solving systems of linear equations

I fail to run/solve the linear equ. (Ax=B) form as below: x1+x2=10; x1-x2=2; if u can help; % it is fine with x=A\B;

clc; clear all;
syms u 
A=[1 1; 1 -1]   
B= [10; 2]
for i=1:2
for j=1:2      
E1=A(i,j)*u - B(i)                                         end
[uarray(j)] =solve(E1)   
end
uarray=double(uarray)

% I am getting=>[0,10],[0,-2] % I am getting=> 0 -2

Answer 1

[John D'Errico]

But what you did was not a correct solution to the linear system of equations.

The solution to the linear system of equations is most simply done via backslash. No symbolic solve is even necessary. I think you understood that from your comment, but are really looking to use the symbolic toolbox with solve.

A=[1 1; 1 -1]; 
B= [10; 2];

X = A\B
ans =
     6
     4

So you want to solve the linear system of equations: A*X = B. As you see, the solution is the vector [6;4]. Is that correct?

[A*X,B]
ans =
    10    10
     2     2

Yes, it is correct. Of course, you could have used solve to do the work too.

syms x1 x2
X = solve(A*[x1;x2] == B)
X = 
  struct with fields:

    x1: [1×1 sym]
    x2: [1×1 sym]
    
X.x1
ans =
6

X.x2
ans =
4

Or, if you want to use solve, but do so moredirectly, without converting the problem to one with matrix coefficients, we could do...

syms x1 x2
[x1,x2] = solve(x1+x2 == 10,x1-x2 == 2)
x1 =
6
x2 =
4

As you see by either approach, solve agrees with backslash. It even agrees with pencil and paper. If

x1 + x2 = 10

x1 - x2 = 2

Then just add the two equations together. We get

2*x1 = 12

Which clearly implies x1=6. Then you can go back and recover x2 from either of those equations. Thus...

6 + x2 = 10

So x2 = 4 drops out.