anonymous
  • anonymous
I am confused on how to go about doing this. Linear algebra question for homework
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
misty1212
  • misty1212
HI!!
anonymous
  • anonymous
Hello

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misty1212
  • misty1212
one way to do it would be to multiply the matrix on the right by the inverse of the matrix on the left
misty1212
  • misty1212
\[AB=C\iff A=CB^{-1}\]
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
let me try that i'll respond once I have done that
misty1212
  • misty1212
inverse of a 2 by 2 matrix is easy to find right?
anonymous
  • anonymous
yes
misty1212
  • misty1212
\[\begin{pmatrix} x & y \\ z & v \end{pmatrix}^{-1}=\frac{1}{D}\begin{pmatrix} v & -y \\ -z & x \end{pmatrix} \]
misty1212
  • misty1212
course you can always cheat too, not really necessary
misty1212
  • misty1212
https://www.wolframalpha.com/input/?i=%7B%7B2%2C4%7D%2C%7B2%2C5%7D%7D%5E%28-1%29
anonymous
  • anonymous
so the inverse would be \[\left[\begin{matrix}2.5 & -2 \\ -0.5 & 0.5\end{matrix}\right]\] right?
anonymous
  • anonymous
or have i done it wrong?
misty1212
  • misty1212
looks good
misty1212
  • misty1212
oh no
misty1212
  • misty1212
the -5 is wrong
misty1212
  • misty1212
that should be -1
anonymous
  • anonymous
-5?
misty1212
  • misty1212
first get \[\begin{pmatrix} 5& -4 \\ -2 & 2 \end{pmatrix}\]
misty1212
  • misty1212
the divide all by 2
anonymous
  • anonymous
oh wait i know what u mean lol, that was my mistake
anonymous
  • anonymous
i thought the 2 were 1s
anonymous
  • anonymous
lol
misty1212
  • misty1212
\[\begin{pmatrix} 2.5& -2 \\ -1 & 1 \end{pmatrix}\]
misty1212
  • misty1212
i think that is right
anonymous
  • anonymous
yes, that is, so now I just have to multiply that by the resultant in the question correct?
misty1212
  • misty1212
should work i got this https://www.wolframalpha.com/input/?i=%7B%7B2%2C7%7D%2C%7B0%2C5%7D%7D.%7B%7B2.5%2C-2%7D%2C%7B-1%2C1%7D%7D
anonymous
  • anonymous
ok, ill do it, and see if i got that on paper
misty1212
  • misty1212
ok good luck!\[\color\magenta\heartsuit\]
anonymous
  • anonymous
yup i got the same answer, Thank you so much for your help. :D

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