Linear systems solved with Gaussian elimination

The first column has the pivot $1$ at the top, so with $R_2\to R_2-2R_1$ and $R_3\to R_3-3R_1$ the entries below it are made zero; the two operations can be performed in the same step because both use only the row $R_1$, which remains unchanged. The multiplication $R_2\to -\frac{1}{3}{R}_2$ brings the second pivot to $1$ and avoids fractions in the next step, where $R_3\to R_3+7R_2$ makes the entry $-7$ zero. The resulting matrix is in echelon form with three pivots, one for each unknown, therefore no unknown is free. Back substitution starts from the last equation, $-2z=-2$, and goes up to the first. The system is determined with solution $(2,-1,1)$, and the check in the third equation gives $6+1+2=9$.

After the first step the second and third rows coincide, so $R_3\to R_3-R_2$ produces a zero row, which corresponds to the equation $0=0$ and is discarded. Two pivots remain, in the columns of $x$ and $y$, while the column of $z$ has no pivot; therefore $z$ is a free unknown and we set $z=t$. The second equation, $y-z=1$, gives $y=1+t$, and the first gives $x=3-y-2z=2-3t$. The system is underdetermined with ${\infty }^1$ solutions, one for each $t\in \mathbb{R}$. For $t=0$ one obtains the solution $(2,1,0)$, which satisfies the second equation: $4+3+0=7$.

After the first step the rows $R_2$ and $R_3$ have the same coefficients of the unknowns and different constant terms, $0$ and $2$. The subtraction $R_3\to R_3-R_2$ makes the coefficients zero and leaves the constant term $2$, so the last row has the form $(000\mid 2)$ and corresponds to the equation $0=2$, which no triple satisfies. Since the reduced system is equivalent to the original one, the system has no solutions: it is inconsistent. The reason can also be read from the original equations: the sum of the first and third gives $2x+y-z=4$, whereas the second requires $2x+y-z=2$.

The first step makes the first column zero below the pivot and also makes the second column zero, because the coefficients of $x_2$ are twice those of $x_1$ in every equation. The pivot of the second row is therefore in the column of $x_3$, and $R_3\to R_3-2R_2$ produces a zero row. There are two pivots, in the columns of $x_1$ and $x_3$, while the columns of $x_2$ and $x_4$ have none; therefore the free unknowns are $x_2=s$ and $x_4=t$, and the solutions are ${\infty }^{4-2}={\infty }^2$. The second equation, $x_3+x_4=2$, gives $x_3=2-t$, and the first, after substitution, gives $x_1=3-2s-2t$. Separating the constant terms from those in $s$ and in $t$, the solutions are written as the particular solution $(3,0,2,0)$ plus a combination of two fixed vectors.