chrisplusian
  • chrisplusian
Linear algebra... Please see attachments
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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chrisplusian
  • chrisplusian
1 Attachment
chrisplusian
  • chrisplusian
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chrisplusian
  • chrisplusian
I am pretty sure these answers are correct. My question...

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chrisplusian
  • chrisplusian
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
steve816
  • steve816
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?
steve816
  • steve816
nvm, so you take linear algebra.
thomas5267
  • thomas5267
Nope! The answer is wrong. I feel sad for you.
zzr0ck3r
  • zzr0ck3r
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.
thomas5267
  • thomas5267
The first question is right. The second is not.
thomas5267
  • thomas5267
The 2x2 matrix in the second question is invertible so you are good to go.
chrisplusian
  • chrisplusian
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
chrisplusian
  • chrisplusian
I really don't know what I should have seen that would clue me in on what approach to take working on this problem
chrisplusian
  • chrisplusian
any advice is greatly appreciated
chrisplusian
  • chrisplusian
@zzr0ck3r could you illustrate what you mean?
thomas5267
  • thomas5267
\[ \mathbf{A}=\begin{pmatrix}2&1\\5&3\end{pmatrix}\\ \mathbf{A}^{-1}\text{ exists.}\\ \mathbf{AY}=\begin{pmatrix}1&0&-1&0\\5&0&2&-4\end{pmatrix}\\ \mathbf{A}^{-1}\mathbf{AY}=\mathbf{A}^{-1}\begin{pmatrix}1&0&-1&0\\5&0&2&-4\end{pmatrix}\\ \mathbf{Y}=\mathbf{A}^{-1}\begin{pmatrix}1&0&-1&0\\5&0&2&-4\end{pmatrix}\\ \]
chrisplusian
  • chrisplusian
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?
zzr0ck3r
  • zzr0ck3r
what he said^^^
thomas5267
  • thomas5267
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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