 5 \\ 1 & - 2 \end{matrix})$ .
Multiplying gives $(\begin{matrix} 4 \times 2 + 5 \times 3 \\ 1 \times 2 - 2 \times 3 \end{matrix}) = (\begin{matrix} 23 \\ - 4 \end{matrix})$ .
So the image is the point $(23, - 4)$ .
The image of $P$ under $M = (\begin{matrix} 4 & 5 \\ 1 & - 2 \end{matrix})$ is $(11, 6)$ . How do you find $P$ ?
Call $P$ the point $(x, y)$ and multiply, which gives $4 x + 5 y = 11$ and $x - 2 y = 6$ .
Solving those simultaneously gives $P = (4, - 1)$ .
How do you build a transformation matrix from the unit square?
Find where $A (1, 0)$ and $C (0, 1)$ move to, then use those two images as the columns of the matrix.
The origin never moves, so those two points alone are enough to fix the whole matrix.
Complete the matrix for a rotation of $90 °$ anticlockwise about the origin:
$(\begin{matrix} 0 & ______ \\ 1 & 0 \end{matrix})$
The completed matrix is:
$(\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$
$A (1, 0)$ moves to $(0, 1)$ and $C (0, 1)$ moves to $(- 1, 0)$ , giving those two columns.
What matrix represents a rotation of $180 °$ about the origin?
The matrix is $(\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix})$ , which is $- I$ .
Both $A$ and $C$ end up on the negative axes, so both columns are negated.
What matrix represents a rotation of $270 °$ anticlockwise about the origin?
The matrix is $(\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})$ , since $270 °$ anticlockwise is the same as $90 °$ clockwise.
$A$ moves to $(0, - 1)$ and $C$ moves to $(1, 0)$ .
Complete the reflection matrices:
Reflection in the $x$ -axis: $(\begin{matrix} 1 & 0 \\ 0 & ______ \end{matrix})$
Reflection in the $y$ -axis: $(\begin{matrix} ______ & 0 \\ 0 & 1 \end{matrix})$
The completed matrices are:
Reflection in the $x$ -axis: $(\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$
Reflection in the $y$ -axis: $(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix})$
Reflecting in an axis leaves the point on that axis where it is and negates the other one.
What matrices represent reflection in $y = x$ and in $y = - x$ ?
Reflection in $y = x$ is $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ , which swaps the coordinates over.
Reflection in $y = - x$ is $(\begin{matrix} 0 & - 1 \\ - 1 & 0 \end{matrix})$ , which swaps them and changes both signs.
True or False?
$(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ is the identity matrix.
False.
The 1s are on the wrong diagonal, so this reflects in $y = x$ rather than leaving points alone.
The identity has its 1s running from top-left to bottom-right.
What matrix represents an enlargement of scale factor $k$ about the origin?
The matrix is $(\begin{matrix} k & 0 \\ 0 & k \end{matrix})$ , which is $k I$ .
$A$ moves to $(k, 0)$ and $C$ to $(0, k)$ , and negative scale factors work the same way.
Describe the transformation given by $(\begin{matrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{matrix})$ .
The matrix is $\frac{1}{4} I$ , so it represents an enlargement of scale factor $\frac{1}{4}$ about the origin.
Recognising a multiple of the identity is what tells you a transformation is an enlargement.
Complete the combined-transformation rule:
If a point is transformed first by $P$ and then by $Q$ , the single matrix is $M =______$ .
The completed rule is:
If a point is transformed first by $P$ and then by $Q$ , the single matrix is $M = QP$ .
The first transformation is written last, so the order is the reverse of what you might expect.
True or False?
For combined transformations it does not matter which order you multiply the matrices.
False.
Matrix multiplication is not commutative, so $QP$ and $PQ$ generally describe different transformations.
Getting the order wrong sends the point somewhere else entirely, rather than just looking different.
$B$ reflects in the $y$ -axis and $C$ reflects in the $x$ -axis. What single matrix is $B$ followed by $C$ ?
The single matrix is $CB$ , and multiplying gives $(\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix})$ .
Note the order: $C$ comes first in the product even though it acts second.
What single transformation is a reflection in the $y$ -axis followed by one in the $x$ -axis?
The result is a rotation of $180 °$ about the origin, which is also an enlargement of scale factor $- 1$ .
Both descriptions correspond to the same matrix, $- I$ .
Why does a combined transformation come from multiplying rather than adding?
Because each transformation is applied by multiplying the point, so doing two of them means multiplying twice.
Writing $Q (P v)$ as $(QP) v$ is exactly what associativity allows.
How can you check a claimed equivalence between transformations?
Write each transformation as a matrix and multiply them in the correct order.
If the product equals the matrix for the single transformation claimed, the two really are the same.

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