# 07 Pre-Class Assignment: Transformation Matrix¶

## 1. The Inverse Matrix (aka $$A^{-1}$$)¶

For some (not all) square matrices $$A$$, there exists a special matrix called the Inverse Matrix, which is typically written as $$A^{-1}$$ and when multiplied by $$A$$ results in the identity matrix $$I$$:

$A^{-1}A = AA^{-1} = I$

Some properties of an Inverse Matrix include:

1. $$(A^{-1})^{-1} = A$$

2. $$(cA)^{-1} = \frac{1}{c}A^{-1}$$

3. $$(AB)^{-1} = B^{-1}A^{-1}$$

4. $$(A^n)^{-1} = (A^{-1})^n$$

5. $$(A^\top)^{-1} = (A^{-1})^\top$$ here $$A^\top$$ is the tranpose of the matrix $$A$$.

If you know that $$A^{-1}$$ is an inverse matrix to $$A$$, then solving $$Ax=b$$ is simple, just multiply both sides of the equation by $$A^{-1}$$ and you get:

$A^{-1}Ax = A^{-1}b$

If we apply the definition of the inverse matrix from above we can reduce the equation to:

$Ix = A^{-1}b$

We know $$I$$ times $$x$$ is just $$x$$ (definition of the identity matrix), so this further reduces to:

$x = A^{-1}b$

To conclude, solving $$Ax=b$$ when you know $$A^{-1}$$ is really simple. All you need to do is multiply $$A^{-1}$$ by $$b$$ and you know $$x$$.

DO THIS: Find a Python numpy command that will calculate the inverse of a matrix and use it invert the following matrix A. Store the inverse in a new matirx named A_inv

import numpy as np
import sympy as sym
sym.init_printing(use_unicode=True) # Trick to make matrixes look nice in jupyter

A = np.matrix([[1, 2, 3], [4, 5, 6], [7,8,7]])

sym.Matrix(A)

---------------------------------------------------------------------------
ModuleNotFoundError                       Traceback (most recent call last)
<ipython-input-1-f30c35d3bf8d> in <module>
1 import numpy as np
----> 2 import sympy as sym
3 sym.init_printing(use_unicode=True) # Trick to make matrixes look nice in jupyter
4
5 A = np.matrix([[1, 2, 3], [4, 5, 6], [7,8,7]])

ModuleNotFoundError: No module named 'sympy'

#put your answer to the above question here.


Lets check your answer by multiplying A by A_inv.

A * A_inv

np.allclose(A*A_inv, [[1,0,0],[0,1,0],[0,0,1]])


QUESTION: What function did you use to find the inverse of matrix $$A$$?

### How do we create an inverse matrix?¶

From previous assignments, we learned that we could string together a bunch of Elementary Row Operations to get matrix ($$A$$) into it’s Reduced Row Echelon form. We now know that we can represent Elementary Row Operations as a sequence of Elementary Matrices as follows:

$E_n \dots E_3 E_2 E_1 A = RREF$

If $$A$$ reduces to the identity matrix (i.e. $$A$$ is row equivalent to $$I$$), then $$A$$ has an inverse and its inverse is just all of the Elementary Matrices multiplied together:

$A^{-1} = E_n \dots E_3 E_2 E_1$

Consider the following matrix.
$$$A = \left[ \begin{matrix} 1 & 2 \\ 4 & 6 \end{matrix} \right]$$$

A = np.matrix([[1, 2], [4,6]])


It can be reduced into an identity matrix using the following elementary operators

Words

Elementary Matrix

Adding -4 times row 1 to row 2.

$$$E_1 = \left[\begin{matrix}1 & 0 \\ -4 & 1 \end{matrix}\right]$$$

Adding row 2 to row 1.



E_2 = \left[ \begin{matrix} 1 & 1 \ 0 & 1 \end{matrix} \right] $$$| | Multiplying row 2 by$$-\frac{1}{2}$$.|$$$$E_3 = \left[ \begin{matrix} 1 & 0 \\ 0 & -\frac{1}{2} \end{matrix} \right]$$$ |

E1 = np.matrix([[1,0], [-4,1]])
E2 = np.matrix([[1,1], [0,1]])
E3 = np.matrix([[1,0],[0,-1/2]])


We can just check that the statment seems to be true by multiplying everything out.

E3*E2*E1*A


DO THIS: Combine the above elementary Matrices to make an inverse matrix named A_inv

# Put your answer to the above question here.


DO THIS: Verify that A_inv is an actual inverse and chech that $$AA^{-1} = I$$.

# Put your code here.


QUESTION: Is an invertible matrix is always square? Why or why not?

QUESTION: Is a square matrix always invertible? Why or why not?

QUESTION: Describe the Reduced Row Echelon form of a square, invertible matrix.

QUESTION: Is the following matrix in the Reduced Row Echelon form?

$\begin{split} \left[ \begin{matrix} 1 & 2 & 0 & 3 & 0 & 4 \\ 0 & 0 & 1 & 3 & 0 & 7 \\ 0 & 0 & 0 & 0 & 1 & 6 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right] \end{split}$

QUESTION: If the matrix shown above is not in Reduced Row Echelon form. Name a rule that is violated?

QUESTION: What is the size of the matrix described in the previous QUESTION?

• $$4 \times 6$$

• $$6 \times 4$$

• $$3 \times 6$$

• $$5 \times 3$$

QUESTION: Describe the elementary row operation that is implemented by the following matrix

$\begin{split} \left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \end{split}$

## 2. Transformation Matrix¶

Consider the following set of points:

%matplotlib inline
import matplotlib.pylab as plt

x = [0.0,  0.0,  2.0,  8.0, 10.0, 10.0, 8.0, 4.0, 3.0, 3.0, 4.0, 6.0, 7.0, 7.0, 10.0,
10.0,  8.0,  2.0, 0.0, 0.0, 2.0, 6.0, 7.0,  7.0,  6.0,  4.0,  3.0, 3.0, 0.0]
y = [0.0, -2.0, -4.0, -4.0, -2.0,  2.0, 4.0, 4.0, 5.0, 7.0, 8.0, 8.0, 7.0, 6.0,  6.0,
8.0, 10.0, 10.0, 8.0, 4.0, 2.0, 2.0, 1.0, -1.0, -2.0, -2.0, -1.0, 0.0, 0.0]

plt.plot(x,y, color='green');
plt.axis('equal');


We can rotate these points around the origin by using the following simple set of equations:

$x \cos(\theta) - y \sin(\theta) = x_{rotated}$
$x \sin(\theta) + y \cos(\theta) = y_{rotated}$

This can be rewritten as the following system of matrices:

$\begin{split} \left[ \begin{matrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] = \left[ \begin{matrix} x_{rotated}\\ y_{rotated} \end{matrix} \right] \end{split}$

We can rotate the points around the origin by $$\pi/4$$ (i.e. $$45^o$$} as follows:

import numpy as np
import sympy as sym
sym.init_printing(use_unicode=True) # Trick to make matrices look nice in jupyter

points = np.matrix([x,y])

angle = np.pi/4
R = np.matrix([[np.cos(angle), -np.sin(angle)], [np.sin(angle), np.cos(angle)]]);
sym.Matrix(R)

p=R*points

plt.plot(p.T,p.T);
plt.axis('equal');

#print(p.T)


We can even have a little fun and keep applying the same rotation over and over again.

# Apply R and plot 8 times
for i in range(0,8):
p = R * p
plt.plot(p.T,p.T);

plt.axis('equal');


QUESTION: In the above code what does the T call in p.T do?

## 3. Assignment wrap-up¶

Assignment-Specific QUESTION: If the matrix shown above is not in Reduced Row Echelon form. Name a rule that is violated?

QUESTION: Summarize what you did in this assignment.

QUESTION: What questions do you have, if any, about any of the topics discussed in this assignment after working through the jupyter notebook?

QUESTION: How well do you feel this assignment helped you to achieve a better understanding of the above mentioned topic(s)?

QUESTION: What was the most challenging part of this assignment for you?

QUESTION: What was the least challenging part of this assignment for you?

QUESTION: What kind of additional questions or support, if any, do you feel you need to have a better understanding of the content in this assignment?

Written by Dr. Dirk Colbry, Michigan State University 