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Mathematics
General Math
Factoring Matrices with Elementary Row Operations
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[QUOTE="cbarker1, post: 6582983, member: 663258"] [B]TL;DR Summary:[/B] I am working on reviewing some Linear Algebra for a Graduate course in the Spring. I thought I did it correctly when I finished. But I looked in the book a different answer. I used my calculator to check the book answer and gives the correct matrix. Dear Everybody, I have some trouble with this problem: Finding a sequence of elementary matrix for this matrix A. Let ##A=\begin{bmatrix} 4 & -1 \\ 3& -1\end{bmatrix}##. I first used the ##\frac{1}{4}R1##-> ##R1##. So the ##E_1=\begin{bmatrix} \frac{1}{4} & 0 \\ 0& 1\end{bmatrix}##. So the matrix ##A= \begin{bmatrix}1 & \frac{-1}{4} \\ 3& -1\end{bmatrix}## we can use ##-3R1+R2->R2##. ##A''= \begin{bmatrix}1 & \frac{-1}{4} \\ 0& \frac{-1}{4}\end{bmatrix}## and ##E_2=\begin{bmatrix} 1 & 0 \\ -3& 1\end{bmatrix}##. We multiply ##4R2->R2##,##A'''= \begin{bmatrix} 1 & \frac{-1}{4} \\ 0& 1\end{bmatrix}## and ##E_3=\begin{bmatrix} 1 & 0 \\ 0& 4\end{bmatrix}##. Then we multiply 1/4 to row 2 and add to row 1,##A''''= \begin{bmatrix}1 & 0 \\ 0& 1\end{bmatrix}## and ##E_4=\begin{bmatrix} 1 & \frac{1}{4} \\ 0& 1\end{bmatrix}##. So ##A={E_1}^{-1}{E_2}^{-1}{E_3}^{-1}{E_4}^{-1}##. But in the book's answer key, it said that ##A={E_2}^{-1}{E_3}^{-1}{E_4}^{-1}##. I am confused as to why the book's answer is different from mine. I understand that the sequence is not unique. Here is the study guide's answer as well. [/QUOTE]
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Mathematics
General Math
Factoring Matrices with Elementary Row Operations
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