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Readings for this topic (Recommended in bold)¶

Pre-Class Assignment: Transformation Matrix¶

Assignment Overview¶

The Inverse Matrix
Transformation Matrix
Assignment wrap-up

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:

$(A^{-1})^{-1} = A$
$(cA)^{-1} = \frac{1}{c}A^{-1}$
$(AB)^{-1} = B^{-1}A^{-1}$
$(A^n)^{-1} = (A^{-1})^n$
$(A^\top)^{-1} = (A^{-1})^\top$ 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

In [1]:

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)

Out[1]:

$\displaystyle \left[\begin{matrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 7\end{matrix}\right]$

In [2]:

#put your answer to the above question here.

Lets check your answer by multiplying A by A_inv.

In [4]:

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 $A$?

put your answer to the above question here

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

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] $$ |

In [6]:

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.

In [7]:

E3*E2*E1*A

Out[7]:

matrix([[1., 0.],
        [0., 1.]])

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

In [8]:

# Put your answer to the above question here.

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

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

Put your answer to the above question here.

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

Put your answer to the above question here.

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

Put your answer to the above question here.

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

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

Put your answer to the above question here.

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

Put your answer to the above question here.

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

$4 \times 6$
$6 \times 4$
$3 \times 6$
$5 \times 3$

Put your answer to the above question here.

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

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

Put your answer to the above question here.

2. Transformation Matrix¶

Consider the following set of points:

In [10]:

%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');

Out[10]:

$\displaystyle \left( -0.5, \ 10.5, \ -4.7, \ 10.7\right)$

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:

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

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

In [11]:

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

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

In [12]:

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

Out[12]:

$\displaystyle \left[\begin{matrix}0.707106781186548 & -0.707106781186547\\0.707106781186547 & 0.707106781186548\end{matrix}\right]$

In [13]:

p=R*points

plt.plot(p[0].T,p[1].T);
plt.axis('equal');

#print(p[0].T)

Out[13]:

$\displaystyle \left( -6.363961030678928, \ 9.19238815542512, \ -2.121320343559643, \ 13.435028842544403\right)$

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

In [14]:

# Apply R and plot 8 times
for i in range(0,8):
    p = R * p
    plt.plot(p[0].T,p[1].T);

plt.axis('equal');

Out[14]:

$\displaystyle \left( -14.000714267493642, \ 14.000714267493642, \ -14.000714267493638, \ 14.000714267493642\right)$

✅ QUESTION: In the above code what does the T call in p[0].T do?

Put your answer here

3. Assignment wrap-up¶

Please fill out the form that appears when you run the code below. You must completely fill this out in order to receive credit for the assignment!

Direct Link to Google Form

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