Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Buddhayourlord&savoir

  • 2 years ago

Are matrices A and B inverses? Look in comments to see matrices.

  • This Question is Closed
  1. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375487302704:dw|

  2. itsmylife
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    just multiply them ,if your gona get identity matrix then yes they are inverse else no

  3. itsmylife
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[A*A^{-1}=I\]

  4. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What do you mean by "identity matrix"

  5. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    identical?

  6. jdoe0001
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\bf I = \left[ \begin{matrix} 1&0\\ 0&1 \end{matrix}\right]\)

  7. Euler271
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And for 2x2 matrices only: \[A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\] \[A^{-1} = \frac{ 1 }{ detA } \left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]\]

  8. itsmylife
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375487630099:dw|

  9. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    therefore it is not an inverse since its not all the same?

  10. itsmylife
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375487648610:dw|

  11. jdoe0001
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \( A \times B = \left[ \begin{matrix} 1&0\\ 0&1 \end{matrix}\right] = \textit{I} \implies B = A^{-1}; A = B^{-1}\)

  12. Euler271
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    watch out with that cross symbol ^_^

  13. jdoe0001
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol

  14. jdoe0001
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \(A \cdot B \) :)

  15. Euler271
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hehe

  16. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this isnt explaining anything at all guys :L

  17. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i get the equation but how do i know if it is or is not?

  18. jdoe0001
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well, you have just have to multiply them

  19. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Multiply what? AxB as a whole? The numbers in A with each other? Im sorry im feeling pretty hopeless, i just need a more in-depth explanation please ;/

  20. primeralph
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes. Inverses.

  21. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    5*-7=-35 -18*2=-36 therefore the answer is no? right?

  22. primeralph
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I just told you they are inverses.

  23. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, okay.

  24. Buddhayourlord&savoir
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks man

  25. primeralph
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You're welcome.

  26. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy