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chrisplusian

  • one year ago

Linear algebra... Please see attachments

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  1. chrisplusian
    • one year ago
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  2. chrisplusian
    • one year ago
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  3. chrisplusian
    • one year ago
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    I am pretty sure these answers are correct. My question...

  4. chrisplusian
    • one year ago
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    Is there a better way to approach the problem? In my linear algebra class so far we have studied: Matirx algebra,Matrix inverse, powers of matrix,transpose of matrix, vectors in R^n

  5. steve816
    • one year ago
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    Wow, you're super neat! Sorry I can't help you because I haven't learned about matrices yet... but what class do you learn matrices in?

  6. steve816
    • one year ago
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    nvm, so you take linear algebra.

  7. thomas5267
    • one year ago
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    Nope! The answer is wrong. I feel sad for you.

  8. zzr0ck3r
    • one year ago
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    There are many ways to do this problem and the method you took will work but sort of defeats the purpose of linear algebra. I would find the inverse of the 2x2 matrix and multiply both sides, on the left, by that.

  9. thomas5267
    • one year ago
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    The first question is right. The second is not.

  10. thomas5267
    • one year ago
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    The 2x2 matrix in the second question is invertible so you are good to go.

  11. chrisplusian
    • one year ago
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    Honestly this class is really hard for me, the way the professor teaches is by reading the proofs (verbatim) from the book, and then displaying a look that shows he has somehow managed to manifest some secret meaning. The problem is that only he gets it. Our book is a publication from overseas so I can't find a solution guide to check what I am doing, so I am kind of looking here for help

  12. chrisplusian
    • one year ago
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    I really don't know what I should have seen that would clue me in on what approach to take working on this problem

  13. chrisplusian
    • one year ago
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    any advice is greatly appreciated

  14. chrisplusian
    • one year ago
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    @zzr0ck3r could you illustrate what you mean?

  15. thomas5267
    • one year ago
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    \[ \mathbf{A}=\begin{pmatrix}2&1\\5&3\end{pmatrix}\\ \mathbf{A}^{-1}\text{ exists.}\\ \mathbf{AY}=\begin{pmatrix}1&0&-1&0\\5&0&2&-4\end{pmatrix}\\ \mathbf{A}^{-1}\mathbf{AY}=\mathbf{A}^{-1}\begin{pmatrix}1&0&-1&0\\5&0&2&-4\end{pmatrix}\\ \mathbf{Y}=\mathbf{A}^{-1}\begin{pmatrix}1&0&-1&0\\5&0&2&-4\end{pmatrix}\\ \]

  16. chrisplusian
    • one year ago
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    So what I should have done is generically named the two given matrices, used property of matrices to solve for Y and then do the given operations to find Y?

  17. zzr0ck3r
    • one year ago
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    what he said^^^

  18. thomas5267
    • one year ago
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    Not exactly. I named \(\mathbf{A}=\begin{pmatrix}2&1\\5&3\end{pmatrix}\) because I find \(\begin{pmatrix}2&1\\5&3\end{pmatrix}^{-1}\) ugly. Since \(\begin{pmatrix}2&1\\5&3\end{pmatrix}^{-1}\) exists, we could do what I wrote above. Not all matrix has an inverse. For example, \(\begin{pmatrix}1&0\\0&0\end{pmatrix}\) does not have an inverse.

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