MATLAB: solving symbolic matrix

Question

I've got three matrixes, one of which contains the unknowns. Two of them are multiplied by each other and result the last one.

A1 * A2 = A3

So, I've got this code in MATLAB:

syms A1 A2 A3 B C D F
k1=1; k2=2, b=3, a=4

A1 = [ -exp(i*k1*b) exp(-k2*b) exp(k2*b) 0; i*k1*exp(i*k1*b) k2*exp(-k2*b) -k2*exp(k2*b) 0; 0 -exp(-k2*a) -exp(k2*a) (exp(-i*k1*a) + exp(i*k1*a)); 0 -k2*exp(-k2*a) k2*exp(k2*a) i*k1*(exp(-i*k1*a) - exp(i*k1*a)) ]
A2 = [ B; C; D; F ]
A3 = [ exp(-i*k1*b) ; i*k1*exp(-i*k1*b) ; 0 ; 0 ]

I want to solve what's the result of B, C, D and F. I know I've got to use function solve but I got problem with the syntax all the time.

Thanks for your time and reply...

rayryeng rayryeng · Accepted Answer · 2014-11-17T22:39:59

This question has no need for any symbolic mathematics. Your A1 and A3 matrices are defined numerically. All you have to do to find the entries in A2 is to invoke the inverse operator (\) between A1 and A3. Specifically, if you are given:

A1 * A2 = A3   ,

to find A2, you would do:

A2 = A1^{-1} * A3

You would find the inverse of A1 and multiply this with A3. You can easily do this in MATLAB by the inverse operator:

A2 = A1 \ A3;

Therefore, just do this:

k1=1; k2=2; b=3; a=4;

A1 = [ -exp(i*k1*b) exp(-k2*b) exp(k2*b) 0; i*k1*exp(i*k1*b) k2*exp(-k2*b) -k2*exp(k2*b) 0; 0 -exp(-k2*a) -exp(k2*a) (exp(-i*k1*a) + exp(i*k1*a)); 0 -k2*exp(-k2*a) k2*exp(k2*a) i*k1*(exp(-i*k1*a) - exp(i*k1*a)) ]
A3 = [ exp(-i*k1*b) ; i*k1*exp(-i*k1*b) ; 0 ; 0 ]
A2 = A1 \ A3;

A2 will compute B, C, D and F for you. Just refer to the first, second, third and fourth elements of A2 respectively.

MATLAB: solving symbolic matrix

1 Answers