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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

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  1. frx Group Title
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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\[B=A ^{-1}=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\]

    • one year ago
  2. frx Group Title
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    But is there any other solution than the inverse?

    • one year ago
  3. frx Group Title
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    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
  4. Zarkon Group Title
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    \(A\) is not invertable...so how can \(B=A^{-1}\)

    • one year ago
  5. frx Group Title
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    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
  6. Zarkon Group Title
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    \(Det(A)=0\) so \(A^{-1}\) does not exist

    • one year ago
  7. frx Group Title
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    Ok, but is there any other solution than B equals the zerovector?

    • one year ago
  8. Zarkon Group Title
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    yes..there are infinitly many solutions

    • one year ago
  9. Zarkon Group Title
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    you have 2 of the solutions

    • one year ago
  10. frx Group Title
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    So how do I show that it has infinity many solutions?

    • one year ago
  11. Zarkon Group Title
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    show that \(B=\left[\begin{matrix}4 & 2 \\ 2 & 1\end{matrix}\right]\) is a solution

    • one year ago
  12. Zarkon Group Title
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    then \(B\cdot t\) is a solution for all real numbers \(t\)

    • one year ago
  13. frx Group Title
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    So B can be the same as what the inverse of A would have been if it existed?

    • one year ago
  14. Zarkon Group Title
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    it is not really the same

    • one year ago
  15. Zarkon Group Title
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    remember you have to divide by the det(A) for the shortcut way to find the inverse

    • one year ago
  16. frx Group Title
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    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
  17. frx Group Title
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    I= Identity matrix

    • one year ago
  18. Zarkon Group Title
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    yes

    • one year ago
  19. frx Group Title
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    Thank you for your help! :)

    • one year ago
  20. Zarkon Group Title
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    np

    • one year ago
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