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I'm just starting to learn linear algebra and Gaussian elimination to solve systems of linear equations. In the school I used to solve systems of equations by substitution, e.g. express $Y$ via $X$ and then substitute this expression instead of $Y$. Then find $X$ and then find $Y$ from the first substitution. I'm curious why do we need Gaussian elimination, reduced row echelon forms in order to solve systems of linear equations if "old" approaches work too?

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If you start with a simple example, like this one

$$\left(\begin{array}{cc|c} 1 & 1 &2 \\ 0 & 1 & 3 \end{array} \right),\iff \left\{\begin{array}{cccc} x&+&y&=2\\ &&y&=3 \end{array} \right.$$

You can see that substitution yields, $x+3=2,$ so $x=-1.$ Let $R_1$ be row 1 and $R_2$ be row 2, then

$$\left(\begin{array}{cc|c} 1 & 1 &2 \\ 0 & 1 & 3 \end{array} \right)\quad \overset{\implies}{R_1-R_2=R_1} \quad \left(\begin{array}{cc|c} 1 & 0 &-1 \\ 0 & 1 & 3 \end{array} \right)\iff \left\{\begin{array}{c} x&=-1\\ y&=~~~3 \end{array} \right.$$ So you can see that we get the same result. The crucial point here is that elimination is a more general form of substitution, where you don't need to know what the variable is explicitly before substituting it into your equations.

So, no you don't "need" gaussian elimination. It's an algorithm which simplifies the process of reducing the interdependence of systems, and allows you to make variable choices in the easiest manner possible. On the other hand, you should do your best to understand Gaussian elimination, as it is considered the default method for solving these types of problems. Solutions generated by using this technique are canonical, so to speak, which will help you understand certain conventions in linear algebra better which probably draw roots from this method.

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    $\begingroup$ Thanks, Justin, makes sense $\endgroup$
    – Valentin V
    Apr 27 '17 at 14:26

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