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Consider this expression:

Eliminate[a  x + b  y == 0 && c  x + d  y == 0, {x, y}]

Mathematica returns True, which is correct, but I would like to assume that $(x, y) \neq (0, 0),$ and get the determinant condition. Is there some way to do this?

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  • $\begingroup$ Igor, If you mean to have ad-bc==0 as the determinant condition, would you accept/be able to work with ad==bc or any variation there-of? $\endgroup$ – CA Trevillian Jan 29 at 5:16
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    $\begingroup$ @CATrevillian yes, that would be fine, and, in fact, adding another equation $x+y==1$ does return this condition, but it is a bit kludgy ;) $\endgroup$ – Igor Rivin Jan 29 at 13:51
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I assume that by determinant condition you mean

$$b c - a d=0$$

Right?

eqns = a*x + b*y == 0 && c*x + d*y == 0; 
Resolve[Exists[{x}, eqns]]

produces the above result

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  • $\begingroup$ It's buried in a sea of logical statements, though, but it does give a good hint to a path to take. $\endgroup$ – CA Trevillian Jan 29 at 15:30
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Because x and y are two unknowns, but there are only two equations. You can only eliminate one unknowns:

Eliminate[a*x + b*y == 1 && c*x + d*y == 2, {x}]
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Trying

Eliminate[ a x + b y == 0 && c x + d y == 0 && (x != 0 || y != 0), {x, y}]

gives

(a==(b c)/d&&d!=0)||(b==(a d)/c&&c!=0)||(c==0&&d==0)

Which is not quite what one would want, but...

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Resolve[Exists[{x, y}, 
a x + b y == 0 && c x + d y == 0 && {x, y} != 0 && a d != 0 && b c != 0]]

Gives

a != 0 && b != 0 && c != 0 && d != 0 && b c - a d == 0

Whereas

Resolve[Exists[{x, y}, 
a x + b y == 0 && c x + d y == 0 && x != 0 && y != 0
&& a != 0 && b != 0 && c != 0 && d != 0]]

Gives

b c - a d == 0 && a != 0 && b != 0 && c != 0 && d != 0

But I can't seem to have it give

a d - b c == 0

Hope this helps, Igor!

You might be able to brute-force it with some assumptions, but I can't seem to find the right combination at the moment.

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