Invariation Questions
Slides in this set (7)
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- 1. M[x][y] = [x][y] , and M is [1 0][4 1] . Find the line of invariant points.
- 2. M[x][y] = [x][y] , and M is [-1 5][4 -9] . Find the line of invariant points.
- 3. Find an invariant line of the matrix [10 -6][-6 5]
- 4a. Show that the transformation [-4 1][3 -2] has no lines of invariant points
- b. Find equations of invariant lines through the origin
- 5. Show y=-1/2x+3 is an invariant line for the reflection in the line y=2x
- 6. Show the matrix [4 0][0 4] has one invariant point an an infinite number of invariant lines through the origin
- 7. Find all invariant points and lines of the transformation [2 -1][-1 2]
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- 1. M[x][y] = [x][y] , and M is [1 0][4 1] . Find the line of invariant points.
- [1 0][4 1] x [x][y] = [x][4x+y] , and the question tells us this must equal [x][y] , and we can say 4x+y=y, so 4x=0 and therefore x must equal 0. This means the line of invariant points is x=0.
- 2. 2. M[x][y] = [x][y] , and M is [-1 5][4 -9] . Find the line of invariant points.
- [-1 5][4 9] x [x][y] = [-x+5y][4x-9y] and the questions tells us these both must equal x and y, so we can say -x+5y=x and 4x-9y=y. For the equation -x+5y=x , we can rewrite this as 5y=2x (this can be seen easier if you set both equations to equal 0 first) and therefore the line of invariant points is y=2/5x
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- 3. Find an invariant line of the matrix [10 -6][-6 5]
- [10 -6][-6 5] x [x][mx+c] = [10x-6mx+6c][-6x+5mx+5c] , which we sub into y'=mx'+c to get : -6x+5mx+5c=m(10x-6mx+6c)+c. To tidy this up we expand the right side and factor out x when we can to get : x(-6+5m)+5c = x(10m-6m^2)+6mc+c.
- Now we equate the x coefficients to get: -6+5m = 10m-6m^2, which we rearrange to equal 0 to get the quadratic (3m+2)(2m-3), telling us must equal either -2/3 or 3/2.
- Next we equate the constant values to get: 5c=6mc+c. We sub in the m values to get 5c=6(3/2)c+c and 5c=6(-2/3)c+c. For m=3/2 we get 5c=-8m, telling us c=0, and for m=-2/3 we get c=c, telling us y=-2/3 is an invariant line for all c.
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- 4a. Show that the transformation [-4 1][3 -2] has no lines of invariant points
- b. Find equations of invariant lines through the origin
- a). We multiply the matrix [-4 1][3 -2] by [x][y] to get [-4x+y]=[x] and [3x-2y]=[y] and solve as a simultaneous equation (firstly set equal to 0) to get x=0 and y=0. This means (0,0) is an invariant point, but it also means there are no lines of invariant points.
- b). Multiply the matrix [-4 1][3 -2] by [x][mx+c] and sub into y'=mx' (we can ignore c as question states through origin) to get 3x-2mx = m(-4x+mx). Now expand and factor out x wherever possible to get x(3-2m) = x(-4+m^2).
- Equate x coefficients : 3-2m = -4m+m^2 , rearrange to equal 0, so m^2-2m-3=0 , solve quadratic: (m-3)(m+1), so the invariant lines are y=3x and y=-x.
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- 5. Show y=-1/2x+3 is an invariant line for the reflection in the line y=2x
- We know the reflection matrix is [cos2θ sin2θ][sin2θ -cos2θ] and we know y=2x, so tanθ=2. We can do θ=tan-1(2) to get θ=63.4, and then multiply by 2 to get 2θ=126.8. We can sub this into the reflection matrix formula to get [-3/5 4/5][4/5 3/5].
- Now we take the line y=-1/2x+3 and rewrite it as the matrix [x][-1/2x+3] (as y=mx+c can be rewritten as (x,mx+c)). We can multiply our reflection matrix by this to get: [-3/5x-2/5x+12/5][4/5x-3/10x+9/5]. We can simplify this down to [-x+12/5][1/2x+9/5].
- Next we sub this into y'=-1/2x'+3, so : 1/2x+9/5 = -1/2(-x+12/5)+3 , then 1/2x+9/5 = 1/2x-6/5+3, so 1/2x+9/5 = 1/2x+9/5 , therefore y=-1/2x+3 is an invariant line.
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- 6. Show the matrix [4 0][0 4] has one invariant point an an infinite number of invariant lines through the origin
- One variant point: [4 0][0 4] x [x][y] = [4x][4y] . 4x = x , so x must equal 0. 4y = y so y must also equal 0. Therefore the single invariant point is (0,0) .
- An infinite number of invariant lines: As the question states through the origin, we can just multiply the matrix [4 0][0 4] by [x][mx] to get [4x][4mx].
- Next we sub this into y'=mx' , so 4mx = m(4x) . This means 4mx = 4mx and therefore as these are the same y'=mx' meaning all lines through the origin are invariant lines.
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- 7. Find all invariant points and lines of the transformation [2 -1][-1 2]
- Points: [2 -1][-1 2] x [x][y] = [2x-y][-x+2y]. So 2x-y = x and -x+2y=y. These can both be rearranged to get y=x, which is the line of invariant points.
- Lines: [2 -1][-1 2] x [x][mx+c] = [2x-mx-c][-x+2mx+2c]. We sub this into y'=mx'+c to get -x+2mx+2c=m(2x-mx-c)+c. Here we expand and factor out x wherever possible to get: x(-1+2m)+2c=x(2m-m^2)-mc+c.
- Now we equate the x coefficients : -1+2m=2m-m^2 , rearrange to get the quadratic m^2=1 , so therefore m=+/- 1
- Next we equate constant values : 2c=-mc+c , so mc=-c. Now we sub in the values of m, so (1)c=-c, so c=-c therefore c=0. (-1)c=-c , so c=c, which is true for all c. Therefore all lines of y=-x+c are invariant lines.