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Further Pure 1. Lesson 4 – Solving Simultaneous equations. Simultaneous equations. 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.
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Further Pure 1 Lesson 4 – Solving Simultaneous equations
Simultaneous equations • 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 • Solve the following simultaneous equations using matrices. 5x + 4y = 22 3x + 5y = 21 • First write the equations using matrix multiplication. • Now if you multiply the left hand side by the inverse of the matrix then:
Simultaneous equations • The inverse of M is • In matrix form • x = 2, & y = 3
Simultaneous equations • 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.
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.
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.
Example 1 • Solve the following simultaneous equations using matrices. 3x – y = 4 (1) 6x – 2y = 8 (2) • First write the equations using matrix multiplication. • 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 • Solve the following simultaneous equations using matrices. 3x – y = 4 (1) 6x – 2y = 12 (2) • First write the equations using matrix multiplication. • 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 • 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: • 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 • The inverse of M is • Notice that for this particular example the determinant came out to be 1. ( you can see this as the answers are whole numbers). • So • Therefore x = 3, y = -1 & z = -2
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 anti-clockwise? (0,0) has mapped to itself. This is known as an invariant point under the transformation T. Invariant points
Invariant points • 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. • In a reflection, which points map to themselves?
Example • Find the invariant points 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. • This gives us 2x – y = x x = y • Both equations give y = x • So any point on the line y = x will map to itself under T.