Linear algebra... Please see attachments

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Linear algebra... Please see attachments

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I am pretty sure these answers are correct. My question...

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