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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
 one year ago
 one year ago
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

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frxBest ResponseYou've already chosen the best response.0
\[\left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right]\left[\begin{matrix}a &b \\ b & c\end{matrix}\right]=\left[\begin{matrix}0 &0 \\ 0& 0\end{matrix}\right]\] \[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}1 &2\\ 2 & 4\end{matrix}\right]=\left[\begin{matrix}0 & 0\\ 0 &0\end{matrix}\right]\] So all I can figure is that for AB=BA=0 B can be\[B=A ^{1}=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\]
 one year ago

frxBest ResponseYou've already chosen the best response.0
But is there any other solution than the inverse?
 one year ago

frxBest ResponseYou've already chosen the best response.0
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?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
\(A\) is not invertable...so how can \(B=A^{1}\)
 one year ago

frxBest ResponseYou've already chosen the best response.0
It's not I guess, A is not invertable if AX=0 is the definition I remember right now, is that right?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
\(Det(A)=0\) so \(A^{1}\) does not exist
 one year ago

frxBest ResponseYou've already chosen the best response.0
Ok, but is there any other solution than B equals the zerovector?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
yes..there are infinitly many solutions
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
you have 2 of the solutions
 one year ago

frxBest ResponseYou've already chosen the best response.0
So how do I show that it has infinity many solutions?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
show that \(B=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\) is a solution
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
then \(B\cdot t\) is a solution for all real numbers \(t\)
 one year ago

frxBest ResponseYou've already chosen the best response.0
So B can be the same as what the inverse of A would have been if it existed?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
it is not really the same
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
remember you have to divide by the det(A) for the shortcut way to find the inverse
 one year ago

frxBest ResponseYou've already chosen the best response.0
Oh I get it, it can't be the inverse since the definition says that A^1*A=I isn't that right?
 one year ago

frxBest ResponseYou've already chosen the best response.0
Thank you for your help! :)
 one year ago
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