What does a negative determinant tell you about a transformation?

The orientation of the shape has been reversed, which is what a reflection does. The change in size still depends only on how big the determinant is, so the minus sign carries information about orientation rather than about area.

When a 3D rotation is described as anticlockwise about an axis, from where is that direction being judged?

From a viewpoint out on the positive part of that axis, looking back towards the origin. Without that convention the very same rotation could equally be called clockwise, which is why a 3D rotation is always described by naming both the axis and the direction.

Define a line of invariant points .

A line of invariant points is a line on which every single point is mapped to itself. The mirror line of a reflection is the standard example.

Why is it easy to miss that the two invariant-point equations are really the same equation?

One of them can be an awkward multiple of the other, so the coefficients look unrelated until you divide through. Comparing the ratios of the coefficients is the quickest way to see that the two equations carry the same information.

What is the difference between a line of invariant points and an invariant line ?

On a line of invariant points every point stays exactly where it is. On an invariant line each point is allowed to move, but it must land somewhere else on the same line , so it is only the line as a whole that is unchanged.

A LevelFurther MathsEdexcelCore PureFlashcardsMatricesTransformations using Matrices

Transformations using Matrices (Edexcel A Level Further Maths: Core Pure): Flashcards

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FrontTransformations using a Matrix
How is the image of the point $(x, y)$ under a transformation matrix found?

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How is the image of the point $(x, y)$ under a transformation matrix found?
Write the point as the position vector $[\begin{matrix} x \\ y \end{matrix}]$ and multiply it on the left by the transformation matrix.
So $[\begin{matrix} x' \\ y' \end{matrix}] = [\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} x \\ y \end{matrix}]$ gives the coordinates of the image.
How can a whole shape be transformed in a single multiplication?
Put the vertices side by side as the columns of a position matrix, then multiply that on the left by the transformation matrix.
Each column of the answer is the image of the corresponding vertex, so a quadrilateral needs a $2 \times 4$ position matrix.
The image of a point under a transformation matrix $T$ is known. How do you get back to the original point?
Multiply the image by the inverse matrix $T^{- 1}$ , since that undoes whatever $T$ did.
So from $[\begin{matrix} x' \\ y' \end{matrix}] = T [\begin{matrix} x \\ y \end{matrix}]$ you get $[\begin{matrix} x \\ y \end{matrix}] = T^{- 1} [\begin{matrix} x' \\ y' \end{matrix}]$ .
Complete the geometric meaning of the determinant of a transformation matrix:
For a $2 \times 2$ matrix the area of the image is the area of the object multiplied by the $______$ value of the determinant, and for a $3 \times 3$ matrix that same number is the $______$ scale factor instead.
The completed statement is: for a $2 \times 2$ matrix the area of the image is the area of the object multiplied by the absolute value of the determinant, and for a $3 \times 3$ matrix that same number is the volume scale factor instead.
In symbols this is $area of image = |\det A| \times area of object$ .
What does a negative determinant tell you about a transformation?
The orientation of the shape has been reversed, which is what a reflection does.
The change in size still depends only on how big the determinant is, so the minus sign carries information about orientation rather than about area.
True or False?
A transformation whose matrix is singular can still be reversed.
False.
A singular matrix has determinant $0$ and therefore no inverse, and the inverse is the only thing that reverses a transformation.
Its area scale factor is $0$ as well, so the image of any shape is squashed down to no area at all.
The determinant of a $2 \times 2$ transformation matrix has size less than $1$ . What happens to areas?
Every area is reduced, because the area scale factor is the size of the determinant and that is less than $1$ .
A determinant whose size is greater than $1$ enlarges areas in exactly the same way.
A transformation matrix contains an unknown, and the image area is known to be twice the object area. How is the unknown found?
Set the absolute value of the determinant equal to $2$ and solve the equation that results.
Because it is an absolute value, that splits into two cases, $\det = 2$ and $\det = - 2$ , so there can be more than one valid answer.
How can a $2 \times 2$ transformation matrix be found by looking at where just two points go?
Find the images of $(1, 0)$ and $(0, 1)$ : they become the first and second columns of the matrix.
So if $(1, 0)$ maps to $(a, c)$ and $(0, 1)$ maps to $(b, d)$ , the matrix is $[\begin{matrix} a & b \\ c & d \end{matrix}]$ .
Complete the $2 \times 2$ matrices for an enlargement and for a stretch, both with scale factor $k$ :
$enlargement : [\begin{matrix} k & 0 \\ 0 & ______ \end{matrix}] and stretch parallel to the x -axis : [\begin{matrix} ______ & 0 \\ 0 & 1 \end{matrix}]$
The completed matrices are:
$enlargement : [\begin{matrix} k & 0 \\ 0 & k \end{matrix}] and stretch parallel to the x -axis : [\begin{matrix} k & 0 \\ 0 & 1 \end{matrix}]$
An enlargement scales both directions, while a stretch scales only one, which is why the other entry stays at $1$ .
A reflection in the line $y = (\tan θ) x$ has matrix $[\begin{matrix} \cos 2 θ & \sin 2 θ \\ \sin 2 θ & - \cos 2 θ \end{matrix}]$ . Which angle gives a reflection in the $y$ -axis, and what matrix results?
Take $θ = 90 °$ , since that is the angle the $y$ -axis makes with the positive $x$ -axis.
Then $\cos 2 θ = - 1$ and $\sin 2 θ = 0$ , giving the matrix $[\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}]$ .
An anticlockwise rotation of $θ$ about the origin has matrix $[\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ . What is the matrix for a clockwise rotation of the same angle?
Replace $θ$ by $- θ$ , which gives $[\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$ .
Only the two sine entries change sign, because cosine is unaffected by a change of sign in the angle.
A transformation with matrix $M$ is applied first and then one with matrix $N$ . What single matrix does the same job?
The single matrix is $N M$ , with the transformation done first written on the right.
The matrices are applied right to left, exactly as composite functions are, because each new transformation multiplies on the left of what is already there.
True or False?
A reflection followed by a rotation gives the same image as that rotation followed by the same reflection.
False.
Matrix multiplication is not commutative, so $N M$ and $M N$ are usually different matrices.
The order in which the two transformations are carried out therefore changes where the shape ends up.
A transformation with matrix $T$ is applied five times over. What single matrix represents that, and what has happened to areas?
The combined matrix is $T^{5}$ .
Each application multiplies the area by $|\det T|$ , so after five applications the original area has been multiplied by ${|\det T|}^{5}$ .
How does transforming a point in 3D differ from transforming one in 2D?
The position vector has three entries rather than two, so the transformation matrix is $3 \times 3$ rather than $2 \times 2$ .
The method itself is unchanged: multiply the position vector, or a position matrix holding several points, on the left by the transformation matrix.
Complete the $3 \times 3$ matrices for reflections in two of the coordinate planes:
$in x = 0 : [\begin{matrix} ______ & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] and in z = 0 : [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & ______ \end{matrix}]$
The completed matrices are:
$in x = 0 : [\begin{matrix} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] and in z = 0 : [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}]$
Only the direction perpendicular to the mirror plane is reversed, so exactly one $1$ on the leading diagonal turns into $- 1$ .
What does the matrix for an anticlockwise rotation about the $z$ -axis look like, and why does it take that shape?
It is $[\begin{matrix} \cos θ & - \sin θ & 0 \\ \sin θ & \cos θ & 0 \\ 0 & 0 & 1 \end{matrix}]$ , with the $2 \times 2$ rotation matrix sitting in the top left corner.
A rotation about the $z$ -axis leaves the $z$ coordinate alone, which is why the last row and last column are those of the identity matrix.
When a 3D rotation is described as anticlockwise about an axis, from where is that direction being judged?
From a viewpoint out on the positive part of that axis, looking back towards the origin.
Without that convention the very same rotation could equally be called clockwise, which is why a 3D rotation is always described by naming both the axis and the direction.
Define an invariant point of a transformation.
An invariant point is one that is mapped to itself by the transformation, so it stays exactly where it started.
In matrix form that means the position vector $x$ satisfies $T x = x$ .
For a transformation with matrix $[\begin{matrix} a & b \\ c & d \end{matrix}]$ how do you find the invariant points?
Set $[\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} x \\ y \end{matrix}]$ and compare the two rows.
That gives the simultaneous equations $(a - 1) x + b y = 0$ and $c x + (d - 1) y = 0$ , which are then solved for $x$ and $y$ .
True or False?
Every linear transformation of the plane leaves the origin invariant.
True.
Multiplying any matrix by $[\begin{matrix} 0 \\ 0 \end{matrix}]$ gives $[\begin{matrix} 0 \\ 0 \end{matrix}]$ straight back.
So the origin always comes out as a solution, and the real question is whether there are any others.
Define a line of invariant points.
A line of invariant points is a line on which every single point is mapped to itself.
The mirror line of a reflection is the standard example.
When you solve the invariant-point equations, what tells you there is a whole line of invariant points rather than just one?
The two equations reduce to the same equation, so instead of a single solution there are infinitely many.
That one remaining equation is the equation of the line, such as $y = 3 x$ .
Why is it easy to miss that the two invariant-point equations are really the same equation?
One of them can be an awkward multiple of the other, so the coefficients look unrelated until you divide through.
Comparing the ratios of the coefficients is the quickest way to see that the two equations carry the same information.
What is the difference between a line of invariant points and an invariant line?
On a line of invariant points every point stays exactly where it is.
On an invariant line each point is allowed to move, but it must land somewhere else on the same line, so it is only the line as a whole that is unchanged.
When hunting for an invariant line $y = m x + c$ , complete the general point and its image:
$[\begin{matrix} x \\ ______ \end{matrix}] maps to [\begin{matrix} x' \\ ______ \end{matrix}]$
The completed pair is:
$[\begin{matrix} x \\ m x + c \end{matrix}] maps to [\begin{matrix} x' \\ m x' + c \end{matrix}]$
The values of $m$ and $c$ are the same in both, because the image is a different point lying on the same line.
Once the general point and its image have been set up, how do you actually find $m$ and $c$ ?
Multiply out both rows, then use the first row to eliminate $x'$ from the second.
What is left is an equation in $x$ alone, and comparing the coefficients of $x$ and the constant terms gives equations for $m$ and $c$ .
There may be more than one solution, and each one is a separate invariant line.

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