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In that document, we will take advanrage of solving a little puzzle to explore everything about the solution of a system of two linear equations in two real variables.

We will first see how to solve such a system with the method of substitution/elimination.

Then we will explore end solve the equivalent matrix equation, manually and in Python, with the packages ‘numpy’ and ‘numpy.linalg’.

Here is a little puzzle that we will aim to solve.

The hypotheses of the puzzle are the following ones:

1. The double of the age of the captain plus the number of children of my aunt is 89.

2. The age of the captain plus the triple of the number of children of my aunt is 52.

And the questions the puzzle asks are the following ones:

1. How old is the captain?

2. How many children does my aunt have?

Let’s denote:

1. \( x\) the age of the captain,

2. and \( y\) the number of children of my aunt.

Then the hypotheses are transformed the following way:

1. “The double of the age of the captain plus the number of children of my aunt is 89" becomes:

   \[ \begin{equation} 2x+y=89 \end{equation} \]

   (1)

2. “The age of the captain plus the triple of the number of children of my aunt is 52” becomes:

   \[ \begin{equation} x+3y=52 \end{equation} \]

   (2)

The questions of the puzzle are then transformed the following way:

1. “How old is the captain?” becomes "What is the value of \( x\) ?"

2. “How many children does my aunt have?” becomes "What is the value of \( y\) ?"

The puzzle is thus transformed into the following problem:

Find which real numbers fulfil the equations 1 and 2 together.

That means solving the following system of 2 linear equations of 2 real variables:

The equation 1 is equivalent to the linear equation in \( x\) :

Its solution depends on \( y\) and is: \( x=\frac{89-y}{2}= \frac{89}{2} -\frac{y}{2} \) .

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

In the equation 2, we eliminate \( x\) by replacing it by its value given by the equation 5.

The equation 2 is thus equivalent to:

If we multiply both members of the equality by \( 2\) , we obtain:

The equation 7 is equivalent to the affine equation in \( y\) :

Its solution is: \( y=\frac{104-89}{5}=\frac{15}{5}=3 \) .

Consequently, the solution of the system 3 in \( y\) is \( y=3\) and the answer to the second quetion of the puzzle is:

My aunt has \( 3\) children.

In the equation 5, we aliminate \( x\) by replacing it by its value \( 3\) .

The equation 5 is thus equivalent to:

If we set \( \frac{1}{2}\) into factor, we obtain: \( x=\frac{1}{2}(89-3)=86\times \frac{1}{2}=43 \) .

Consequently, the solution of the system 3 in \( x\) is \( x=3\) and the answer to the first quetion is:

The captain is \( 43\) years old.

We may define the column vectors: \( X=\begin{bmatrix}x\\ y\end{bmatrix}\) and \( B=\begin{bmatrix}89\\ 52\end{bmatrix}\) , as well as the square matrix: \( A=\begin{bmatrix}2&1\\ 1&3\end{bmatrix}\) .

The product \( AX\) of the matrix \( A\) and the column vector \( X\) is a column vector with two elements composed of:

1. its first element is the sum of the products of the elements of the first row of \( A\) by the elemens of \( X\)

2. its second element is the sum of the products of the elements of the second row of \( A\) by the elemens of \( X\)

So that it is equal to

The elements of the column vector \( AX\) are thus the first members of the linear system 3.

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

Section 4.2.1

Section 4.2.2

We may multiply the matrix \( A^{-1}\) by the column vector \( B=\begin{bmatrix} 89\\ 52 \end{bmatrix}\) :

\( A^{-1}B=\begin{bmatrix} \frac{3}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{2}{5} \end{bmatrix} \begin{bmatrix}89\\ 52\end{bmatrix}=\begin{bmatrix} \frac{3\times 89-52}{5}\\ \\ \frac{2\times 52-89}{5} \end{bmatrix} \)

If we calculate further, we obtain: \( A^{-1}B= \begin{bmatrix} \frac{3\times 89-52}{5}\\ \\ \frac{2\times 52-89}{5} \end{bmatrix}= \begin{bmatrix} \frac{215}{5}\\ \\ \frac{15}{5} \end{bmatrix}= \begin{bmatrix} 43\\ 3 \end{bmatrix} \)

The elements of the column vector \( A^{-1}B\) are the element of the solution of the system 3, that is equivalent to the matrix system \( AX=B\) .

To define vectors in Python, we first use the command:

from numpy import *

to import all the functions of ‘numpy’ package.

The simpler way to create a vector is listing its elements.

For instance:

V=array([1, 2])

creates the line vector \( V=\begin{bmatrix}1&2\end{bmatrix}\) .

To define matrices in Python, we first use the command:

from numpy import *

to import all the functions of ‘numpy’ package.

The simpler way to create a matrix is listing its elements line by line.

For instance:

M=array([[1, 2], [3, 4]])

creates the line vector \( M=\begin{bmatrix}1&2\\ 3&4\end{bmatrix}\) .

An important fact about the multiplication of a matrix and a vector in Python is that the vector is dealed as a column vector for the multiplication.

Then the result, that is a column vector, is displayed as a line vector.

Namely, let’s define a 2x2 matrix \( M\) and two vectors \( V\) and \( W\) with 2 elements in Python:

from numpy import *
	M=array([[1, 2], [3, 4]])
	V=array([1, 2])
	W=M@V

Then the line vector \( W=\begin{bmatrix} u&v \end{bmatrix}\) is so that the column vector \( W'=\begin{bmatrix} u\\ v \end{bmatrix}\) is the product of the matrix \( M=\begin{bmatrix}1&2\\ 3&4\end{bmatrix}\) and the column vector \( V'=\begin{bmatrix}1\\ 2\end{bmatrix}\) :

\( W'=\begin{bmatrix}1&2\\ 3&4\end{bmatrix}\begin{bmatrix}1\\ 2\end{bmatrix}= \begin{bmatrix}1\times 1 + 2\times 2\\ 3\times 1+4\times 2\end{bmatrix} =\begin{bmatrix}5\\ 11\end{bmatrix}\) , so that \( W=\begin{bmatrix}5&11\end{bmatrix}\) .

We need both ‘numpy’ and ‘numpy.linalg’ packages:

from numpy import *
	from numpy.linalg import *

Let’s now defin two matrices \( M\) and \( InvM\) :

M=array([[1, 2], [3, 4]])
	InvM=inv(M)

Then \( InvM\) is the inverse of the matrix \( M=\begin{bmatrix}1&2\\ 3&4\end{bmatrix}\) :

\( InvM=M^{-1}=\begin{bmatrix}1&2\\ 3&4\end{bmatrix}^{-1} =\frac{1}{1\times 4-3\times 2} \begin{bmatrix}4&-2\\ -3&1\end{bmatrix} =\begin{bmatrix}\frac{4}{-2}&\frac{-2}{-2}\\ \\ \frac{-3}{-2}&\frac{1}{-2}\end{bmatrix} = \begin{bmatrix}-2&0.5\\ 1&-1.5\end{bmatrix}\)

The following script ‘LinearSystem.py’ defines and solves the matrix equation \( AX=B\) , with \( A=\begin{bmatrix}2&1\\ 1&3\end{bmatrix}\) and \( B=\begin{bmatrix}89\\ 52\end{bmatrix}\) .

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' 
'''