frank0520
  • frank0520
A=-3 1 1 | 1 -3 1|  1 1 -3 Calculate A(A+5I) and use the result to find the inverse of A.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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frank0520
  • frank0520
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?
jim_thompson5910
  • jim_thompson5910
I'm getting \[\Large A(A+5I) = \begin{bmatrix}-4 & 0 & 0\\0 & -4 & 0\\0 & 0 & -4\\ \end{bmatrix}\] as well
jim_thompson5910
  • jim_thompson5910
I don't agree with your inverse though. I'm getting a different matrix for \(\Large A^{-1}\)

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jim_thompson5910
  • jim_thompson5910
To check, the product of \(\Large A\) and \(\Large A^{-1}\) (in either order) should be equal to \[\Large I_3 = \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\]
frank0520
  • frank0520
@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.
jim_thompson5910
  • jim_thompson5910
You have the correct inverse
jim_thompson5910
  • jim_thompson5910
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
frank0520
  • frank0520
@jim_thompson5910 Thanks for the help! I'm not sure either, maybe its just worded wrong.
jim_thompson5910
  • jim_thompson5910
you're welcome

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