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SamIam

  • 4 years ago

Help please :D

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  1. SAMIAM
    • 4 years ago
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  2. SAMIAM
    • 4 years ago
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    I need help with part c

  3. razor99
    • 4 years ago
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    srry

  4. SAMIAM
    • 4 years ago
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    mth3v4 Can you help me plz????

  5. anonymous
    • 4 years ago
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    im not very good with marices but ill "try"

  6. SAMIAM
    • 4 years ago
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    okkiedokieee

  7. anonymous
    • 4 years ago
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    it will probably take time pulling books

  8. SAMIAM
    • 4 years ago
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    lol Thanx

  9. razor99
    • 4 years ago
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    did u told me that question

  10. SAMIAM
    • 4 years ago
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    sorry?

  11. anonymous
    • 4 years ago
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    sorries :(

  12. anonymous
    • 4 years ago
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    dont know this 1

  13. razor99
    • 4 years ago
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    r u ok mth3v4

  14. anonymous
    • 4 years ago
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    I apologize, I didn't know you wanted me to help with this question. I'm taking a look now.

  15. SAMIAM
    • 4 years ago
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    thanks jemurray. I was busy fighting with someone on os

  16. SAMIAM
    • 4 years ago
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    So it was fine lol

  17. anonymous
    • 4 years ago
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    Okay, these are just asking for relatively simple verifications. Let me show you how to do the first one.

  18. razor99
    • 4 years ago
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    Jemurray3 and SamIam are typing replies…

  19. SAMIAM
    • 4 years ago
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    ok i did the first two parts but let me see what u got i really need part c

  20. SAMIAM
    • 4 years ago
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    Razor are you ok?

  21. anonymous
    • 4 years ago
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    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.

  22. SAMIAM
    • 4 years ago
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    Just processing it give me a sec :D

  23. SAMIAM
    • 4 years ago
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    OHHHH GOTTTT ITTTTT Thankx jemurray :D

  24. anonymous
    • 4 years ago
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    No problem :)

  25. SAMIAM
    • 4 years ago
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    R u gonna be around fro much longer?

  26. anonymous
    • 4 years ago
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    For a little while

  27. SAMIAM
    • 4 years ago
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    ok i may be back :D

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