Create a piece of code that takes an Augmented matrix (Concatenating A and B) and produces the linear combination of the system in terms of the free variables if there are non-basic columns in the reduced row echelon matrix. The free variables will be given the variables t1, t2, ..., tn for the nth free variable. The inputs must match the expected outputs below. The code should be able to output the linear combination of the system in terms of the free variables regardless of the size of the matrix, for example, a 1 x 10, 5 x 5, or 7 x 3. If there are no non basic columns then print "Matrix A has One Solution". Make sure the output from the code looks like this as an example:

Test Cases

1) Input:

A = [4 8; 1 7; 2 4]

B = [12;8;6]

2)

Input:

A = [2 3 4 56 6 8;1 2 4 5 6 7;3 6 12 15 18 21;4 6 8 112 12 16]

B = [9;8;24;18]

3)

Inputs:

A = [3 4 5;1.5 2 2.5]

B = [1;0.5]

4)

34t135t231t3t1t2t3=t134100+t235010+t331001 Output: 4128741286 1) Given the RREF of A is 100010110 this gives the general solution is =x1x2x3=t1t1t1=t1111 213432664412856515112661812872116982418 2) Given the RREF of A is 100001004400974600660056006700 this gives the general solution is - x1x2x3x4x5x6x7=4t197t2+6t3+5t4+6t54t1+46t26t36t47t5t1t2t3t4t5=t14410000+t2974601000+t36600100+t45600010+t56700001 (31.54252.5121) [10340350310]utionis=34t135t231t3t1t2t3=t134100+t235010+t331001 Input: A=214325452541636B=000 Expected Output: Code form: t1[ 7;6;1;0;0]+t2[5;3;0;1;1] x1x2x3x4x5=7t16t1+t1t2t25t23t2=t176100+t253011