i want to solve a set of homogeneous linear equation

A = (n,n) :- a (n,n) order of matrix which i get from previous calculations

B = [1 , x1 , x2 , x3 , .......... xn]' :- vector in which 1st element is 1 and rest all are unknown of (n,1) order

C = [0 , 0 , 0 , 0 ]' :- null vector of (n,1) order

i want solution to A*B = C

that will give me values of B vectors

Answer 1

[Dyuman Joshi]

Note - Symbolic Toolbox required

Note that you might not get a solution for x depending upon the values of A.

One such example would be - A is an Identity matrix, any order greater than 1; or in this particular case, magic() of any odd order

%Random example
A=magic(6);
n=size(A,1); 
syms x [n-1 1]
B=[1;x];
sol=solve(A*B==0,x)

sol = struct with fields:
    x1: 1
    x2: -1/2
    x3: -1
    x4: -1
    x5: 1/2

[Shardul]

Hi Dyuman thanks for responding,

I used your code but it it giving the following answer in solution [0x1 sym] , how to get a numerical value ?

P =1.0e+04*[6.6064,-3.5642,0,0;-3.5642,6.6064,-3.5642,0;0,-3.5642,6.6064,-3.5642;0,0,-3.5642,3.2624]

P = 4×4

66064 -35642 0 0 -35642 66064 -35642 0 0 -35642 66064 -35642 0 0 -35642 32624

na = size(P,1);
syms u [na-1 1]
Ba = [1; u];
sol = solve(P*Ba==0,u)

sol = struct with fields:
    u1: [0×1 sym]
    u2: [0×1 sym]
    u3: [0×1 sym]

[Dyuman Joshi]

As I said earlier - "Note that you might not get a solution for x depending upon the values of A."

For the values you have for A, there is no solution for x.

P = 1.0e+04*[6.6064,-3.5642,0,0;-3.5642,6.6064,-3.5642,0;0,-3.5642,6.6064,-3.5642;0,0,-3.5642,3.2624];
na = size(P,1);
syms u [na-1 1]
Ba = [1; u];
%Let's solve for each equation corresponding to each row
y = P*Ba

y = 

%1st equation, get the value of u1
U1 = solve(y(1)==0)

U1 = 

%2nd equation, use u1 to get the value of u2
U2 = solve(subs(y(2),u1,U1)==0,u2)

U2 = 

Now we have two equations remaining to get the value of u3, and the values of u1 and u2 will only be correct if we get the same u3 from

%value of u3 from 3rd equation
U3_1 = solve(subs(y(3),[u1 u2],[U1 U2])==0,u3)

U3_1 = 

%value of u3 from 4th equation
U3_2 = solve(subs(y(4),u2,U2)==0,u3)

U3_2 = 

As you can see, the values of u3 are not the same and hence there are no unique values for which the equation P*Ba=0 is satisfied.

The empty output (i.e. 0x1 sym) denote that there is no solution to the equation.

[Shardul]

OK, Thanks for your help.

Answer 2

[Torsten]

Or use the following code to produce an optimal solution in the least-squares sense:

A =1.0e+04*[6.6064,-3.5642,0,0;-3.5642,6.6064,-3.5642,0;0,-3.5642,6.6064,-3.5642;0,0,-3.5642,3.2624]

A = 4×4

66064 -35642 0 0 -35642 66064 -35642 0 0 -35642 66064 -35642 0 0 -35642 32624

C = -A(:,1);
B = A(:,2:end)\C

B = 3×1

1.8535 2.4356 2.6609

A*[1;B]

ans = 4×1

0.2555 0.4735 0.6221 0.6797