it appears that this is

= x a,b,c + y b,b,c + z c,b,c - d,b,c ;
a′,b′,c′ b′,b′,c′ c′,b′,c′ d′,b′,c′
a″,b″,c″ b″,b″,c″ c″,b″,c″ d″,b″,c″

viz. the second and third terms each vanishing, it is

= x a,b,c - d,b,c .
a′,b′,c′ d′,b′,c′
a″,b″,c″ d″,b″,c″

But if the linear equations hold good, then the first column of the original determinant is = 0, and therefore the determinant itself is = 0; that is, the linear equations give

x a,b,c - d,b,c = 0;
a′,b′,c′ d′,b′,c′
a″,b″,c″ d″,b″,c″

which is the result obtained above.

We might in a similar way find the values of y and z, but there is a more symmetrical process. Join to the original equations the new equation

αx + βy + γz = δ;

a like process shows that, the equations being satisfied, we have