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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

Mathematics
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\[\left[\begin{matrix}1 & -2 \\ -2 & 4\end{matrix}\right]\left[\begin{matrix}a &b \\ b & c\end{matrix}\right]=\left[\begin{matrix}0 &0 \\ 0& 0\end{matrix}\right]\] \[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}1 &-2\\ -2 & 4\end{matrix}\right]=\left[\begin{matrix}0 & 0\\ 0 &0\end{matrix}\right]\] So all I can figure is that for AB=BA=0 B can be\[B=A ^{-1}=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\]
But is there any other solution than the inverse?
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?

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Other answers:

\(A\) is not invertable...so how can \(B=A^{-1}\)
It's not I guess, A is not invertable if AX=0 is the definition I remember right now, is that right?
\(Det(A)=0\) so \(A^{-1}\) does not exist
Ok, but is there any other solution than B equals the zerovector?
yes..there are infinitly many solutions
you have 2 of the solutions
So how do I show that it has infinity many solutions?
show that \(B=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\) is a solution
then \(B\cdot t\) is a solution for all real numbers \(t\)
So B can be the same as what the inverse of A would have been if it existed?
it is not really the same
remember you have to divide by the det(A) for the shortcut way to find the inverse
Oh I get it, it can't be the inverse since the definition says that A^-1*A=I isn't that right?
I= Identity matrix
yes
Thank you for your help! :)
np

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