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I am struggling to find a simple form to establish a linear combination of rows 1 and 2 in to transform it into Echelon form on row 3:

\begin{bmatrix} 1&3/2&1/2\\ 0&1&1\\ 2&8&13\end{bmatrix}

This is being multiplied by a vector \begin{bmatrix} a\\ b\\ c\end{bmatrix} which yields the result

\begin{bmatrix} 9/4\\ -1/2\\ 2\end{bmatrix}

I tried many combinations but have been struggling to find an easy/quick way to get this done. Any recommendations of methods to quickly resolve this?Elimination and substitution is the method I'm using, but it seems counter intuitive with many trial and errors sometimes...

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  • $\begingroup$ Is it the original matrix that is multiplied by the unknown vector that you’re trying to solve for or the reduced matrix? If the former, then the last part of omer’s answer is incorrect: you have to apply the same row operations to that constant result vector, too. $\endgroup$ – amd Oct 12 '19 at 23:55
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    $\begingroup$ I recommend proceeding methodically instead of trying to take short-cuts or jumping back and forth from row to row until you’re fluent with this stuff. You’re much less likely to make an error or overlook something that way. $\endgroup$ – amd Oct 12 '19 at 23:57
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I hope I understand your problem right. It wasn't very clear to me but I'll do my best.

We start with basic row operations as follows:

$$ \left[\begin{array}{ccc}{1} & {3 / 2} & {1 / 2} \\ {0} & {1} & {1} \\ {2} & {8} & {13}\end{array}\right] \quad \quad \underset{R_{3}: R_{3}-2 R_{1}}{\longrightarrow}\left[\begin{array}{ccc}{1} & {3 / 2} & {1 / 2} \\ {0} & {1} & {1} \\ {0} & {5} & {12}\end{array}\right] \quad \underset{R_{3}: R_{3}-5 R_{2}}{\longrightarrow}\left[\begin{array}{ccc}{1} & {3 / 2} & {1 / 2} \\ {0} & {1} & {1} \\ {0} & {0} & {7}\end{array}\right] $$

Now if I understand correctly, we want to solve for $(a,b,c)$ to get: $$ \left[\begin{array}{ccc}{1} & {3 / 2} & {1 / 2} \\ {0} & {1} & {1} \\ {0} & {0} & {7}\end{array}\right]\left[\begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right]=\left[\begin{array}{c}{9 / 4} \\ {-1 / 2} \\ {2}\end{array}\right] $$

Multiplying gives the following system of equations which should be pretty straightforward to solve:

$$ \begin{aligned} a+\frac{3}{2} b+\frac{1}{2} c &=\frac{9}{4} \\ b+c &=-\frac{1}{2} \\ 7 c &=2 \end{aligned} $$

Can you finish?

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  • $\begingroup$ Great-It's pretty clear. That was I was doing. But I felt like going back and forth on each row sometimes. But if that's the way it is.... I heard that Elimination + Back substitution is computationally effective, so I'll just go with it $\endgroup$ – Batmaths Oct 12 '19 at 13:45
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Are you trying to do this with one step? It can easily be done with two row operations. First subtract twice the first row from the third row. That gives $\begin{bmatrix}1 & 3/2 & 1/2 \\ 0 & 1 & 1 \\ 0 & 3 & 12\end{bmatrix}$ and is equivalent to multiplying by $\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{bmatrix}$.

Then subtract 3 times the second row from the third row. That gives $\begin{bmatrix}1 & 3/2 & 1/2 \\ 0 & 1 & 1 \\ 0 & 0 & 9\end{bmatrix}$ and is equivalent to multiplying by $\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -3 & 1 \end{bmatrix}$.

Doing the two together, as one step, is equivalent to multiplying by $\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -3 & 1\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & -3 & 1 \end{bmatrix}$

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