Question

Elementary matrices: finding A^(-1)

prove that A=[1 2; 3 4] is nonsingular and write it as a product of elementary matrices.

answer: A =[1 0; 3 1][1 0; 0 -2][1 2; 0 1]

I don't get how you're supposed to do this question.. I got [-2 1; 3/2 -1/2] as the inverse but I don't know what to do after.. can someone please help me?

Answer 1

\[\det\left[\begin{matrix}1&2\\3&4\end{matrix}\right]=-2\neq0\]so that is the first part, which I know you probably already had

Answer 2

we also have that if\[A=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\]is invertible, then its inverse is given by\[A^{-1}=\frac1{\det A}\left[\begin{matrix}d&-b\\-c&a\end{matrix}\right]\]

Answer 3

For the product of elementary matrices bit, I think it's easiest if you find the inverse of your matrix by finding the rref, while applying the same changes at the same time to the identity matrix. Now you have your inverse, and you know what row operations you did. If you did it this way, could you quickly write out the row operations for me?

Answer 4

@KingGeorge
r2-r1 -> r2 [1 2; 0 -2]
r1+r2 ->r1 [1 0; 0 -2]
(-1/2)r2 -> r2 [1 0; 0 1]
Do you mean this?

Answer 5

@TuringTest I still don't know what to do after finding the inverse. I don't know how to "break down" the product(?) of the elementary matrices like the answer

Answer 6

Yes, that stuff. Because the way you find the elementary matrices is from the row operations. Although shouldn't that first one be r2-3r1?

Answer 7

my method will not work backwards, so I lead you astray....
I'm sorry, you need to do it KingGeorge's way