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frx

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

  2. frx
    • 2 years ago
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    But is there any other solution than the inverse?

  3. frx
    • 2 years ago
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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?

  4. Zarkon
    • 2 years ago
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    \(A\) is not invertable...so how can \(B=A^{-1}\)

  5. frx
    • 2 years ago
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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?

  6. Zarkon
    • 2 years ago
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    \(Det(A)=0\) so \(A^{-1}\) does not exist

  7. frx
    • 2 years ago
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    Ok, but is there any other solution than B equals the zerovector?

  8. Zarkon
    • 2 years ago
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    yes..there are infinitly many solutions

  9. Zarkon
    • 2 years ago
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    you have 2 of the solutions

  10. frx
    • 2 years ago
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    So how do I show that it has infinity many solutions?

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

  12. Zarkon
    • 2 years ago
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    then \(B\cdot t\) is a solution for all real numbers \(t\)

  13. frx
    • 2 years ago
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    So B can be the same as what the inverse of A would have been if it existed?

  14. Zarkon
    • 2 years ago
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    it is not really the same

  15. Zarkon
    • 2 years ago
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    remember you have to divide by the det(A) for the shortcut way to find the inverse

  16. frx
    • 2 years ago
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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?

  17. frx
    • 2 years ago
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    I= Identity matrix

  18. Zarkon
    • 2 years ago
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    yes

  19. frx
    • 2 years ago
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    Thank you for your help! :)

  20. Zarkon
    • 2 years ago
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    np

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