Error using symengine and matrix must be square

While typing the following I get this error:

• error using symengine
• matrix must be square

A= [(2.*s +2) - (2.*s + 1) -1 -(2.*s +1)*(9.*s +1) -4.*s -1 -4.*s*(4.*s +1 +1./s)];

>> B= [I1;I2;I3];

>> C=[V;0;0];

>> B= inv(A).*C;

Error using symengine

Matrix must be square.

Where does the error comes from and how can I solve it?

[Ameer Hamza]

A is 1x6 matrix. What do you expect the output of inv(A) to be?

[Jordy Mopewou]

A= [(2*s +2) - (2*s + 1) -1 -(2*s +1) (9*s +1) -4*s -1 -4*s*(4*s +1 +1.s)];

At first when I typed this I had the following as error:

Error: Invalid expression. When calling a function or indexing a variable, use

parentheses. Otherwise, check for mismatched delimiters.

So it is from here that I got into more trouble

[Ameer Hamza]

Can you share the mathematical form of the equation you are trying to solve?

[Jordy Mopewou]

Answer 1

[Star Strider]

MATLAB will not automatically parse ‘A’ into separate elements. It is necessary to use appropriate delimiters to separate the individual elements (,) to define the elements and the rows (;).

Try this:

syms I1  I2 I3 V s 
A= [(2.*s +2), - (2.*s + 1), -1; -(2.*s +1), (9.*s +1), -4.*s; -1, -4.*s, (4.*s +1 +1./s)];
B= [I1;I2;I3];
C=[V;0;0];
Eq = C == A * B;
[Isln] = solve(Eq, [I1,I2,I3]);
I1 = simplify(Isln.I1, 'Steps',100)
I2 = simplify(Isln.I2, 'Steps',100)
I3 = simplify(Isln.I3, 'Steps',100)

producing:

I1 =
(V*(20*s^3 + 13*s^2 + 10*s + 1))/(24*s^4 + 30*s^3 + 17*s^2 + 16*s + 1)
 
I2 =
(V*(s + 1)*(8*s^2 + 2*s + 1))/(24*s^4 + 30*s^3 + 17*s^2 + 16*s + 1)
 
I3 =
(V*s*(8*s^2 + 13*s + 1))/(24*s^4 + 30*s^3 + 17*s^2 + 16*s + 1)

EDIT — (10 Sep 2020 at 15:52)

To calculate them as impedances:

Z1 = simplify(V/I1, 'Steps',100)
Z2 = simplify(V/I2, 'Steps',100)
Z3 = simplify(V/I3, 'Steps',100)

producing:

Z1 =
(24*s^4 + 30*s^3 + 17*s^2 + 16*s + 1)/(20*s^3 + 13*s^2 + 10*s + 1)
 
Z2 =
3*s + (8*s^2 + 13*s + 1)/((s + 1)*(8*s^2 + 2*s + 1))
 
Z3 =
(24*s^4 + 30*s^3 + 17*s^2 + 16*s + 1)/(s*(8*s^2 + 13*s + 1))

.

Answer 2

[Ameer Hamza]

An alternate solution is to use the inv() or mldivide (\) operator to solve the equation. Since this seems to be the lalpalce transform of a circuit, the following also shows how to get the time-domain solution for a specific voltage signal (v).

syms s V(s) t

A = [(2*s+1) -(2*s+1) -1;
     -(2*s+1) (9*s+1) -4*s;
     -1 -4*s (4+1+1/s)];
 
C = [V; 0; 0];
I = A\C; % same as inv(A)*B but more efficient

% Let v is a constant signal, i.e., v=1
V1 = 1/s;
I1 = subs(I, V, V1);
i(t) = ilaplace(I1, s, t);

tv = 0:0.01:1;
iv = double(subs(i, t, tv));
plot(tv, iv.');

[Jordy Mopewou]

Thank you