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I was wondering how to calculate the solution of the following augmented matrix, given that $x_1$ and $x_2$ need to be free variables.

$\begin{bmatrix}1 & 2 & 3 & 1 & 3\\ 1 & 1 & 1 & 1 & 5 \end{bmatrix}$

Thus the answer has to be in the form: $\begin{cases} x_3 &= \ldots + \dots x_1 + \ldots x_2 \\ x_4 &= \ldots + \dots x_1 + \ldots x_2 \\ x_1, x_2 & free\end{cases} $

The given answer is $\begin{cases} x_3 &= -1 - \frac{1}{2} x_2 \\ x_4 &= 6 - x_1 - \frac{1}{2} x_2 \\ x_1, x_2 & free\end{cases} $

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An echelon form of this matrix is $$ \begin{bmatrix}1 & 1 & 1 & 1 & 5 \\ 0 & 1 & 2 & 0 & -2 \end{bmatrix} $$ suggesting that two variables should be free (because only two leading entries out of four columns), but you are at liberty to use the equations to express $x_{3}$ and $x_{4}$ in terms of $x_{1}$ and $x_{2}$ if you wish.

We have from row 2: $$ x_{2} + 2x_{3} = - 2 \, \Rightarrow \, x_{3} = -1 - \frac{1}{2}x_{2} $$ and from row 1: \begin{align*} x_{1} + x_{2} + x_{3} + x_{4} & = 5 \\ x_{1} + x_{2} - 1 - \frac{1}{2} x_{2} + x_{4} & = 5 \\ x_{1} + \frac{1}{2} x_{2} + x_{4} & = 6 \\ x_{4} & = 6 - x_{1} - \frac{1}{2}x_{2} \end{align*} Looking at the bigger picture, it doesn't matter which variables we choose to be free. The solutions generated will all be the same anyway.

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Do you mean that you intend to solve the underdetermined system:

$$\begin{cases}1x_1 + 2x_2 +3x_3 + 1x_4 &=& 3\\ 1x_1 + 1x_2 +1x_3 + 1x_4 &=& 5 \end{cases} \ \ (1) ? $$

for which you have the (exact) solution (I have checked) $$\begin{cases} x_3 &= -1 - \frac{1}{2} x_2 \\ x_4 &= 6 - x_1 - \frac{1}{2} x_2 \end{cases} \ \ \ (2)$$

and you desire an explanation ?

The principal idea is to select $x_3$ and $x_4$ as the "main" unknowns. You keep them on the LHS, and transfer the rest (with $x_1$ and $x_2$ considered as parameters) in the RHS. You have now the system with 2 unknows :

$$\begin{cases}3x_3 + x_4 &=& A \\ x_3 + x_4 &=&B \end{cases} \ \ (3) \ \ \text{with} \ \ A=3 - x_1 - 2x_2 \ \ and \ \ B=5 -x_1 -x_2 $$

Now you solve system (3) with Cramer formulas, to get formulas (2).

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