solving binary linear equation in matlab -

consider binary linear equation of form x*a = b. want solve x, efficiency way avoiding using x = b*inv(a) , instead using x= b/a.but command results not in binary form. tried command x = mod(b/a ,2)but still result not in binary. how fix this?

example `

x = 1     0     1     1     0     0     0     1     1     0` 

and matrix

`0     1     0     1     0     0     1     1     1     1  0     1     1     1     1     1     1     1     0     1  0     0     1     0     1     1     0     1     1     1  1     0     0     1     1     1     1     1     1     1  1     1     0     0     1     1     0     0     0     0  0     1     1     1     0     1     1     0     1     0  0     0     1     1     0     0     0     1     0     0  1     1     1     1     0     1     0     1     1     1  1     0     1     0     1     1     1     0     1     1  1     1     1     0     0     0     1     1     0     0`  full rank. 

then

>> b = mod (x*a,2) 

b =

 1     0     1     1     1     0     1     0     1     1 

to find x, getting

>> k = b / 

k =

columns 1 through 6

1.3750   -0.5000   -0.7500   -0.7500    0.8750   -0.5000 

columns 7 through 10

1.8750   -0.5000    2.1250   -0.7500 

or if using modulus 2, result is

>> k = mod (b / a,2) 

k =

columns 1 through 6

1.3750    1.5000    1.2500    1.2500    0.8750    1.5000 

columns 7 through 10

1.8750    1.5000    0.1250    1.2500 

so ,how can x in same binary form? , way matrices in class double not galois field

this example @ mathworks shows how perform boolean matrix inversion matlab , believe answers question.

i have not quite gotten working perfectly, believe need use combination of mod() , logical() example:

a=logical(a); b=(mod(x*a,2)); inversea= ~a; k=mod(b*inversea,2) 

this gives

k=[1 1 0 0 1 1 0 1 0 0] 

which not x, think if play around logical function , logical operations in conjunction mod() should able work