## anonymous one year ago Linear algebra question.... see attachment please

1. anonymous

2. IrishBoy123

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3. anonymous

Can you see the attachment?

4. thomas5267

Yes it is possible.

5. anonymous

Let's call the matrix on the RHS $$M$$, so you have ${\bf{AB}}={\bf{M}}$ If $$|{\bf{A}}|\neq0$$, you can find its inverse $${\bf{A}}^{-1}$$, which is pretty straightforward since it's 2x2. Then ${\bf{B}}={\bf{A}}^{-1}{\bf{M}}$ Then you can find the inverse of $$\bf{B}$$, provided it's not singular.

6. anonymous

7. thomas5267

Both $$\bf{AB}$$ and $$\bf{A}$$ have full rank.

8. anonymous

^^^ that was my attempt but it did not work out

9. thomas5267

$\det(\mathbf{A})=5-(-3)(-3)=5-9=-4\neq5+9=13$

10. anonymous

I see that now!!! is my approach correct?

11. thomas5267

Yes I think.

12. anonymous

Ok thank you

13. thomas5267

$\mathbf{A}^{-1}\text{ and }\mathbf{AB}^{-1}\text{ exist.}\\ \mathbf{A}^{-1}\mathbf{AB}=\mathbf{B}\\ \mathbf{B}^{-1}=\left(\mathbf{A}^{-1}\mathbf{AB}\right)^{-1}=\left(\mathbf{AB}\right)^{-1}\mathbf{A}\\ \left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)\mathbf{B}=\left(\mathbf{AB}\right)^{-1}\left(\mathbf{A}\mathbf{B}\right)=\mathbf{I}\\ \mathbf{B}\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)=\mathbf{A}^{-1}\mathbf{AB}\left(\left(\mathbf{AB}\right)^{-1}\mathbf{A}\right)=\mathbf{A}^{-1}\left(\mathbf{AB}\left(\mathbf{AB}\right)^{-1}\right)\mathbf{A}=\mathbf{I}\\ \therefore \mathbf{B}^{-1}=\left(\mathbf{AB}\right)^{-1}\mathbf{A}$

14. anonymous

^^^ I wish I could see it that way.

15. anonymous

If I could give you another medal for that answer I would lol