Create a random 4X4 matrix

>>A=rand(4) % the numbers are not going to agree with the ones here.

type
>>help rand

A =
    0.2760    0.1190    0.5853    0.5060
    0.6797    0.4984    0.2238    0.6991
    0.6551    0.9597    0.7513    0.8909
    0.1626    0.3404    0.2551    0.9593

create a 4X1 vector:

>>x = rand(4,1)
x =
    0.5472
    0.1386
    0.1493
    0.2575

create a linear combination of columns of A call this b:

>>b = A*x
b =
    0.3852
    0.6545
    0.8331
    0.4213

check sizes of arrays/vectors:
>>size(A)
ans =
     4     4
>>size(b)
ans =
     4     1
>>size(x)
ans =
     4     1

form an augmented matrix:

>>A_augmented=[A,b]
A_augmented =
    0.2760    0.1190    0.5853    0.5060    0.3852
    0.6797    0.4984    0.2238    0.6991    0.6545
    0.6551    0.9597    0.7513    0.8909    0.8331
    0.1626    0.3404    0.2551    0.9593    0.4213
>>size(A_augmented)
ans =
     4     5

type
>>help rref

use the rref command to find reduced echelon form.

>>row_echelon=rref(A_augmented)
row_echelon =
    1.0000         0         0         0    0.5472
         0    1.0000         0         0    0.1386
         0         0    1.0000         0    0.1493
         0         0         0    1.0000    0.2575

let x_re be the last column of the row echelon. It should agree with x:
>>x_re = row_echelon(:,5)
x_re =
    0.5472
    0.1386
    0.1493
    0.2575

type:
>>help norm

check how close x_re is from x:
>>norm(x_re-x)
ans =
   5.9140e-16

check the linear independence of the columns of A:
>>rref(A)
ans =
     1     0     0     0
     0     1     0     0
     0     0     1     0
     0     0     0     1

create a vector that's a linear combination of columns of A:

>>b4=A(:,4)+2*A(:,1)
b4 =
    1.0580
    2.0585
    2.2011
    1.2845

form a new matrix B:
>>B=[A,b4]
B =
    0.2760    0.1190    0.5853    0.5060    1.0580
    0.6797    0.4984    0.2238    0.6991    2.0585
    0.6551    0.9597    0.7513    0.8909    2.2011
    0.1626    0.3404    0.2551    0.9593    1.2845

check to see that indeed the last column reflects the
structure of b4 as being the 4th column of A + 2*1st column of A:
>>rref(B)
ans =
    1.0000         0         0         0    2.0000
         0    1.0000         0         0   -0.0000
         0         0    1.0000         0   -0.0000
         0         0         0    1.0000    1.0000
---------------------------------------------
create a 3X2 matrix:
>>A=rand(3,2)
A =
    0.8407    0.2435
    0.2543    0.9293
    0.8143    0.3500
 replace the 2nd column by 56X the first column.
 the columns are no longer linearly independent.
>>A(:,2)=56*A(:,1)
A =
    0.8407   47.0802
    0.2543   14.2398
    0.8143   45.6000

use rref to show the new A does not have linearly
independent columns:
>>rref(A)
ans =
     1    56
     0     0
     0     0