chapter4.md

title	description
Basic Matrix Operations	undefined

Matrix Equality

type: TabExercise

xp: NaN

key: 5bcf5ade9a

In these exercises we will explore the concept of matrix equality and how it is implemented in Python.

Two matrices are equal if all three of the following conditions are met:

• Each matrix has the same number of rows.
• Each matrix has the same number of columns.
• Corresponding elements within each matrix are equal.

---

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...

Is this the same as the mathematical definition of matrix equality?

@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! Note that this is not the same as the mathematical definition of matrix equality. Re-read the defintion above to see why.')

---

type: NormalExercise

xp: NaN

key: 1553b7299f

@instructions This example uses the np.isEqual() method to compare two matrices.

• run the script and examine the output
• modify the array $b$ so that np.equal(a,b) is TRUE

@hint

@sample_code

import numpy as np

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

is_Equal =  np.equal(a,b)

print( is_Equal )

@solution

import numpy as np

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

is_Equal =  np.equal(a,b)

print( is_Equal )

@sct

Ex().check_object('is_Equal').has_equal_value()
success_msg('Great job! This is the same as the mathematical definition matrix equality.')

---

Transpose of a Matrix

type: TabExercise
key: 256c8afbac
lang: python
xp: 100

@pre_exercise_code

@sample_code

---

Sub Heading

type: NormalExercise
xp: 100
key: b805f1f556

@instructions

Use the transpose method on $a$

@sample_code

## Load numpy on first use
import numpy as np

## Define an array A
a = np.array([[1, 2], [3, 4]])
print(a)

b = ___.___()
print(b)

@solution

## Load numpy on first use
import numpy as np

## Define an array A
a = np.array([[1, 2], [3, 4]])
print(a)

b = a.transpose()
print(b)

@hint

@sct

---

Sub Heading 2

type: NormalExercise
xp: 100
key: 1291abb827

@instructions Alternatively, you can use the

@sample_code

## Load numpy on first use
import numpy as np

## Define an array A
a = np.array([[1, 2], [3, 4]])
print(a)

b = ___.___()
print(b)

@solution

## Load numpy on first use
import numpy as np

## Define an array A
a = np.array([[1, 2], [3, 4]])
print(a)

b = a.T
print(b)

@hint

@sct

---

Solving Matrix Equations

type: TabExercise
key: 25e0f1445a
lang: python
xp: 100

@pre_exercise_code

@sample_code

---

The long way

type: NormalExercise
lang: python
xp: 100
skills: 2
key: c5511f767e

Online resources:

- https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/matrices-elimination/v/matrices-reduced-row-echelon-form-1

Consider a set of linear equations:

\begin{equation*} \begin{array}{cccccc} a_{11}x_1+&a_{12}x_2+&\ldots&+a_{1n}x_n&=&b_1\ a_{21}x_1+&a_{22}x_2+&\ldots&+a_{2n}x_n&=&b_2\ &\vdots&&\vdots&=&\vdots\ a_{n1}x_1+&a_{n2}x_2+&\ldots&+a_{nn}x_n&=&b_n \end{array} \end{equation*}

Which we can write as $\mathbf{A}\mathbf{x}=\mathbf{b}$. Then $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}$.

Problems of this type can be solved to find x using: np.linalg.solve(A,b)

@instructions 'Solving' matrices is a basic concept in linear algebra.

Let's take a problem $\mathbf{Ax}=\mathbf{b}$.

where the variables are:

$\mathbf{A} = \left[ \begin{array}{ccc} 2& 1 & 1\ 1 & 3 & 0\ 1 & 0 & 0 \end{array} \right]$ , $\mathbf{x} = \left[ \begin{array}{ccc} a\ b\ c \end{array} \right]$ , $\mathbf{b} = \left[ \begin{array}{ccc} 4\ 5\ 6 \end{array} \right]$

This is a standard matrix problem where we want to solve for $\mathbf{x}$.

The standard approach to solve this would be:

$\mathbf{A^{-1}Ax}=\mathbf{x}=\mathbf{A^{-1}b}$

So the solution for $\mathbf{x}$ can be found by taking the inverse of $\mathbf{A}$ and dotting it with $\mathbf{b}$.

@hint

@sample_code

import numpy as np

# define matrix A using Numpy arrays 
A = np.array([[2, 1, 1], [1, 3, 2], [1, 0, 0]]) 

#define matrix b 
b = np.array([4, 5, 6]) 

x = np.linalg.inv( ___ ).dot( ___ )

print("Solution:\n", x)

@solution

import numpy as np

# define matrix A using Numpy arrays 
A = np.array([[2, 1, 1], [1, 3, 2], [1, 0, 0]]) 

#define matrix b 
b = np.array([4, 5, 6]) 

x = np.linalg.inv(A).dot(b)

print("Solution:\n", x)

@sct

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

---

Using linalg.solve()

type: NormalExercise

xp: 100

key: 632f323c4b

Whilst this can be implemented simply in Python, taking the inverse of matrices is computationally expensive and may not be feasible.

A range of methods exist for solving these types of problem - the most standard approach is to use x = np.linalg.solve(A, b). This is much more efficient.

@instructions Solve the matrix equation $\mathbf{Ax}=\mathbf{b}$ for $\mathbf{x}$ using lining.solve()

@hint You can look up the help page by typing ?np.linalg.solve() in the iPython shell

@pre_exercise_code

import numpy as np

@sample_code

# define matrix A using Numpy arrays 
A = np.array([[2, 1, 1], [1, 3, 2], [1, 0, 0]]) 

#define matrix b 
b = np.array([4, 5, 6]) 

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

# linalg.solve is the function of NumPy to solve a system of linear scalar equations 
print("Solution:\n", x )

@solution

# define matrix A using Numpy arrays 
A = np.array([[2, 1, 1], [1, 3, 2], [1, 0, 0]]) 

#define matrix b 
b = np.array([4, 5, 6]) 

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

# linalg.solve is the function of NumPy to solve a system of linear scalar equations 
print("Solution:\n", x )

@sct

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