I'm getting the inv error code saying it must be squared

I can't find where my error is, can some one help please

My code is :

E1=3;
E2=2;
Z1= j*10e-03;
Z2= 1.5e03;
Z3= -j* 0.1e-06;
Z4= j*1e-03;
Z5= -j*1e-06;
Z6= 10e03;
Z= [(Z1+Z2+Z3) (-Z3); (-Z3) (Z3+Z4+Z5); (Z5+Z6) (-Z5)]

Z =

1.0e+04 *

0.1500 + 0.0000i 0.0000 + 0.0000i

0.0000 + 0.0000i 0.0000 + 0.0000i

1.0000 - 0.0000i 0.0000 + 0.0000i

E= [E1;0;-E2]

E =

3

0

-2

I = inv(Z)*E

Error using inv

Matrix must be square.

Answer 1

[Walter Roberson]

Avoid using inv(); it is not as mathematically stable as the alternatives.

format long g
E1=3;
E2=2;
Z1= j*10e-03;
Z2= 1.5e03;
Z3= -j* 0.1e-06;
Z4= j*1e-03;
Z5= -j*1e-06;
Z6= 10e03;
Z= [(Z1+Z2+Z3) (-Z3); (-Z3) (Z3+Z4+Z5); (Z5+Z6) (-Z5)]

Z = 
                       1500 +             0.0099999i                          0 +                 1e-07i
                          0 +                 1e-07i                          0 +             0.0009989i
                      10000 -                 1e-06i                          0 +                 1e-06i

E= [E1;0;-E2]

E = 3×1

3 0 -2

I = Z\E

I = 
      -0.000151589225999787 -  2.93415473517444e-10i
       3.15167210679577e-06 +     0.161724791615002i

Z*I

ans = 
         -0.227383855169225 -  1.95599999618422e-06i
      -0.000161546894344196 +  3.13304634487832e-09i
          -1.51589242172266 -  2.93399999427634e-06i

You wil notice that this is not really a good match for E -- but it is the version of I that give the least-squared error compared to E. You have an over-determined system, so you cannot get a perfect fit.

[Daniel Salazar]

Got it, thank you for the help!

Answer 2

[Voss]

inv takes the inverse of a square matrix (that is, a matrix with size n-by-n for some n). In this case you are trying to take the inverse of the matrix Z, but Z is not square (it's 3-by-2), so you get that error. Check that your definition of Z is correct.