rsst123
  • rsst123
**Will Medal**
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rsst123
  • rsst123
anonymous
  • anonymous
use your calculator
rsst123
  • rsst123
without calculator

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More answers

zzr0ck3r
  • zzr0ck3r
do you know how to find eigen values and eigen vectors? Do you know what diagonalization is ?
zzr0ck3r
  • zzr0ck3r
there are other ways, so I need to know what you are doing in class if not.
rsst123
  • rsst123
yes isn't it A=SDS^-1?
zzr0ck3r
  • zzr0ck3r
Correct. But you first need to find the Eigen values and vectors so that you can construct \(S\)
rsst123
  • rsst123
ok so after finding eigen values and vectors what would i do to find the matrix?
zzr0ck3r
  • zzr0ck3r
Ok if we call your 2x2 matrix given above A and the matrix of eigen vectors S. Then you will find D by D = \(SAS^{-1}\). Now once you have D you just take the square root of the entries (they will only be non 0 on the diagonal). You with me?
rsst123
  • rsst123
oh ok i see, I would get a diagonal matrix and then I can just square root the diagonal is that correct?
zzr0ck3r
  • zzr0ck3r
then \(\sqrt{A} = S^{-1}\sqrt{D}S\)
zzr0ck3r
  • zzr0ck3r
yes
rsst123
  • rsst123
ok perfect thank you!
zzr0ck3r
  • zzr0ck3r
np
amoodarya
  • amoodarya
\[\left[\begin{matrix}4 & 3 \\ 2 & 3\end{matrix}\right]=\left[\begin{matrix}-1 & \frac{3}{2} \\ 1 & 1\end{matrix}\right]\times \left[\begin{matrix}1 & 0 \\ 0 & 6 \end{matrix}\right] \times \left[\begin{matrix}\frac{-2}{5} & \frac{3}{5} \\ \frac{2}{5} & \frac{2}{5} \end{matrix}\right]=\\=S D S^{-1}\\\sqrt{A}==S \left[\begin{matrix}\sqrt{1} & 0 \\ 0 & \sqrt{6}\end{matrix}\right] S^{-1}\]

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