Download FP1 simultaneous equations lesson 4

Further Pure 1
Lesson 4 –
Solving Simultaneous equations
Simultaneous equations
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 You have already learnt how to solve
simultaneous equations at GCSE and AS level.
 We can now add a new method using Matrices.
 You can use matrices to solve n number of
simultaneous equations with n unknowns.
 The larger n the more useful matrices become.
 Have a look at the example below.
3x + 2y + z = 5
x + 4y – 2z = 3
2x + y + z = 3
 We will solve this later on in the lesson.
Simultaneous equations
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 Solve the following simultaneous equations
using matrices.
5x + 4y = 22
3x + 5y = 21
 First write the equations using matrix
multiplication.  5 4  x   22 

    
 3 5  y   21
 Now if you multiply the
left hand side by the
inverse of the matrix then:
 x   x1 
M    
 y   y1 
1  x 
1  x1 
M M   M  
y
 y1 
x
1  x 1 
   M  
y
 y1 
Simultaneous equations
 The inverse of M is
1  5  4


M 
13   3 5 
1
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Re member
1  d  c

M  
  b a 
1
 In matrix form
x
1  5
  

 y  13   3
 x = 2, & y = 3
 4  22 
1  26   2 

 

   
5  21 13  39   3 
Simultaneous equations
Wiltshire
 When you solve simultaneous equations in
2D there are 3 possibilities.



There are two lines that meet at one unique
point.
There are two parallel lines that never meet.
There are two parallel lines that overlap.
Simultaneous equations
 There are two lines that
meet at one unique point.
 Here the det = 0 and the
matrix will be non-singular.
 This means that an inverse
exists.
 There will be one solution
as shown.
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Simultaneous equations
 There are two parallel lines
that never meet.
 Here the det = 0 and the
matrix will be singular.
 This means that an inverse
does not exist.
 There will be no solution
as shown.
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Simultaneous equations
 There are two parallel lines
that overlap.
 Here the det = 0 and the
matrix will also be singular.
 This means that an inverse
does not exist.
 However you can see that
in this case there will be
infinitely many solutions as
shown.
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Example 1
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 Solve the following simultaneous equations using matrices.
3x – y = 4
(1)
6x – 2y = 8
(2)
 First write the equations using matrix multiplication.
 3  1  x   4 

    
 6  2  y   8 
 From this we can see that the determinant of the matrix is equal





to zero.
This tells us that either the lines are distinct parallel or they
overlap.
If you re-arrange (1) you get
y = 3x – 4
If you re-arrange (2) you get
y = 3x – 4
From this you can see that the lines overlap.
There are infinitely many solutions
Let x = λ, then y = 3λ - 4
Example 2
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 Solve the following simultaneous equations using matrices.
3x – y = 4
(1)
6x – 2y = 12
(2)
 First write the equations using matrix multiplication.
 3  1  x   4 

    
 6  2  y  12 
 From this we can see that the determinant of the matrix is equal to




zero.
This tells us that either the lines are distinct parallel or they
overlap.
If you re-arrange (1) you get
y = 3x – 4
If you re-arrange (2) you get
y = 3x – 6
From this you can see that they are distinct parallel and that no
solution exists.
Three Simultaneous equations
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 Now lets look at the example from the beginning of the
lesson with the 3 equations.
3x + 2y + z = 5
x + 4y – 2z = 3
2x + y + z = 3
 This can be written in matrix form like so:
 3 2 1  x   5 

   
 1 4  2  y    3 
 2 1 1  z   3 

   
 To solve this equation we need to know the inverse of M.
 In FP1 we will not learn the specific method for finding the
inverse of a 3 × 3 matrix.
 However there are questions that are structured to help you
find the inverse of a 3 × 3 matrix in the textbooks and exams.
 For this example I have used a graphical calculator.
Three Simultaneous equations
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 The inverse of M is
 6 1  8


1
M   5 1
7 
  7 1 10 


 Notice that for this particular example the determinant came out
to be 1. ( you can see this as the answers are whole numbers).
 So
x  6
  
y   5
z  7
  
 Therefore
1
1
1
 8  5   3 
  

7  3     1 
 3   2
10 
  

x = 3, y = -1 & z = -2
Invariant points
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 An invariant point is any point that maps to itself under a
transformation.
 Look at the rotation below.
 What has happened to the co-
ordinate (0,0) under the
transformation rotate 90o anticlockwise?
 (0,0) has mapped to itself.
 0 - 1 0   0 

    
 1 0  0   0 
 This is known as an invariant point
under the transformation T.
 0 - 1

T  
1 0 
Invariant points
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 Explain why the origin is always an invariant
point in any transformation that can be
represented by a matrix.
 Because the transformation uses multiplication,
and multiplying by zero is zero.
 There are lots of points that map to themselves
under matrix transformations.
 6 5  2   2 

    
 2 3   2    2 
 In a reflection, which points map to
themselves?
Example
 Find the invariant points





Wiltshire
 2 - 1

T  
1 0 
under the transformation
given by the matrix:
Under this transformation the co-ordinate (x,y) would
map to (x,y).
Write this as a matrix
multiplication.
 2 - 1 x   x 
This gives us
2x – y = x

    
 1 0  y   y 
x=y
 2x  y   x 
Both equations give y = x

   
So any point on the line y = x
 x  y
will map to itself under T.