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I've been working on some row reduction problems in linear algebra, and I came across a specific exercise where I'm having trouble identifying the mistake in my calculations. I was hoping some of you could help me out with some error analysis.

Here's the original matrix and the steps I took for row reduction:

$$\begin{aligned} & \left(\begin{array}{ccc|c}2 & 1 & -2 & -3 \\ 0 & 2 & -1 & -1 \\ 1 & 0 & -1 & -1\end{array}\right) \sim\left(\begin{array}{ccc|c}2 & 1 & -2 & -3 \\ 0 & 2 & -1 & -1 \\ 0 & \frac{1}{2} & 0 & -4\end{array}\right) \\ & \left(\begin{array}{ccc|c}2 & 1 & -2 & -3 \\ 0 & 2 & -1 & -1 \\ 0 & 0 & -\frac{1}{4} & -\frac{17}{4}\end{array}\right) \sim\left(\begin{array}{ccc|c}2 & 1 & -2 & -3 \\ 0 & 2 & -1 & -1 \\ 0 & 0 & 1 & 17\end{array}\right)\end{aligned}$$

$$\begin{aligned} & \left(\begin{array}{ccc|c}2 & 1 & -2 & -3 \\ 0 & 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & 1 & 17\end{array}\right) \sim\left(\begin{array}{ccc|c}1 & \frac{1}{2} & -1 & -\frac{3}{2} \\ 0 & 1 & -\frac{1}{2} & -\frac{{ }_2^2}{2} \\ 0 & 0 & 1 & 17\end{array}\right) \\ & \left(\begin{array}{ccc|c}1 & 0 & -\frac{5}{4} & -\frac{7}{4} \\ 0 & 1 & -\frac{1}{2} & \frac{-1}{2} \\ 0 & 0 & 1 & 17\end{array}\right) \sim\left(\begin{array}{ccc|c}1 & 0 & -\frac{5}{4} & -\frac{7}{4} \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 17\end{array}\right) \\ & \sim\left(\begin{array}{ccc|c}1 & 0 & 0 & \frac{78}{4} \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 17\end{array}\right)\end{aligned}$$

However, the correct answer should be:

$$\left(\begin{array}{cccc} 1 & 0 & -1 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & -1 & 1 \end{array}\right) \sim \left(\begin {array}{cccc} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{array}\right)$$

I'm confused about where I went wrong in my row reduction process. It would be greatly appreciated if someone could point out my mistake or provide any insights into why my calculations didn't yield the correct result.

Thanks in advance for your help!

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1 Answer 1

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Is your first move to replace $(3)$ with $\frac12(1) - (3)$? (I'm using $(3)$ to stand for row $3$, etc.)

If so, there's an error in the righmost column: you should have $\frac12(-3)-(-1)=-\frac12$ there, not $-4$.

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