Solve a linear system with more variables than equations
Suppose that, after a series of elementary row operations the augmented matrix of a linear system with variables $x_1$, $x_2$, $x_3$, $x_4$ is transformed into reduced row echelon form as follows:
$$\left(\begin{array}{cccc|c}1 & 0 & 0 & 1 & 0\\0 & 1 & 0 & 2 & 1 \\0 & 0 & 1 & 3 & 0 \end{array}\right)$$.
Can I solve the linear system as below?
Let $t$ be an arbitrary real number. Then solving each linear equation corresponding to the augmented matrix for leading variable and setting $x_4=t$, we get $x_1=-t, x_2=1-2t$, and $x_3=-3t$. Thus the general solution of the linear system is
\begin{align} x_1=-t\\ x_2=1-2t\\ x_3=-3t\\ \end{align}
where t is an arbitrary real number.
Solution 1:
you have the answer already. you have 3 equations and 4 variables and infinitely many solutions.
Solution 2:
What you have done is a decomposition. Given an underconstrained system $A x =b$ you can split the variables to dependent and independent components such that
$$ x = T x_{dep} + U x_{ind} $$
in your example
$$ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \begin{pmatrix} x_4 \end{pmatrix} $$
Now you reduce the system with
$$ \begin{aligned} A T x_{dep} + A U x_{ind} & = b \\ A T x_{dep} & = b - A U x_{ind} \\ x_{dep} &= (A T)^{-1} \left(b - A U x_{ind}\right) \end{aligned} $$
The final values are
$$ x = T (A T)^{-1} b + \left({\bf 1_{4×4}} - T (A T)^{-1} A\right) U x_{dep} \\ x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix}0\\1\\0\end{pmatrix} + \begin{bmatrix} 0&0&0&-1 \\0&0&0&-2 \\ 0&0&0&-3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \begin{pmatrix} x_4 \end{pmatrix} $$ $$ x = \begin{pmatrix} -x_4 \\ 1-2 x_4 \\ -3 x_4 \\ x_4 \end{pmatrix} $$