SamIam
  • SamIam
Help please :D
Mathematics
  • 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!
SamIam
  • SamIam
1 Attachment
SamIam
  • SamIam
I need help with part c
razor99
  • razor99
srry

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SamIam
  • SamIam
mth3v4 Can you help me plz????
anonymous
  • anonymous
im not very good with marices but ill "try"
SamIam
  • SamIam
okkiedokieee
anonymous
  • anonymous
it will probably take time pulling books
SamIam
  • SamIam
lol Thanx
razor99
  • razor99
did u told me that question
SamIam
  • SamIam
sorry?
anonymous
  • anonymous
sorries :(
anonymous
  • anonymous
dont know this 1
razor99
  • razor99
r u ok mth3v4
anonymous
  • anonymous
I apologize, I didn't know you wanted me to help with this question. I'm taking a look now.
SamIam
  • SamIam
thanks jemurray. I was busy fighting with someone on os
SamIam
  • SamIam
So it was fine lol
anonymous
  • anonymous
Okay, these are just asking for relatively simple verifications. Let me show you how to do the first one.
razor99
  • razor99
Jemurray3 and SamIam are typing replies…
SamIam
  • SamIam
ok i did the first two parts but let me see what u got i really need part c
SamIam
  • SamIam
Razor are you ok?
anonymous
  • anonymous
Oh, okay, well if you did the first two I won't bother :) Watch now: \[ A^{-1}\cdot A = \frac{1}{5}(2I - A) \cdot A = \frac{1}{5}(2I\cdot A - A^2) \] but since \[A^2 -2IA + 5I = 0\] as per the assumption, then \[2IA - A^2 = 5I\] so plugging that in yields \[\frac{1}{5}(5I) = I \] So that is in fact the proper inverse matrix.
SamIam
  • SamIam
Just processing it give me a sec :D
SamIam
  • SamIam
OHHHH GOTTTT ITTTTT Thankx jemurray :D
anonymous
  • anonymous
No problem :)
SamIam
  • SamIam
R u gonna be around fro much longer?
anonymous
  • anonymous
For a little while
SamIam
  • SamIam
ok i may be back :D

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