## frx Group Title Let A be the matrix $A=\left[\begin{matrix}1 & -2\\ -2 &4\end{matrix}\right]$ decide all the 2x2 matrices B such that AB=BA=0, where 0 is the zeromatrix one year ago one year ago

1. frx Group Title

$\left[\begin{matrix}1 & -2 \\ -2 & 4\end{matrix}\right]\left[\begin{matrix}a &b \\ b & c\end{matrix}\right]=\left[\begin{matrix}0 &0 \\ 0& 0\end{matrix}\right]$ $\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}1 &-2\\ -2 & 4\end{matrix}\right]=\left[\begin{matrix}0 & 0\\ 0 &0\end{matrix}\right]$ So all I can figure is that for AB=BA=0 B can be$B=A ^{-1}=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]$

2. frx Group Title

But is there any other solution than the inverse?

3. frx Group Title

I guess it could be $\left[\begin{matrix}1 & -2 \\ -2 & 4\end{matrix}\right]\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]=\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]$ $\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]\left[\begin{matrix}1 & -2 \\ -2 & 4\end{matrix}\right]=\left[\begin{matrix}0 & 0 \\ 0& 0\end{matrix}\right]$ but so the two solutions for AB=BA=0 should be B=A^-1 and B=0, am I right?

4. Zarkon Group Title

$$A$$ is not invertable...so how can $$B=A^{-1}$$

5. frx Group Title

It's not I guess, A is not invertable if AX=0 is the definition I remember right now, is that right?

6. Zarkon Group Title

$$Det(A)=0$$ so $$A^{-1}$$ does not exist

7. frx Group Title

Ok, but is there any other solution than B equals the zerovector?

8. Zarkon Group Title

yes..there are infinitly many solutions

9. Zarkon Group Title

you have 2 of the solutions

10. frx Group Title

So how do I show that it has infinity many solutions?

11. Zarkon Group Title

show that $$B=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]$$ is a solution

12. Zarkon Group Title

then $$B\cdot t$$ is a solution for all real numbers $$t$$

13. frx Group Title

So B can be the same as what the inverse of A would have been if it existed?

14. Zarkon Group Title

it is not really the same

15. Zarkon Group Title

remember you have to divide by the det(A) for the shortcut way to find the inverse

16. frx Group Title

Oh I get it, it can't be the inverse since the definition says that A^-1*A=I isn't that right?

17. frx Group Title

I= Identity matrix

18. Zarkon Group Title

yes

19. frx Group Title

Thank you for your help! :)

20. Zarkon Group Title

np