chapter2.md

title	description
Solving Simultaneous Equation	In this chapter you will explore how to solve sets of simultaneous equations by solving matrices

Solving two simultaneous equations

type: TabExercise

xp: NaN

key: a32c80fcb2

In this exercise we will explore how to visualise and solve two simultaneous equations using numpy.

In the next section, we will go on to explore what it means for two simultaneous equations to have no solution and finally generalise this approach to more than two equations.

numpy has been imported as np and matplotlib.pyplot as plt

@pre_exercise_code

import numpy as np
import matplotlib.pyplot as plt

---

type: NormalExercise

xp: 100

key: a50d12ec7d

@instructions Let's start by looking at what happens if we use the equality operator == to compare two matrices.

Create two matrices...

@hint

@sample_code

import numpy as np

a = np.array( [[1,2],[3,4]] )
b = np.array( [[1,3],[2,4]] )

print( a==b )

@solution

import numpy as np

a = np.array( [[1,2],[3,4]] )
b = np.array( [[1,3],[2,4]] )

print( a==b )

@sct

Ex().check_object('a').has_equal_value()
Ex().check_object('b').has_equal_value()
success_msg('Great job!')

---

type: NormalExercise

xp: 100

key: 0e46483072

@instructions You should see that these two lines intersect near 0.

Now, look at this page: http://dwightreid.com/blog/2015/09/21/python-how-to-solve-simultaneous-equations/

which I found by googling "python solution to simultaneous equations".

Use this linear algebra method to solve the equations above

@hint

@sample_code

A = np.array([[-5,1], [2,5]])
B = np.array([-2,1])

C = np.linalg.solve(___, ___)

print("Solution:\n",C)

@solution

A = np.array([[-5,1], [2,5]])
B = np.array([-2,1])

C = np.linalg.solve(A, B)

print("Solution:\n",C)

@sct

Ex().check_object('C').has_equal_value()
success_msg('Great job!')

---

type: NormalExercise

xp: 100

key: ac15a94b68

@instructions

Graphical check that you have the correct solution

Now add horizontal and vertical lines corresponding to your solution using plt.axvline() and plt.axhline().

If they intersect with where the two equations cross - you have done it right.

@hint

@sample_code

A = np.array([[-5,1], [2,5]])
B = np.array([-2,1])

C = np.linalg.solve(A, B)

print("Solution:\n",C)


plt.plot(x,y1, label="y1 = 5x-2")
plt.plot(x,y2, label="y2 = 1-2x/5")
plt.xlabel("X")
plt.ylabel("Y")
plt.legend()


plt.axvline(..., ls='--', c='r')
plt.axhline(..., ls='--', c='r')

plt.show()

@solution

A = np.array([[-5,1], [2,5]])
B = np.array([-2,1])

C = np.linalg.solve(A, B)

print("Solution:\n",C)

plt.plot(x,y1, label="y1 = 5x-2")
plt.plot(x,y2, label="y2 = 1-2x/5")
plt.xlabel("X")
plt.ylabel("Y")
plt.legend()


plt.axvline(C[0], ls='--', c='r')
plt.axhline(C[1], ls='--', c='r')

plt.show()

@sct

Ex().check_object('C').has_equal_value()
success_msg('Great job!')

---

Simultaneous equations with no solution

type: TabExercise

xp: NaN

key: e9cb524f11

An important concept for understanding how to solve simultaneous equations is to be able to diagnose when and why equations do not have a solution.

@pre_exercise_code

import numpy as np
import matplotlib.pyplot as plt

---

type: NormalExercise

xp: NaN

key: 8d994fdb1b

@instructions Plot the lines $x+y=1$ and $ x+y=2$ over the range $x \in [-10,10]$ and consider why these simultaneous equations have no solution.

@hint

@sample_code

import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(___,___)
plt.plot( x, 1-x, 'b', x, 2-x, 'g.')

plt.show()

@solution

import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(-2,2)
plt.plot( x, 1-x, 'b', x, 2-x, 'g.')

plt.show()

@sct

Ex().check_object('x').has_equal_value()
success_msg('Great job!')

---

type: NormalExercise

xp: NaN

key: 706f2fe98c

@instructions

@hint

@sample_code

A = np.array([[___,___],[___,___]]) 
b = np.array([[___],[___]])

x = np.linalg.solve(A,b)

print("Solution:\n",x )

@solution

A = np.array([[1,1],[1,1]]) 
b = np.array([[1],[2]])

x = np.linalg.solve(A,b)

print("Solution:\n",x )

@sct

Ex().check_object('x').has_equal_value()
success_msg('Great job!')

---

Geophysical Examples

type: TabExercise

xp: NaN

key: e9645e0fd0

---

type: NormalExercise

xp: NaN

key: be19e96b1c

@instructions In the Measurement Theory course you will have seen that one way in which acceleration (and thus gravity) can be measured is to make measurements of the time and distance of a falling mass.

The height of the mass as a function of time, initial position, initial velocity and acceleration due to gravity is given by:

$ z(t) = z_{0} + v _{0} t + \frac{1}{2} g t^2 $

Let's assume that we do not know the initial height ( $ z_{0} $ ) , the initial velocity ( $ v_{0} $ ) and the acceleration due to gravity ($g$).

However, we have measurements of the position ($z(t)$) at 3 different times.

Time, $t$	Position, $z(t)$
0.1	0.25
0.2	0.5
0.3	0.85

• Insert these values into the equation above to create a set of 3 simultaneous equations.

• From these simultaneous equations, set up the matrix problem as you did in the previous section where you are solving for $ \mathbf{x} = ( z_{0}, v_{0}, g)$

• Now solve for $z_0$, $v_0$ and $g$ in exactly the same way you did in the previous section.

@hint We have 3 equations:

$0.25=Z_o+ 0.1 V_o +\frac{1}{2} g 0.1^2$

$0.50=Z_o+ 0.2 V_o +\frac{1}{2} g 0.2^2$

$0.85=Z_o+ 0.3 V_o +\frac{1}{2} g 0.3^2$

We can write this in the form $\mathbf{Ax}=\mathbf{y}$

$\mathbf{x}=(Z_o, V_o, g)$

$\mathbf{y}=(0.25, 0.50, 0.85)$

$\mathbf{A}= \left[ \begin{array}{ccc} 1& 0.1 & 0.005\ 1 & 0.5 & 0.020\ 1 & 0.3 & 0.045 \end{array} \right]$

@sample_code

import numpy as np

A = np.array([[1,0.1,0.005], [1,0.2,0.020], [1, 0.3, 0.045]])
y = np.array([0.25, 0.50, 0.85])
x = np.linalg.solve(A, y)

print( "Initial height = ", x[0] )
print( "Initial velocity = ", x[1] )
print( "g = ", x[2] )

@solution

import numpy as np

A = np.array([[1,0.1,0.005], [1,0.2,0.020], [1, 0.3, 0.045]])
y = np.array([0.25, 0.50, 0.85])
x = np.linalg.solve(A, y)

print( "Initial height = ", x[0] )
print( "Initial velocity = ", x[1] )
print( "g = ", x[2] )