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    Table of contents

    First published on Monday, Jan 6, 2025 and last modified on Monday, Jan 6, 2025 by Fabienne Chaplais.

    Linear Algebra in the Euclidian Space: Section 3 test Corrected
    2022

    Fabienne Chaplais Mathedu SAS

    Keywords: Linear Systems

    1 Solve a \( 2\times 2\) System

    \paragraph{} Solve the following system by the method of substitution/elimination.

    \[\begin{equation} \left\{ \begin{matrix} 4x&+&2y&=&-11\\ 3x&-&5y&=&-\frac{23}{2} \end{matrix} \right. \end{equation}\]

    (1)

    Make exact calculations with fractions from end to end.

    1.1 Solve the first Equation in \( x\)

    1.1.1 Solution

    The first equation is equivalent to the linear equation in \( x\) :

    \[\begin{equation} 4x=-11-2y \end{equation}\]

    (2)

    Its solution depends on \( y\) and is:

    \( x=\frac{-11-2y}{4}= -\frac{11}{4} -\frac{y}{2} \) .

    Consequently, the solution of the first equation in \( x\) is:

    \[\begin{equation} x= -\frac{11}{4} -\frac{1}{2}y \end{equation}\]

    (3)

    1.2 Replace \( x\) by its Value in the Second Equation

    Tips: Distribute the factor \( 2\) and then multiply both members of the equality by \( 4\) . Don’t hesitate to use the calculator, but only to help fraction computations.

    1.2.1 Solution

    In the second equation, we eliminate \( x\) by replacing it by its value given by the equation 3.

    The second equation is thus equivalent to:

    \[\begin{equation} 3\left(-\frac{11}{4} -\frac{1}{2}y\right) -5y=-\frac{23}{2} \end{equation}\]

    (4)

    If we distribute the factor \( 3\) , we obtain:

    \[\begin{equation} -\frac{33}{4} -\frac{3}{2}y -5y=-\frac{23}{2} \end{equation}\]

    (5)

    If we multiply both members of the equality by \( 3\) , we obtain:

    \[\begin{equation} -33-6y-20y=-46 \end{equation}\]

    (6)

    The equation 6 is equivalent to the affine equation in \( y\) :

    \[\begin{equation} 26y+33=46 \end{equation}\]

    (7)

    Its solution is: \( y=\frac{46-33}{26}=\frac{13}{26}=\frac{1}{2} \) .

    Consequently, the solution of the system 1 in \( y\) is:

    \[\begin{equation} y=\frac{1}{2} \end{equation}\]

    (8)

    1.3 Replace y by its value in in the equation giving \( x\) as a function of \( y\)

    1.3.1 Solution

    In the equation 3, we eliminate \( y\) by replacing it by its value \( 10\) .

    The equation 3 is thus equivalent to:

    \[\begin{equation} x=-\frac{11}{4}-\frac{1}{2}\times\frac{1}{2} \end{equation}\]

    (9)

    If we continue the calculation, we obtain: \( x=-\frac{11}{4}-\frac{1}{4}=-\frac{12}{4}=-3 \) .

    Consequently, the solution of the system 1 in \( x\) is

    \[\begin{equation} x=-3 \end{equation}\]

    (10)

    1.4 Give the solution of the linear system

    The solution is a couple \( (x,y)\) . In particular, check the solution.

    The solution of the linear system is the couple of real numbers is \( (x,y)=(-3,\frac{1}{2})\) .

    This is because a solution must be \( (-3,\frac{1}{2})\) , and \( (x,y)=(-3,\frac{1}{2})\) is a solution as:

    1. \( 4x+2y=4\times(-3)+2\times\frac{1}{2}=-12+1=-11\) ,

    2. and \( 3x-5y=3\times(-3)-5\times\frac{1}{2}=-9-\frac{5}{2}=\frac{-18-5}{2}=-\frac{23}{2}\) .

    2 The Matrix View

    2.1 The linear system as a matrix equation

    \paragraph{} Find a matrix equation \( AX=B\) equivalent to the system.

    2.1.1 Define relevant column vectors \( X\) and \( B\) , and square matrix \( A\)

    TIP: \( 3x-5y=3x+(-5)y\) .

    2.1.1.1 Solution

    We define the column vectors: \( X=\begin{bmatrix}x\ y\end{bmatrix}\) and \( B=\begin{bmatrix}-11\ ­\frac{23}{2}\end{bmatrix}\) , as well as the square matrix: \( A=\begin{bmatrix}4&2\ 3&-5\end{bmatrix}\) .

    2.1.2 Multiply the square matrix \( A\) by the column vector \( X\)

    2.1.2.1 Solution

    The product \( AX\) of the matrix \( A\) and the column vector \( X\) is equal to:

    \[\begin{equation} AX=\begin{bmatrix}4&2\\ 3&-5\end{bmatrix} \begin{bmatrix}x\\ y\end{bmatrix} =\begin{bmatrix}4x+2y\\ 3x-5y\end{bmatrix} \end{equation}\]

    (11)

    2.1.3 Deduce that the system is equivalent to the matrix vector \( AX=B\)

    2.1.3.1 Solution

    The elements of the column vector \( AX\) are the first members of the linear system:

    \( \left\{ \begin{matrix} 4x&+&2y&&=&-11\  3x&-&5y&&=&-\frac{23}{2} \end{matrix} \right. \)

    Moreover, the elements of the column vector \( B\) are the second membre of that system.

    Consequently, the linear system above is equivalent to the matrix equation \( AX=B\) .

    2.2 Invert the Matrix \( A\)

    2.2.1 Calculate the determinant of the matrix \( A\)

    2.2.1.1 Solution

    The determinant of the matrix \( A=\begin{bmatrix}4&2\ 3&-5\end{bmatrix}\) is:

    \(\det(A)=4\times (-5) - 3\times 2 = -20-6 = -26 \).

    2.2.2 Calculate the inverse of the matrix \( A\)

    Let the result as fractions of positive integers of their opposite.

    TIP: Define a matrix \( AA\) the following way:

    1. The diagonal elements of \( A\) are exchanged to obtain the diagonal elements of \( AA\) .

    2. The anti-diagonal elements of \( A\) are opposed to obtain the anti-diagonal elements of \( AA\) .

    \( AA\) is equal to:

    \[\begin{equation} AA=\begin{bmatrix}-5&-2\\ -3&4\end{bmatrix} \end{equation}\]

    (12)

    Section 2.2.2.1

    The inverse of the matrix \( A\) is the product of the inverse of \( \det(A)\) and \( AA\) :

    \[\begin{equation} A^{-1}=\frac{1}{-26}\begin{bmatrix}-5&-2\\ -3&4\end{bmatrix} \end{equation}\]

    (13)

    Let’s enter the factor \( \frac{1}{5}\) in the matrix \( AA\) element by element:

    \[\begin{equation} A^{-1}= \begin{bmatrix} \frac{-5}{-26}&\frac{-2}{-26}\\ \\ \frac{-3}{-26}&\frac{4}{-26} \end{bmatrix} \end{equation}\]

    (14)

    so that:

    \[\begin{equation} A^{-1}= \begin{bmatrix} \frac{5}{26}&\frac{1}{13}\\ \\ \frac{3}{26}&-\frac{2}{13} \end{bmatrix} \end{equation}\]

    (15)

    2.3 Solve the system the matrix way

    2.3.1 Multiply the matrix \( A^{-1}\) by the column vector \( B\) .

    Calculate with fractions from end to end, but don’t hesitate to use the calculator to help you.

    2.3.1.1 Solution

    Let’s multiply the matrix \( A^{-1}\) by the column vector \( B\) :

    \( A^{-1}B= \begin{bmatrix} \frac{5}{26}&\frac{1}{13}\  \  \frac{3}{26}&-\frac{2}{13} \end{bmatrix} \begin{bmatrix} -11\ -\frac{23}{2} \end{bmatrix}= \begin{bmatrix} \frac{5}{26}\times(-11)+ \frac{1}{13}\times\left(-\frac{23}{2}\right)\  \  \frac{3}{26}\times(-11) - \frac{2}{13}\times\left(-\frac{23}{2}\right) \end{bmatrix} \)

    Let’s calculate further, we obtain: \( A^{-1}B= \begin{bmatrix} \frac{-55 -23}{26}\  \  \frac{-33+46}{26} \end{bmatrix}= \begin{bmatrix} \frac{78}{26}\  \  \frac{13}{26} \end{bmatrix}= \begin{bmatrix} -3\  \frac{1}{2} \end{bmatrix} \)

    2.3.2 Deduce the matrix way to solve the system

    The solution of the matrix eqution \( AX=B\) is the column vector \( A^{-1}B\) .

    Consequently, the solution of the system of linear equations:

    \( \left\{ \begin{matrix} 4x&+&2y&&=&-3\  3x&-&5y&&=&-\frac{23}{2} \end{matrix} \right. \)

    is the couple \( (x,y)=(-3,\frac{1}{2})\) of elements of the column vector \( A^{-1}B\) .

    3 Solve the matrix Equation in Python

    In Python, the fractions may be entered as divisions (1/2 for \( \frac{1}{2}\) ), and are displayed as (approximate) decimal numbers. Moreover, the solution \( (x,y)=(-3,0.5)\) may be approximated as well because of roundoff errors.

    3.1 IMPORTANT

    Do it in a script, in order to test your program!

    3.2 SCRIPT

    Update the following script ‘LinearSystem.py’ to define and solve the matrix equation \( AX=B\) , with \( A=\begin{bmatrix}4&2\ 3&-5\end{bmatrix}\) and \( B=\begin{bmatrix}-11\ -\frac{23}{2}\end{bmatrix}\) .

    from numpy import *
from numpy.linalg import * 

X0=array([43,3])
print('X0=',X0)

'''
It is a line vector 
'''

A=array([[2,1],[1,3]])
print('A=\n',A)

'''
It is a matrix with 2 rows and 2 columns 
'''
B0=A@X0

'''
The column vector B0' is the product of the matrix A and the column vector 
X0'
'''

print('B0=',B0)

```
We have reconstructed the system of two linear equations and two 
variables equivalent to the matric equation AX=B, with the solution (43,3).
```

B=array([89,52])
InvA=inv(A) 
print('The inverse of A is\n',InvA) 

X=InvA@B

'''
The column vector X' is the product of the inverse of A and the column 
vector B'
'''

print('X=',X)

'''
The column vector X' is the solution of the matrix equation AX'=B' 
'''

    3.3 Solution

    Here is the updated script:

    from numpy import *
from numpy.linalg import * 

X0=array([-3,1/2])
print('X0=',X0)

'''
It is a line vector 
'''

A=array([[4,2],[3,-5]])
print('A=\n',A)

'''
It is a matrix with 2 rows and 2 columns 
'''
B0=A@X0

'''
The column vector B0' is the product of the matrix A and the column vector 
X0'
'''

print('B0=',B0)

```
We have reconstructed the system of two linear equations and two 
variables equivalent to the matric equation AX=B, with the solution 
(-3,1/2). Note that -11.5=-23/2.
```

B=array([-11,-23/2])
InvA=inv(A) 
print('The inverse of A is\n',InvA) 

X=InvA@B

'''
The column vector X' is the product of the inverse of A and the column 
vector B'
'''

print('X=',X)

'''
The column vector X' is the solution of the matrix equation AX'=B' 
'''

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