Solving for a matrix

Hello all,

I have a matrix and need to invert it( Ax=b) and solve for x. However, I know one aspect of x and only need the others.

   K = [7 -8 1 0 0 0 0; -8 16 -8  0 0 0 0; 1 -8 14 -8 1 0 0; 0 0 -8 16 -8 0 0; 0 0 1 -8 14 -8 1; 0 0 0 0 -8 16 -8; 0 0 0 0 1 -8 7];

syms f1 f2 f3 f4 f5 f6 f7 u1 u2 u3 u4 u5 u6 u7
F=[f1; f2; f3; f4; f5; f6; f7];
U=[u1; u2; u3; u4; u5; u6; u7];
eqn = F==K*U;
L = subs(eqn,[f1 f2 f3 f4 f5 f6 f7 u7], [0 0 0 0 0 0 -1 0])

This does give me the vector in terms of unknowns but doesn't actually solve for unknowns; is there a way to code it to do this? I tried U=K\F and it says the system is inconsistent however to my understanding I should be able to get numerical answers for each u aspect.

[Matt J]

So why is this being done symbolically if you want numerical answers?

[James Bader]

I did not know another way to enter a system of equations with several unknown variables; it was recommended to me to use symbolic variables.

Answer 1

[Bruno Luong]

>  K = [7 -8 1 0 0 0 0; -8 16 -8  0 0 0 0; 1 -8 14 -8 1 0 0; 0 0 -8 16 -8 0 0; 0 0 1 -8 14 -8 1; 0 0 0 0 -8 16 -8; 0 0 0 0 1 -8 7]

K =

       7    -8     1     0     0     0     0
      -8    16    -8     0     0     0     0
       1    -8    14    -8     1     0     0
       0     0    -8    16    -8     0     0
       0     0     1    -8    14    -8     1
       0     0     0     0    -8    16    -8
       0     0     0     0     1    -8     7

>> cond(K)

ans =

     2.7552e+16

>> rank(K)

ans =

       6

>> size(K)

ans =

       7     7

>>

Essentially that shows your linear system is ill-posed; meaning it either has no solution, and by chance it has (pick the right rhs) it has infinity number of solutions.