# Invariant Points

#### Information

An Invariant Point is a point in which even after transformation the point remains in the same place (same value)

An example of this is (When two matrices are placed together with no operator [symbol +-x÷] they are multiplied):

6 | 5 |

2 | 3 |

2 |

-2 |

=

2 |

-2 |

This is an invariant point as the matrix has changed into itself

The origin (point 0,0) will always be an invariant point as no multiplication will change its value

#### Finding Invariant Points

In order to find if a point will be invariant you must find that both:

x = x' and

y = y'

If these two requirments are met then those points are invariant as both the x and y value remain after transformation.

2 | -1 |

1 | 0 |

x |

y |

=

x' |

y' |

From this matrices equation we are seeing what points are variant using this matrix

By multiplying out the matrices you find that:

Equation 1: 2x - y = x'

Equation 2: x = y'

We also know that for the point to be invariant:

Equation 3: x = x'

Equation 4: y = y'

This is because the points must be the same before and after the transformation

Substituting Equation 3 into Equation 1 we find:

x = y

Also Substituting Equation 4 into Equation 2 gives us:

x = y

This means that wherever the value of x is equal to the value of y the point will be invariant we can test by substituting values into the matrix where x = y

For example if x=2 then y=2 such as below:

2 | -1 |

1 | 0 |

2 |

2 |

=

2 |

2 |

When multiplying out the matrices we do find that the same point that goes into the transformation comes out

#### The Relationship between the Determinant and the Area

In order to find the area of a translated shape we first need to know the original area of the shape

For example the area of a 1X1 sqaure has the area of 1 and the plots:

Q(0,0), R(1,0), P(0,1) and S(1,1)

We can represent all these points with the matrix (In the order QRPS e.g.Q is represented by the first column [red]):

0 | 1 | 0 | 1 |

0 | 0 | 1 | 1 |

If we then translate the matrix such as shown below:

3 | 1 |

1 | 2 |

0 | 1 | 0 | 1 |

0 | 0 | 1 | 1 |

=

0 | 3 | 1 | 4 |

0 | 1 | 2 | 3 |

The new locations of the points are represented by a dash after the previous notation e.g. Q becomes Q'

This means the points Q(0,0), R(1,0), P(0,1) and S(1,1)

Have now been moved to Q'(0,0), R'(3,1), P'(1,2) and S'(4,3)

We can then find the absolute value of the determinant which in turn shows us the area of the new shape

(absolute value means the answer is always positive e.g. |-3| = 3)

|det

3 | 1 |

1 | 2 |

| = |(3x2)-(1x1)| = |6-1| = 5

This means the new parallelgram has an area of 5

#### Example Questions

#### Question 1

If points (x,y) are reflected [R] along the line y=x

followed by a translation [T] where (x,y) is translated to (x+1,y+2)

What are the invariant points?

First we need to find what matrices represent R and T

R=

0 | 1 | 0 |

1 | 0 | 0 |

0 | 0 | 1 |

T=

1 | 0 | 1 |

0 | 1 | 2 |

0 | 0 | 1 |

Next we need to multiply these two matrices together in reverse order so we can see what the full transformation will be:

TR=

1 | 0 | 1 |

0 | 1 | 2 |

0 | 0 | 1 |

0 | 1 | 0 |

1 | 0 | 0 |

0 | 0 | 1 |

=

0 | 1 | 1 |

1 | 0 | 2 |

0 | 0 | 1 |

Now we compare the old points multiplied by the transformation and compare them to the new transformed points

TR=

0 | 1 | 1 |

1 | 0 | 2 |

0 | 0 | 1 |

x |

y |

1 |

=

x' |

y' |

1 |

Multiplying the tranformation by the old points we can see what happens to get to the new moved points

y+1 |

x+2 |

1 |

=

x' |

y' |

1 |

Separating these matrices into two equations gives us:

Equation 1: y + 1 = x'

Equation 2: x + 2 = y'

These equations are not the same therefore they can never be equal meaning there are **no** invariant points.

#### Question 2

If points (x,y) are translated [T] where (x,y) is translated to (x+a,y+b)

followed by a reflection [R] where the line of reflection is the line y=x

What are the invariant points?

First we need to find what matrices represent T and R

T=

1 | 0 | a |

0 | 1 | b |

0 | 0 | 1 |

R=

0 | 1 | 0 |

1 | 0 | 0 |

0 | 0 | 1 |

Next we need to multiply these two matrices together in reverse order so we can see what the full transformation will be:

RT=

0 | 1 | 0 |

1 | 0 | 0 |

0 | 0 | 1 |

1 | 0 | a |

0 | 1 | b |

0 | 0 | 1 |

=

0 | 1 | b |

1 | 0 | a |

0 | 0 | 1 |

Now we compare the old points multiplied by the transformation and compare them to the new transformed points

RT=

0 | 1 | b |

1 | 0 | a |

0 | 0 | 1 |

x |

y |

1 |

=

x' |

y' |

1 |

Multiplying the tranformation by the old points we can see what happens to get to the new moved points

y+b |

x+a |

1 |

=

x' |

y' |

1 |

Separating these matrices into two equations gives us:

Equation 1: y + b = x'

Equation 2: x + a = y'

Remembering that x = x' and y = y' for invariant points the equations become:

Equation 1: y + b = x

Equation 2: x + a = y

If we rearrange such that y is the left side of each equation

Equation 3: y = x - b

Equation 4: y = x + a

Replacing y in equation 3 with equation 4 we get:

Equation 5: x + a = x - b

This means that a = -b where there are invariant points

To test this we input numbers where a = -b and y = x + a

In this example we will use:

a = 2

b = -2

x = 1

y = 1 + 2 = 3

Filling these values into the matrix

y+b |

x+a |

1 |

=

x' |

y' |

1 |

Gives us:

3-2 |

1+2 |

1 |

=

x' |

y' |

1 |

1 |

3 |

1 |

=

x' |

y' |

1 |

Separating the Matrices into equations gives:

Equation 1: 1 = x'

Equation 2: 3 = y'

We used the values of x = 1 and y = 3

Replacing these values gives us:

x = x' and y = y' meaning these are invariant points

Therefore where y = x + a and a = -b these are invariant points

#### Question 3 Part a

If points (x,y) are rotated 90° anti-clockwise [R]

followed by a translation [T] where (x,y) becomes (x+1,y+2)

What are the invariant points?

First we need to find what matrices represent T and R

R=

0 | -1 | 0 |

1 | 0 | 0 |

0 | 0 | 1 |

T=

1 | 0 | 1 |

0 | 1 | 2 |

0 | 0 | 1 |

Next we need to multiply these two matrices together in reverse order so we can see what the full transformation will be:

TR=

1 | 0 | 1 |

0 | 1 | 2 |

0 | 0 | 1 |

0 | -1 | 0 |

1 | 0 | 0 |

0 | 0 | 1 |

=

0 | -1 | 1 |

1 | 0 | 2 |

0 | 0 | 1 |

#### Question 3 Part b

Find the fixed points (invariant points)

Now we compare the old points multiplied by the transformation and compare them to the new transformed points

RT=

0 | -1 | 1 |

1 | 0 | 2 |

0 | 0 | 1 |

x |

y |

1 |

=

x' |

y' |

1 |

Multiplying the tranformation by the old points we can see what happens to get to the new moved points

-y+1 |

x+2 |

1 |

=

x' |

y' |

1 |

Separating these matrices into two equations gives us:

Equation 1: -y + 1 = x'

Equation 2: x + 2 = y'

Remembering that x = x' and y = y' for invariant/fixed points the equations become:

Equation 1: -y + 1 = x

Equation 2: x + 2 = y

If we rearrange such that y is the left side of each equation

Equation 3: y = 1 - x

Equation 4: y = x + 2

Replacing y in equation3 with equation 4 gives:

Equation 5: x + 2 = 1 - x

2x = -1

This means that x = -½

Using Equation 4 we see that:

y = -½ + 2

y = 3/2

x = -1/2 and y = 3/2

#### Question 3 Part b

Find the equation of the image on the line y = 2x - 1?

Using the Equations we found in part b we already know that:

-y + 1 = x'

x + 2 = y'

Therfore solving for y and x as the subject we find:

y = 1 - x'

x = y' - 2

Finally replacing y and x given in the question we find:

(1 - x') = 2(y' - 2) - 1

Solving this comes to:

1 - x' = 2y' - 4 - 1

2y' + x' = 6

This means the new image is 2y' + x' = 6