Linear combination

$R_n = \{\overrightarrow {v_1}, \overrightarrow {v_2}, ...., \overrightarrow {v_n}\}$
$\text{scalar} \; c=\{c_1, c_2,..., c_n\}$
For vector R and scalar c, if the following relationship Equation 7 is defined
$$y=c_1 \overrightarrow {v_1} + c_2 \overrightarrow {v_2} + .... +c_n \overrightarrow {v_n},$$
This is called the linear combination of the vector R and the scalar (weight).

1. Are linear combinations in vectors a₁, a₂ and vector b?
Whether the relationship is established,
$$x_1 a_1 +x_2 a_2 = b \quad x_1, x_2 : scalar(weight)$$
$$a_1 =\left[\begin{array}{r}1\\-2\\-5 \end{array}\right], \qquad a_2=\left[\begin{array}{r}2\\5\\6 \end{array}\right], \qquad b=\left[\begin{array}{r}7\\4\\3\end{array}\right]$$
$$x_1 \left[\begin{array}{r}1\\-2\\-5 \end{array}\right] + x_2 =a_2=\left[\begin{array}{r}2\\5\\6 \end{array}\right]=\left[\begin{array}{r}7\\4\\3\end{array}\right]$$
The above is arranged in the form of matrix product as follows.
$$\left[\begin{array}{r}1&2\\-2&5\\-5&6 \end{array}\right] \left[\begin{array}{r}x_1\\x_2\end{array}\right]= \left[\begin{array}{r}7\\4\\3\end{array}\right]$$
As a result, we are looking for the existence of a solution in the above linear system.

$$\text{Standard Matrix(Coefficient matrix)} =\left[\begin{array}{r}1&2\\-2&5\\-5&6 \end{array}\right]$$
$$\text{Varaible vector}= \left[\begin{array}{r}x_1\\x_2\end{array}\right]$$
$$\text{Constant vector}= \left[\begin{array}{r}7\\4\\3\end{array}\right]$$
In this system, you can calculate the solution of each variable by making the coefficient matrix an identity matrix. To create an identity matrix, you need to calculate the inverse of the coefficient matrix. The inverse matrix is computed only if it is a square matrix. Therefore, the solution is computed using rref of the augment matrix, which combines the coefficient matrix and the constant matrix.
>>> import numpy as np
>>> from sympy import *
>>> a1=np.mat("1;-2;-5")
>>> a2=np.mat("2;5;6")#coefficient matrix = np.c_[a1, a2]
>>> b=np.mat("7;4;-3")# constant matrix
>>> au=Matrix(np.c_[a1,a2,b]) #augment matrix
>>> au
Matrix([
[ 1, 2,  7],
[-2, 5,  4],
[-5, 6, -3]])
>>> au.rref() #rref of augment matrix
(Matrix([
[1, 0, 3],
[0, 1, 2],
[0, 0, 0]]), (0, 1))
From the result,
$$x_1=3, x_2=2$$
As a result, a linear combination of the above three vectors is established.
Again in the above problem b can be represented by a linear combination of $a_1$ and $a_2$ with some scalar. This relationship of linear combination can be expressed in terms of span as follows.
Span$\{\overrightarrow{a_1}, \;  \overrightarrow{a_2} \}$= b

2. Is there a linear combination between vectors a1, a2 and b?
b=Span{a₁, a₂}
>>> a1=np.mat("1;-2;3")
>>> a2=np.mat("5;-13;-3")
>>> b=np.mat("-3;8;1")
>>> a1
matrix([[ 1],
        [-2],
        [ 3]])
>>> a2
matrix([[  5],
        [-13],
        [ -3]])
>>> b
matrix([[-3],
        [ 8],
        [ 1]])
>>> au=Matrix(np.c_[a1,a2,b])
>>> au
Matrix([
[ 1,   5, -3],
[-2, -13,  8],
[ 3,  -3,  1]])
>>> au.rref()
(Matrix([
[1, 0, 0],
[0, 1, 0],
[0, 0, 1]]), (0, 1, 2))
From the rref,
$$x_1=0 \\ x_2=0 \\ 0x_1+0x_2=1$$
However, $0x_1+0x_2 \neq 1.$
These systems are called inconsistent systems and do not have a span relationship between the three vectors.

3. Column vectors (a₁, a₂, a₃) in Matrix A and vector b
If W=span{a₁, a₂, a₃}, is b included in W?
$A=\left[\begin{array}{rrr}2&0&6\\-1&8&5\\1&-2&1\end{array}\right]\\
b=\left[\begin{array}{r}4\\1\\-4 \end{array}\right]$
W is the result of linear combination of vectors a₁, a₂, a₃. That is, the solution exists on the following systems.
$$\left[\begin{array}{rrr}2&0&6\\-1&8&5\\1&-2&1\end{array}\right]\left[\begin{array}{r}x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{r}w_1\\w_2\\w_3\end{array}\right]$$
If the vector b is one of W, then the solution must be in the case of assigning the vector b to the right-hand side of the above system.

>>> A=np.array([[1,0,-4],[0,3,-2],[-2, 6, 3]])
>>> b=np.mat("4;1;-4")
>>> au=Matrix(np.c_[A,b])
>>> au
Matrix([
[ 1, 0, -4,  4],
[ 0, 3, -2,  1],
[-2, 6,  3, -4]])
>>> au.rref()
(Matrix([
[1, 0, 0, -4],
[0, 1, 0, -1],
[0, 0, 1, -2]]), (0, 1, 2))
$x_1=-4\\ x_2=-1\\x_3=-2$
The above system has a unique solution. Therefore, b is included in W.
Since matrix A is a square matrix, it can be computed using an inverse matrix instead of rref.

>>> import scipy.linalg as LA
>>> np.linalg.det(A)
-2.9999999999999996
>>> LA.solve(A, b)
array([[-4.],
       [-1.],
       [-2.]])