Matlab order of operators

I'm trying to understand a MATLAB code for reserach but seem to be stuck in a basic question regarding the order of the transpose operator and ml divide operator. (please see the code below)

```

C = randi([0 ,1], [3,3])

D = randi([0 ,1], [3,3])

disp("C-(D')*(C\D')")

disp(C-(D')*(C\D'))

```

(you may need to run this multiple times until the inverse exists)

Because the transpose operator is computed before the `\` operator, it seems taht `C\D'` should be equal to finding the `x` s.t. `C=(D')*x`... However, when I subsrtitute `x` as `C\D'` and display `C-(D')*x`, I don't get a zero... could anyone give me a reason why?

Answer 1

[Bruno Luong]

"`C\D'` should be equal to finding the `x` s.t. `C=(D')*x`."

No. Rather

D' = C*x

Your code should be

while true
    C = randi([0 ,1], [3,3]);
    if rank(C) == 3
        break
    end
end
D = randi([0 ,1], [3,3]);
C

C = 3×3

0 1 1 1 1 1 1 0 1

D

D = 3×3

0 1 1 1 0 1 1 1 0

disp("D'-C*(C\D')")

D'-C*(C\D')

disp(D'-C*(C\D'))

     0     0     0
     0     0     0
     0     0     0

Or

C = randi([0 ,1], [3,3]);
while true
    D = randi([0 ,1], [3,3]);
    if rank(D) == 3
        break
    end
end
C

C = 3×3

1 0 1 1 0 1 0 0 1

D

D = 3×3

0 1 1 1 1 1 1 0 1

disp("C-D'*(D'\C)")

C-D'*(D'\C)

disp(C-D'*(D'\C))

     0     0     0
     0     0     0
     0     0     0