I'll use syms here, because it displays things nicely.
B=[0;0;x;0;0;y]
B =
C=[a;b;0;c;d;-2000]
C =
So B and C are VECTORS, not matrices. Now you have a matrix A, that is 6x6, and is completely known. As long as A is entirely known and is of full rank, you can use any number of ways to solve the problem.
For example, you could just use solve. But solve will often be slow. It is trivial to accomplish though. As an example, I need to have the matrix A, in numerical form.
A = randi(9,[6,6])
A =
4 4 8 7 6 8
3 4 7 7 3 2
8 2 7 8 6 8
4 9 7 1 5 5
7 4 9 7 4 2
5 1 5 2 1 3
And now the solution using solve, for this particular matrix:
sol = solve(A*B == C,[a,b,c,d,x,y])
sol =
a: -16000/19
b: -84000/19
c: -42000/19
d: -116000/19
x: -16000/19
y: 14000/19
As easy, we can use equationsToMatrix to write the probmel in terms of linear algebra.
[S,T] = equationsToMatrix(A*B == C,[a,b,c,d,x,y])
S =
T =
So, for that particular matrix A, we now have two matrices, S and T that represnet the linear system. SOlve them, using backslash.
abcdxy = double(S)\double(T)
abcdxy =
1.0e+03 *
-0.8421
-4.4211
-2.2105
-6.1053
-0.8421
0.7368
As you can see with format rat, the solution is the same.
abcdxy
abcdxy =
-16000/19
-84000/19
-42000/19
-116000/19
-16000/19
14000/19
Could we have done it differently yet? Yes. Move the unknowns to the left hand side, and move everything known to the right hand side. You will still have a system of linear equations. The result will be the same.
