Show that if
then Ax=0 has only the trivial solution if and only if ad-bc ≠ 0
Solving the top equation for y gives:
Putting this into the bottom equation:
So either ad - bc = 0 or x = 0 (and hence y = 0).
This is, of course, much easier to do using matrix properties, but I thought it might be informative to show you the "algebraic" version for 2 x 2s.
Thanks I get it. How would this be done using the matrix properties? That's what I tried to use but go messed up.
EDIT: I got messed up because how can you possibly solve for a variable without an augmented matrix?
Well, you can use an augmented matrix- in this case it would be
because you are solving
To "row reduce that you would divide the first row by a, to get a "1" in the first row, first column, and then subtract that new first row, multiplied by c, from the second row:
The next step would be to get a "1" in the second row, second column, by dividing the second row by .
Now do you see why "ad- bc" being 0 or not is important?
To solve the equation
without using an "augmented matrix" or "row reduction", you could find the inverse matrix and multiply both sides by it:
as long as A has an inverse and so we would have that single solution. Of course, if B is the 0 vector, then so that single solution is the 0 or "trivial" solution. Such an equation will have a "non-trivial" solution, then, if and only if A does not have an inverse. How can you determine if a 2 by 2 matrix has an inverse or not?