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anonymous
 4 years 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
anonymous
 4 years 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

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anonymous
 4 years ago
Best 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\[B=A ^{1}=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But is there any other solution than the inverse?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1\(A\) is not invertable...so how can \(B=A^{1}\)

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

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1\(Det(A)=0\) so \(A^{1}\) does not exist

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, but is there any other solution than B equals the zerovector?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1yes..there are infinitly many solutions

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1you have 2 of the solutions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So how do I show that it has infinity many solutions?

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

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1then \(B\cdot t\) is a solution for all real numbers \(t\)

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

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1it is not really the same

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1remember you have to divide by the det(A) for the shortcut way to find the inverse

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you for your help! :)
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