Let A=[[a b][c d]], ad-bc=0, a=b=0; examine A[[x1][x2]]

[warwick]

As opposed to intuitively for the one on the right.

0 + 0 = 0
cx1 + dx2 = 0

Obviously, x1 would have to equal d and x2 equal -c, or vice versa, whatever.

 

[daon]

Re: What's the rigorous way to solve this?

Just solve for x1. x1 = (-d/c)x2. There are infinitely many solutions.

Try x = [1, -c/d]

 

[warwick]

Re: What's the rigorous way to solve this?

Just solve for x1. x1 = (-d/c)x2. There are infinitely many solutions.

Try x = [1, -c/d]

Yeah, I know that. I just wondered if there was a similar way to get x = (d, -c).

 

[daon]

Re: What's the rigorous way to solve this?

Just solve for x1. x1 = (-d/c)x2. There are infinitely many solutions.

Try x = [1, -c/d]

Yeah, I know that. I just wondered if there was a similar way to get x = (d, -c).

Yeah, choose x2 = -c. The author of that solution must have felt it was "nicer."

 

[warwick]

Re: What's the rigorous way to solve this?

Just solve for x1. x1 = (-d/c)x2. There are infinitely many solutions.

Try x = [1, -c/d]

Yeah, I know that. I just wondered if there was a similar way to get x = (d, -c).

Yeah, choose x2 = -c. The author of that solution must have felt it was "nicer."

Ah. Since x2 is a free variable you can do that. lol.