10. 연립일차방정식System of Linear Equations

Solution.

1. The system is equivalent to the single matrix equation $$\begin{pmatrix} 1 & 1 & -1 \\ 4 & 1 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.$$
2. Applying the elementary row operation $R_2 \to R_2 - 4R_1$ to the augmented matrix gives $$\begin{pmatrix} 1 & 1 & -1 & 0 \\ 4 & 1 & -2 & 0 \end{pmatrix} \ \longrightarrow\ \begin{pmatrix} 1 & 1 & -1 & 0 \\ 0 & -3 & 2 & 0 \end{pmatrix}.$$ The second row gives $-3x_2 + 2x_3 = 0$, i.e. $x_2 = \frac{2}{3}x_3$. Setting $x_3 = t$ ($t \in \mathbb{R}$) yields $x_2 = \frac{2}{3}t$, and the first row gives $x_1 = x_3 - x_2 = t - \frac{2}{3}t = \frac{1}{3}t$. Hence every solution has the form $$(x_1, x_2, x_3) = \frac{1}{3}t\,(1, 2, 3), \qquad t \in \mathbb{R},$$ and reparametrizing with $s = t/3$, the solution set is $$\{\, s(1,2,3) : s \in \mathbb{R} \,\} = \operatorname{span}\{(1,2,3)\}.$$ Indeed, substituting $(x_1,x_2,x_3) = (1,2,3)$ gives $1+2-3=0$ and $4(1)+2-2(3)=4+2-6=0$, confirming that it is a solution.

Solution.

1. The system is equivalent to the matrix equation $$A\mathbf{x} = \begin{pmatrix} 1 & 2 & -1 \\ 1 & 1 & 1 \\ 2 & -2 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \\ 4 \end{pmatrix} = \mathbf{b}.$$
2. We find $A^{-1}$ by row-reducing $(A \mid I)$. $$\left(\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 1 & 0 \\ 2 & -2 & 1 & 0 & 0 & 1 \end{array}\right)$$ Applying $R_2 \to R_2 - R_1$, $R_3 \to R_3 - 2R_1$ gives $$\left(\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 1 & 0 \\ 0 & -6 & 3 & -2 & 0 & 1 \end{array}\right),$$ and $R_3 \to R_3 - 6R_2$ gives $$\left(\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 1 & 0 \\ 0 & 0 & -9 & 4 & -6 & 1 \end{array}\right).$$ Scaling the pivots to $1$ via $R_2 \to -R_2$, $R_3 \to -\frac{1}{9}R_3$ gives $$\left(\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ 0 & 1 & -2 & 1 & -1 & 0 \\ 0 & 0 & 1 & -\frac{4}{9} & \frac{2}{3} & -\frac{1}{9} \end{array}\right),$$ and finally applying $R_2 \to R_2 + 2R_3$, $R_1 \to R_1 + R_3$, $R_1 \to R_1 - 2R_2$ in turn gives $$\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{3} & 0 & \frac{1}{3} \\ 0 & 1 & 0 & \frac{1}{9} & \frac{1}{3} & -\frac{2}{9} \\ 0 & 0 & 1 & -\frac{4}{9} & \frac{2}{3} & -\frac{1}{9} \end{array}\right).$$ That is, $$A^{-1} = \begin{pmatrix} \frac13 & 0 & \frac13 \\ \frac19 & \frac13 & -\frac29 \\ -\frac49 & \frac23 & -\frac19 \end{pmatrix}.$$ Therefore $$\mathbf{x} = A^{-1}\mathbf{b} = \begin{pmatrix} \frac13 & 0 & \frac13 \\ \frac19 & \frac13 & -\frac29 \\ -\frac49 & \frac23 & -\frac19 \end{pmatrix}\begin{pmatrix} 5 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} \frac{5}{3}+\frac{4}{3} \\ \frac{5}{9}+\frac{3}{9}-\frac{8}{9} \\ -\frac{20}{9}+\frac{6}{9}-\frac{4}{9} \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ -2 \end{pmatrix},$$ i.e. $x_1=3,\ x_2=0,\ x_3=-2$. Substituting back, $3+2(0)-(-2)=5$, $3+0+(-2)=1$, and $2(3)-2(0)+(-2)=4$ all hold, confirming this solution.

Solution. Since $T^{-1}(1,11) = \{(a,b,c) \in \mathbb{R}^3 : T(a,b,c) = (1,11)\}$, it suffices to solve the system

$$\begin{aligned} a + b &= 1 \\ 2a - c &= 11. \end{aligned}$$

Since there are $3$ unknowns and only $2$ equations, the solution set has one free parameter. Setting $a = t$ ($t \in \mathbb{R}$), the first equation gives $b = 1-t$ and the second gives $c = 2t-11$. Hence

$$T^{-1}(1,11) = \{\, (t,\ 1-t,\ 2t-11) : t \in \mathbb{R} \,\} = (0,1,-11) + t(1,-1,2), \qquad t \in \mathbb{R}.$$

Indeed $T(t, 1-t, 2t-11) = \big(t+(1-t),\ 2t-(2t-11)\big) = (1, 11)$ for every $t$, confirming this parametrization of $T^{-1}(1,11)$.

Proof.

1. Let $L_A : F^n \to F^m$ be the left-multiplication transformation $L_A(x) = Ax$; by definition, $\operatorname{Im}(L_A) = \{L_A(x) : x \in F^n\} = \{Ax : x \in F^n\}$.

   ($\Rightarrow$) If $Ax=b$ has a solution $x_0$, then $Ax_0=b$, so $L_A(x_0)=b$ and hence $b \in \operatorname{Im}(L_A)$.

   ($\Leftarrow$) Conversely, if $b \in \operatorname{Im}(L_A)$, there exists $x_0 \in F^n$ with $L_A(x_0)=b$, i.e. $Ax_0=b$. Then $x_0$ is a solution of $Ax=b$, so $Ax=b$ has a solution.

   Hence $Ax=b$ has a solution if and only if $b \in \operatorname{Im}(L_A)$.

2. By (1), whether $(1,3,-2) \in \operatorname{Im}\,T$ can be decided by checking whether the system $T(a,b,c)=(1,3,-2)$, namely $$\begin{aligned} a + b &= 1 \\ b - 2c &= 3 \\ a + 2c &= -2 \end{aligned}$$ has a solution. Subtracting the third equation from the first gives $b-2c=3$, which is exactly the second equation, so the third equation follows automatically from the first two. Indeed, taking $c=0$, the second equation gives $b=3$ and the first gives $a=1-b=-2$, and the third equation also holds: $a+2c=-2+0=-2$. Thus $(a,b,c)=(-2,3,0)$ satisfies $$T(-2,3,0) = (-2+3,\ 3-2(0),\ -2+2(0)) = (1,3,-2),$$ so $(1,3,-2) \in \operatorname{Im}\,T$.

Solution. The augmented matrix is

$$\left(\begin{array}{ccccc|c} 3 & -1 & 1 & -1 & 2 & 5 \\ 1 & -1 & -1 & -2 & -1 & 2 \\ 5 & -2 & 1 & -3 & 3 & 10 \\ 2 & -1 & 0 & -2 & 1 & 5 \end{array}\right).$$

Swapping $R_1 \leftrightarrow R_2$ to get a pivot of $1$ and applying $R_2 \to R_2-3R_1$, $R_3 \to R_3-5R_1$, $R_4 \to R_4-2R_1$ gives

$$\left(\begin{array}{ccccc|c} 1 & -1 & -1 & -2 & -1 & 2 \\ 0 & 2 & 4 & 5 & 5 & -1 \\ 0 & 3 & 6 & 7 & 8 & 0 \\ 0 & 1 & 2 & 2 & 3 & 1 \end{array}\right).$$

Swapping $R_2 \leftrightarrow R_4$ to get a pivot of $1$ in row $2$ and applying $R_3 \to R_3-3R_2$, $R_4 \to R_4-2R_2$ gives

$$\left(\begin{array}{ccccc|c} 1 & -1 & -1 & -2 & -1 & 2 \\ 0 & 1 & 2 & 2 & 3 & 1 \\ 0 & 0 & 0 & 1 & -1 & -3 \\ 0 & 0 & 0 & 1 & -1 & -3 \end{array}\right),$$

and since the third and fourth rows agree, $R_4 \to R_4-R_3$ turns the fourth row into a zero row, which we discard. Thus the coefficient matrix has rank $3$, with pivots in the $x_1, x_2, x_4$ columns, and $x_3, x_5$ are free. Applying $R_1 \to R_1+2R_3$, $R_2 \to R_2-2R_3$, and then $R_1 \to R_1+R_2$ to reach reduced row echelon form gives

$$\left(\begin{array}{ccccc|c} 1 & 0 & 1 & 0 & 2 & 3 \\ 0 & 1 & 2 & 0 & 5 & 7 \\ 0 & 0 & 0 & 1 & -1 & -3 \end{array}\right),$$

which corresponds to the equations

$$\begin{aligned} x_1 + x_3 + 2x_5 &= 3 \\ x_2 + 2x_3 + 5x_5 &= 7 \\ x_4 - x_5 &= -3. \end{aligned}$$

Setting $x_3 = s$, $x_5 = t$ ($s,t \in \mathbb{R}$) gives $x_1 = 3-s-2t$, $x_2 = 7-2s-5t$, $x_4 = -3+t$, so every solution has the parametric vector form

$$(x_1,x_2,x_3,x_4,x_5) = (3,7,0,-3,0) + s(-1,-2,1,0,0) + t(-2,-5,0,1,1), \qquad s,t \in \mathbb{R}.$$

Check: for $s=t=0$, substituting $(3,7,0,-3,0)$ into the original equations gives $3(3)-7+0-(-3)+2(0)=9-7+3=5$, $3-7-0-2(-3)-0=3-7+6=2$, $5(3)-2(7)+0-3(-3)+3(0)=15-14+9=10$, and $2(3)-7-2(-3)+0=6-7+6=5$, so all four equations hold. Substituting the direction vectors $(-1,-2,1,0,0)$ and $(-2,-5,0,1,1)$ into the corresponding homogeneous system (all right-hand sides set to $0$) also yields $0$ in every equation, confirming the solution set above.

Proof.

Note. The statement's "consistent" appears to be a typo for "inconsistent". We prove the corrected claim: for $(A|b)$ in reduced row echelon form, $Ax=b$ is inconsistent if and only if $(A|b)$ contains a row whose only nonzero entry lies in the last column.

1. Let $(A|b)$ be an $m \times (n+1)$ matrix in reduced row echelon form.

   ($\Leftarrow$) Suppose some row of $(A|b)$ has its only nonzero entry in the last column (the column corresponding to $b$), say this entry is $c \neq 0$. Then this row represents the equation $0\cdot x_1 + \cdots + 0\cdot x_n = c$, i.e. $0 = c$, which no $x$ can satisfy. Hence $Ax=b$ is inconsistent.

   ($\Rightarrow$) Conversely, suppose no row of $(A|b)$ has its only nonzero entry in the last column, i.e. every nonzero row has a leading $1$ (pivot) in one of the first $n$ columns. We construct a solution as follows: for each pivot column of $A$, set the corresponding variable equal to the entry in the last column of the row containing that pivot; set every variable corresponding to a non-pivot (free) column to $0$. Since $(A|b)$ is in reduced row echelon form, each pivot column is a standard basis vector, with a $1$ in the row containing the pivot and $0$ in every other row. Hence the $x$ defined this way satisfies the equation given by every nonzero row, while every zero row gives the trivial equation $0=0$. Thus this $x$ solves $Ax=b$, so $Ax=b$ is consistent. Taking the contrapositive: if $Ax=b$ is inconsistent, then $(A|b)$ has a row whose only nonzero entry lies in the last column.

   Hence the two conditions are equivalent.

2. Row-reduce the augmented matrix $$\left(\begin{array}{cccc|c} 1 & 2 & -1 & 1 & 2 \\ 2 & 1 & 1 & -1 & 3 \\ 1 & 2 & -3 & 2 & 2 \end{array}\right).$$ Applying $R_2 \to R_2-2R_1$, $R_3 \to R_3-R_1$ gives $$\left(\begin{array}{cccc|c} 1 & 2 & -1 & 1 & 2 \\ 0 & -3 & 3 & -3 & -1 \\ 0 & 0 & -2 & 1 & 0 \end{array}\right).$$ The third row represents $-2x_3+x_4=0$; it is neither a zero row nor a row whose only nonzero entry lies in the last column. Reducing fully to reduced row echelon form gives $$\left(\begin{array}{cccc|c} 1 & 0 & 0 & -\frac12 & \frac43 \\ 0 & 1 & 0 & \frac12 & \frac13 \\ 0 & 0 & 1 & -\frac12 & 0 \end{array}\right),$$ which contains no row whose only nonzero entry lies in the last column. By (1), the system is therefore consistent. Indeed, setting $x_4=0$ gives $x_1=\frac43$, $x_2=\frac13$, $x_3=0$, and substituting into the original equations gives $\frac43+\frac23-0+0=2$, $\frac83+\frac13+0-0=3$, and $\frac43+\frac23-0+0=2$, all of which hold, confirming this is a solution.

Solution. Since $A$ and its reduced row echelon form $R$ have the same null space (the solution set of the homogeneous system), the linear relations among the columns of $A$ are exactly the same as the linear relations among the corresponding columns of $R$. In $R$, the pivot columns are the first, second, and fourth, and the remaining two columns are the following combinations of the pivot columns.

$$R\text{'s third column} = 2(R\text{'s first column}) - 5(R\text{'s second column}), \qquad R\text{'s fifth column} = -2(R\text{'s first column}) - 3(R\text{'s second column}) + 6(R\text{'s fourth column}).$$

Writing $a_i$ for the $i$-th column of $A$, the same coefficients give

$$a_3 = 2a_1 - 5a_2, \qquad a_5 = -2a_1 - 3a_2 + 6a_4.$$

1. Since $a_1=(1,-1,3)$, $a_2=(0,-1,1)$, $a_4=(1,-2,0)$ are the first, second, and fourth columns of $A$, $$a_3 = 2(1,-1,3) - 5(0,-1,1) = (2,-2,6)-(0,-5,5) = (2,3,1),$$ $$a_5 = -2(1,-1,3) - 3(0,-1,1) + 6(1,-2,0) = (-2,2,-6)+(0,3,-3)+(6,-12,0) = (4,-7,-9).$$ Hence $$A = \begin{pmatrix} 1 & 0 & 2 & 1 & 4 \\ -1 & -1 & 3 & -2 & -7 \\ 3 & 1 & 1 & 0 & -9 \end{pmatrix}.$$ Check. Applying $R_2\to R_2+R_1$, $R_3\to R_3-3R_1$ gives $$\begin{pmatrix} 1 & 0 & 2 & 1 & 4 \\ 0 & -1 & 5 & -1 & -3 \\ 0 & 1 & -5 & -3 & -21 \end{pmatrix},$$ and $R_3\to R_3+R_2$ gives $$\begin{pmatrix} 1 & 0 & 2 & 1 & 4 \\ 0 & -1 & 5 & -1 & -3 \\ 0 & 0 & 0 & -4 & -24 \end{pmatrix}.$$ Scaling the pivots to $1$ via $R_2 \to -R_2$, $R_3 \to -\frac14 R_3$ gives $$\begin{pmatrix} 1 & 0 & 2 & 1 & 4 \\ 0 & 1 & -5 & 1 & 3 \\ 0 & 0 & 0 & 1 & 6 \end{pmatrix},$$ and finally $R_1 \to R_1 - R_3$, $R_2 \to R_2 - R_3$ gives $$\begin{pmatrix} 1 & 0 & 2 & 0 & -2 \\ 0 & 1 & -5 & 0 & -3 \\ 0 & 0 & 0 & 1 & 6 \end{pmatrix},$$ which matches the given $R$ exactly.
2. Now $a_1=(1,-1,3)$ and $a_2=(0,-1,1)$ are as before, but the third given vector $(1,-2,0)$ is the fifth column of $A$, i.e. $a_5=(1,-2,0)$. Since $a_3=2a_1-5a_2$ depends only on $a_1,a_2$, we still get $a_3=(2,3,1)$ as in (1). However $a_4$ is now unknown, and solving $a_5=-2a_1-3a_2+6a_4$ for $a_4$ gives $$a_4 = \frac16\big(a_5+2a_1+3a_2\big) = \frac16\big((1,-2,0)+(2,-2,6)+(0,-3,3)\big) = \frac16(3,-7,9) = \left(\frac12,\,-\frac76,\,\frac32\right).$$ Hence $$A = \begin{pmatrix} 1 & 0 & 2 & \frac12 & 1 \\ -1 & -1 & 3 & -\frac76 & -2 \\ 3 & 1 & 1 & \frac32 & 0 \end{pmatrix}.$$ Check. Applying $R_2\to R_2+R_1$, $R_3\to R_3-3R_1$ gives $$\begin{pmatrix} 1 & 0 & 2 & \frac12 & 1 \\ 0 & -1 & 5 & -\frac23 & -1 \\ 0 & 1 & -5 & 0 & -3 \end{pmatrix},$$ and $R_3\to R_3+R_2$ gives $$\begin{pmatrix} 1 & 0 & 2 & \frac12 & 1 \\ 0 & -1 & 5 & -\frac23 & -1 \\ 0 & 0 & 0 & -\frac23 & -4 \end{pmatrix}.$$ Scaling the pivots to $1$ via $R_2 \to -R_2$, $R_3 \to -\frac32 R_3$ gives $$\begin{pmatrix} 1 & 0 & 2 & \frac12 & 1 \\ 0 & 1 & -5 & \frac23 & 1 \\ 0 & 0 & 0 & 1 & 6 \end{pmatrix},$$ and finally $R_1 \to R_1 - \frac12 R_3$, $R_2 \to R_2 - \frac23 R_3$ gives $$\begin{pmatrix} 1 & 0 & 2 & 0 & -2 \\ 0 & 1 & -5 & 0 & -3 \\ 0 & 0 & 0 & 1 & 6 \end{pmatrix},$$ which again matches the given $R$ exactly.

Solution. Identifying a $2\times2$ symmetric matrix $\begin{pmatrix}a&b\\b&d\end{pmatrix}$ with its coordinate vector $(a,b,d)\in\mathbb{R}^3$ gives $W \cong \mathbb{R}^3$, so $\dim W = 3$. Labeling the five elements of $S$ in order as $A_1,\ldots,A_5$, their coordinate vectors are

$$A_1 = (0,-1,1), \quad A_2 = (1,2,3), \quad A_3 = (2,1,9), \quad A_4=(1,-2,4), \quad A_5=(-1,2,-1).$$

Form the matrix whose columns are these coordinate vectors, and row-reduce to determine which elements of $S$ form a basis for $W$.

$$\begin{pmatrix} 0&1&2&1&-1\\-1&2&1&-2&2\\1&3&9&4&-1 \end{pmatrix}$$

Swapping the first two rows and negating ($R_1 \leftrightarrow R_2$, $R_1 \to -R_1$) gives

$$\begin{pmatrix} 1&-2&-1&2&-2\\0&1&2&1&-1\\1&3&9&4&-1 \end{pmatrix},$$

and $R_3 \to R_3 - R_1$ gives

$$\begin{pmatrix} 1&-2&-1&2&-2\\0&1&2&1&-1\\0&5&10&2&1 \end{pmatrix},$$

and $R_3 \to R_3 - 5R_2$ gives

$$\begin{pmatrix} 1&-2&-1&2&-2\\0&1&2&1&-1\\0&0&0&-3&6 \end{pmatrix}.$$

The pivots lie in columns $1$, $2$, $4$ (no new pivot appears in column $3$), so $A_1, A_2, A_4$ form a basis for the column space, which is $W$, while $A_3$ and $A_5$ are linear combinations of $A_1,A_2$ and of $A_1,A_2,A_4$, respectively.

To confirm this directly: $A_1,A_2$ are not scalar multiples of each other, hence linearly independent. If $A_4 = xA_1+yA_2$, the first coordinate forces $y=1$ and the second forces $-x+2y=-2 \Rightarrow x=4$, but then the third coordinate would require $x+3y=4+3=7$, not $4$ — a contradiction, so $A_4 \notin \operatorname{span}\{A_1,A_2\}$. Hence $\{A_1,A_2,A_4\}$ is linearly independent, and since $\dim W=3$, it is a basis for $W$.

Therefore the subset of $S$

$$\left\{ \begin{pmatrix} 0 & -1 \\ -1 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix} \right\}$$

is a basis for $W$.

(Note: such a basis subset of $S$ is not unique; for example, $\{A_1,A_2,A_5\}$ is also a basis for $W$.)

Solution. First we find a basis for $V$. Applying $R_2 \to R_2-2R_1$ to the coefficient matrix gives

$$\begin{pmatrix} 1&-1&0&2&-3&1\\2&-1&-1&3&-4&4 \end{pmatrix} \ \longrightarrow\ \begin{pmatrix} 1&-1&0&2&-3&1\\0&1&-1&-1&2&2 \end{pmatrix},$$

and applying $R_1 \to R_1+R_2$ gives the reduced row echelon form

$$\begin{pmatrix} 1&0&-1&1&-1&3\\0&1&-1&-1&2&2 \end{pmatrix}.$$

The coefficient matrix has rank $2$, so the solution space $V$ of this homogeneous system in $6$ unknowns has dimension $6-2=4$. The pivots lie in the $x_1,x_2$ columns, and $x_3,x_4,x_5,x_6$ are free, with

$$x_1 = x_3-x_4+x_5-3x_6, \qquad x_2 = x_3+x_4-2x_5-2x_6.$$

Substituting $(x_3,x_4,x_5,x_6)=(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$ in turn gives the basis for $V$

$$v_1=(1,1,1,0,0,0),\ v_2=(-1,1,0,1,0,0),\ v_3=(1,-2,0,0,1,0),\ v_4=(-3,-2,0,0,0,1).$$

(Check: substituting $v_1$ into the original equations gives $1-1+0-0+0=0$ and $2-1-1+0-0+0=0$, and $v_2,v_3,v_4$ likewise satisfy both equations.)

Now we extend $S=\{s_1,s_2\}=\{(0,-1,0,1,1,0),(1,0,1,1,1,0)\}$. First, $s_1,s_2 \in V$: for $s_1$, $0-(-1)+2(1)-3(1)+0=0$ and $0-(-1)-0+3(1)-4(1)+0=0$; for $s_2$, $1-0+2(1)-3(1)+0=0$ and $2(1)-0-1+3(1)-4(1)+0=0$. So indeed $s_1,s_2\in V$.

To determine whether $\{s_1,s_2,v_3,v_4\}$ is a basis of $V$, row-reduce the $4\times6$ matrix with these four vectors as rows.

$$\begin{pmatrix} 0&-1&0&1&1&0\\1&0&1&1&1&0\\1&-2&0&0&1&0\\-3&-2&0&0&0&1 \end{pmatrix}$$

Swapping the first two rows gives

$$\begin{pmatrix} 1&0&1&1&1&0\\0&-1&0&1&1&0\\1&-2&0&0&1&0\\-3&-2&0&0&0&1 \end{pmatrix},$$

and applying $R_3 \to R_3-R_1$, $R_4 \to R_4+3R_1$ gives

$$\begin{pmatrix} 1&0&1&1&1&0\\0&-1&0&1&1&0\\0&-2&-1&-1&0&0\\0&-2&3&3&3&1 \end{pmatrix},$$

and applying $R_3 \to R_3-2R_2$, $R_4 \to R_4-2R_2$ gives

$$\begin{pmatrix} 1&0&1&1&1&0\\0&-1&0&1&1&0\\0&0&-1&-3&-2&0\\0&0&3&1&1&1 \end{pmatrix},$$

and finally $R_4 \to R_4+3R_3$ gives

$$\begin{pmatrix} 1&0&1&1&1&0\\0&-1&0&1&1&0\\0&0&-1&-3&-2&0\\0&0&0&-8&-5&1 \end{pmatrix}.$$

There are four pivots, with leading entries $1,-1,-1,-8$, so this $4\times6$ matrix has rank $4$, and hence $s_1,s_2,v_3,v_4$ are linearly independent. Since $\dim V=4$, the set $\{s_1,s_2,v_3,v_4\}$ is a basis for $V$.

Thus $S$ extends to the following basis for $V$.

$$\{(0,-1,0,1,1,0),\ (1,0,1,1,1,0),\ (1,-2,0,0,1,0),\ (-3,-2,0,0,0,1)\}.$$

(Note: since we adjoin two more vectors to a space of dimension $\dim V=4$, this extension to a basis is not unique.)