Lesson 4 – Solving Simultaneous equations 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

Invariant points 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 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.