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frank0520

  • one year ago

A=-3 1 1 | 1 -3 1|  1 1 -3 Calculate A(A+5I) and use the result to find the inverse of A.

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  1. frank0520
    • one year ago
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    I got: -4 0 0 | 0 -4 0|  0 0 -4 for the computation. And I got: -1/4 0 0 | 0 -1/4 0|  0 0 -1/4 for the inverse. Is this the correct inverse?

  2. jim_thompson5910
    • one year ago
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    I'm getting \[\Large A(A+5I) = \begin{bmatrix}-4 & 0 & 0\\0 & -4 & 0\\0 & 0 & -4\\ \end{bmatrix}\] as well

  3. jim_thompson5910
    • one year ago
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    I don't agree with your inverse though. I'm getting a different matrix for \(\Large A^{-1}\)

  4. jim_thompson5910
    • one year ago
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    To check, the product of \(\Large A\) and \(\Large A^{-1}\) (in either order) should be equal to \[\Large I_3 = \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\]

  5. frank0520
    • one year ago
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    @jim_thompson5910 I did the inverse of the computation. I also did the inverse of just A and got: \[\left[\begin{matrix}-1/2 & -1/4 & -1/4 \\ -1/4 & -1/2 & -1/4 \\ -1/4 & -1/4 & -1/2\end{matrix}\right]\] I did the product of \[A \ and\ A^-1\] and got the \[I_3\] matrix.

  6. jim_thompson5910
    • one year ago
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    You have the correct inverse

  7. jim_thompson5910
    • one year ago
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    I'm not sure how `Calculate A(A+5I) and use the result to find the inverse of A.` fits into finding the inverse of A though

  8. frank0520
    • one year ago
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    @jim_thompson5910 Thanks for the help! I'm not sure either, maybe its just worded wrong.

  9. jim_thompson5910
    • one year ago
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    you're welcome

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