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anonymous
 4 years ago
Help please :D
anonymous
 4 years ago
Help please :D

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I need help with part c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mth3v4 Can you help me plz????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im not very good with marices but ill "try"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it will probably take time pulling books

razor99
 4 years ago
Best ResponseYou've already chosen the best response.0did u told me that question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I apologize, I didn't know you wanted me to help with this question. I'm taking a look now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks jemurray. I was busy fighting with someone on os

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, these are just asking for relatively simple verifications. Let me show you how to do the first one.

razor99
 4 years ago
Best ResponseYou've already chosen the best response.0Jemurray3 and SamIam are typing replies…

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok i did the first two parts but let me see what u got i really need part c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just processing it give me a sec :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OHHHH GOTTTT ITTTTT Thankx jemurray :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0R u gonna be around fro much longer?
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