Solutions

Solution 2.1

(See Exercise 2.1.) By Definition 2.1, the equation $x+y=3$ is a linear equation in two variables. Solving for $y$ gives $y = 3-x$, so for every $x \in \mathbf{R}$ the pair $(x,\, 3-x)$ is a solution. The solution set is

\[ \{ (x,\, 3-x) \mid x \in \mathbf{R} \}. \]

Geometrically this is a straight line in the $xy$-plane with slope $-1$, passing through the points $(0,3)$ and $(3,0)$:

The equation is not homogeneous (see Definition 2.13), because the right-hand side equals $3 \ne 0$.

Solution 2.2

(See Exercise 2.2.) By Definition 2.1, the equation $x = 3$ is a linear equation.

One variable. Here the only unknown is $x$, and the equation uniquely determines it: $x = 3$. The solution set is therefore

\[ \{3\} \subset \mathbf{R}, \]

a single point on the real line.

Two variables. When we think of the equation as $x + 0 \cdot y = 3$ in the unknowns $x$ and $y$, the variable $y$ does not appear and can take any value in $\mathbf{R}$. For each $y \in \mathbf{R}$ the pair $(3, y)$ is a solution, so the solution set is

\[ \{(3,\, y) \mid y \in \mathbf{R}\} \subset \mathbf{R}^2. \]

Geometrically this is the vertical line $x = 3$ in the $xy$-plane:

Note that the solution set in the two-variable case is much larger than in the one-variable case: adding a variable introduces a whole new dimension of solutions.

Solution 2.3

(See Exercise 2.3.) The augmented matrix associated to the system is (see Definition 2.23)

\[ \left ( \begin{array}{cccc|c} 2 & -1 & 1 & 1 & 1 \\ 0 & 5 & -3 & -5 & -3 \\ 3 & -4 & 3 & 4 & 3 \end{array} \right ). \]

We bring this matrix to reduced row-echelon form using Method 2.31. Subtracting the first row from the third gives $R_3 = (1, -3, 2, 3 \mid 2)$, and swapping this with the first row yields

\[ \left ( \begin{array}{cccc|c} 1 & -3 & 2 & 3 & 2 \\ 0 & 5 & -3 & -5 & -3 \\ 2 & -1 & 1 & 1 & 1 \end{array} \right ). \]

Subtracting $2$ times the first row from the third eliminates $x_1$:

\[ \left ( \begin{array}{cccc|c} 1 & -3 & 2 & 3 & 2 \\ 0 & 5 & -3 & -5 & -3 \\ 0 & 5 & -3 & -5 & -3 \end{array} \right ). \]

The third row equals the second, so subtracting the second row from the third produces a zero row. Dividing the second row by $5$ creates the next leading $1$:

\[ \left ( \begin{array}{cccc|c} 1 & -3 & 2 & 3 & 2 \\ 0 & 1 & -\tfrac{3}{5} & -1 & -\tfrac{3}{5} \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

Adding $3$ times the second row to the first eliminates $x_2$ from the first row and yields the reduced row-echelon form (see Definition 2.27):

\[ \left ( \begin{array}{cccc|c} \underline{1} & 0 & \tfrac{1}{5} & 0 & \tfrac{1}{5} \\[4pt] 0 & \underline{1} & -\tfrac{3}{5} & -1 & -\tfrac{3}{5} \\[4pt] 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

There are two leading ones (underlined), in columns $1$ and $2$, so $x_1$ and $x_2$ are non-free, while $x_3$ and $x_4$ are free variables. Setting $x_3 = s$ and $x_4 = t$ with $s, t \in \mathbf{R}$, the non-zero rows give

\[ \begin{align*} x_1 &= \tfrac{1}{5} - \tfrac{s}{5}, \\ x_2 &= -\tfrac{3}{5} + \tfrac{3s}{5} + t. \end{align*} \]

The solution set is therefore

\[ \left\{ \Bigl(\tfrac{1-s}{5},\; \tfrac{3(s-1)}{5} + t,\; s,\; t\Bigr) \;\middle|\; s, t \in \mathbf{R} \right\}. \]

Solution 2.4

(See Exercise 2.4.) The matrix $A$ has $5$ rows and $7$ columns, so it is a $5\times 7$ matrix (according to the terminology introduced in Definition 2.20). The coefficient part has $6$ columns (corresponding to the six unknowns $x_1,\ldots,x_6$) and one augmented column, so $A$ is used as an augmented matrix for a linear system with $5$ equations in $6$ unknowns (see Definition 2.23).

The entries are $a_{13} = 3$ (row $1$, column $3$) and $a_{31} = 0$ (row $3$, column $1$).

The associated linear system is (see Definition 2.23):

\[ \begin{align*} x_1 + 3x_3 &= 1, \\ x_2 + 2x_3 + 4x_4 + x_5 &= 0, \\ 2x_4 + x_5 &= 2, \\ x_6 &= 3, \\ 0 &= 0. \end{align*} \]

(The fifth equation $0=0$ is trivially satisfied and imposes no constraint.)

The matrix is in row-echelon form (see Definition 2.27): the zero row is at the bottom, and the leading entries in rows $1$–$4$ lie in columns $1,2,4,6$, which are strictly increasing from left to right.

It is not in reduced row-echelon form, because (i) row $3$ has leading entry $2$, not $1$, and (ii) column $4$ contains a nonzero entry ($4$) in row $2$ above the leading entry of row $3$.

We apply the Gaussian algorithm (Method 2.29) to reach reduced row echelon form. First, scale row $3$ by $\tfrac{1}{2}$:

\[ A \xrightarrow{\frac{1}{2}R_3\to R_3} \left ( \begin{array}{cccccc|c} 1 & 0 & 3 & 0 & 0 & 0 & 1 \\ 0 & 1 & 2 & 4 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & \tfrac{1}{2} & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

Then eliminate $x_4$ from row $2$ via $R_2 \to R_2 - 4R_3$:

\[ \xrightarrow{R_2-4R_3\to R_2} \left ( \begin{array}{cccccc|c} \underline{1} & 0 & 3 & 0 & 0 & 0 & 1 \\ 0 & \underline{1} & 2 & 0 & -1 & 0 & -4 \\ 0 & 0 & 0 & \underline{1} & \tfrac{1}{2} & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & \underline{1} & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

This is the reduced row-echelon form: each leading $1$ (underlined) is the only nonzero entry in its column.

The leading $1$s are in columns $1, 2, 4, 6$ (four of them). Hence the pivot variables are $x_1, x_2, x_4, x_6$, and the free variables are $x_3$ and $x_5$.

We read off the solution using Method 2.31. Set $x_3 = s$ and $x_5 = t$ with $s,t\in\mathbf{R}$. Then:

\[ \begin{align*} x_1 &= 1 - 3s, \\ x_2 &= -4 - 2s + t, \\ x_4 &= 1 - \tfrac{t}{2}, \\ x_6 &= 3. \end{align*} \]

The solution set is

\[ \bigl\{\,(1-3s,\; -4-2s+t,\; s,\; 1-\tfrac{t}{2},\; t,\; 3)\;\bigm|\; s,t\in\mathbf{R}\,\bigr\}. \]

Solution 2.5

(See Exercise 2.5.) We apply Method 2.31 to each system. The augmented matrix for the first system is (see Definition 2.23)

\[ \left ( \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 3 & -1 & 2 & 5 \\ 4 & 0 & 1 & 6 \end{array} \right ). \]

Eliminating $x$ from rows $2$ and $3$:

\[ \xrightarrow{R_2-3R_1\to R_2,\;R_3-4R_1\to R_3} \left ( \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & -4 & 5 & 2 \\ 0 & -4 & 5 & 2 \end{array} \right ). \]

The second and third rows are identical, so subtracting row $2$ from row $3$ produces a zero row. Dividing row $2$ by $-4$ creates the next leading $1$:

\[ \xrightarrow{R_3-R_2\to R_3,\;-\frac{1}{4}R_2\to R_2} \left ( \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & -\tfrac{5}{4} & -\tfrac{1}{2} \\ 0 & 0 & 0 & 0 \end{array} \right ). \]

Subtracting row $2$ from row $1$ yields the reduced row-echelon form (see Definition 2.27):

\[ \xrightarrow{R_1-R_2\to R_1} \left ( \begin{array}{ccc|c} \underline{1} & 0 & \tfrac{1}{4} & \tfrac{3}{2} \\[4pt] 0 & \underline{1} & -\tfrac{5}{4} & -\tfrac{1}{2} \\[4pt] 0 & 0 & 0 & 0 \end{array} \right ). \]

The pivot variables are $x$ and $y$; the variable $z$ is free. Setting $z = t$ with $t \in \mathbf{R}$:

\[ \begin{align*} x &= \tfrac{3}{2} - \tfrac{t}{4}, \\ y &= -\tfrac{1}{2} + \tfrac{5t}{4}. \end{align*} \]

The solution set is $\bigl\{\,\bigl(\tfrac{3}{2}-\tfrac{t}{4},\;-\tfrac{1}{2}+\tfrac{5t}{4},\;t\bigr)\bigm|\;t\in\mathbf{R}\,\bigr\}$.

We proceed similarly with the second system. The augmented matrix is

\[ \begin{align*} \left ( \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 3 & -1 & 2 & 0 \\ 1 & 1 & -2 & 2 \end{array} \right ) & \xrightarrow{R_2-3R_1\to R_2,\;R_3-R_1\to R_3} & \left ( \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & -4 & 5 & -3 \\ 0 & 0 & -1 & 1 \end{array} \right ) \\ & \xrightarrow{-R_3\to R_3} & \left ( \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & -4 & 5 & -3 \\ 0 & 0 & 1 & -1 \end{array} \right ) \\ & \xrightarrow{R_2-5R_3\to R_2,\;R_1+R_3\to R_1} & \left ( \begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & -4 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{array} \right ). \end{align*} \]

Scale row $2$ and eliminate $y$ from row $1$:

\[ \xrightarrow{-\frac{1}{4}R_2\to R_2} \left ( \begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -\tfrac{1}{2} \\ 0 & 0 & 1 & -1 \end{array} \right ) \xrightarrow{R_1-R_2\to R_1} \left ( \begin{array}{ccc|c} \underline{1} & 0 & 0 & \tfrac{1}{2} \\[4pt] 0 & \underline{1} & 0 & -\tfrac{1}{2} \\[4pt] 0 & 0 & \underline{1} & -1 \end{array} \right ). \]

All three variables are pivot variables; there are no free variables. The system has the unique solution $(x, y, z) = \bigl(\tfrac{1}{2},\;-\tfrac{1}{2},\;-1\bigr)$.

Solution 2.6

(See Exercise 2.6.) If $a \ne 0$ or $b \ne 0$, then the equation $ax+by = c$ has infinitely many solutions. Indeed, if, say $a \ne 0$, we can subtract $by$ and divide by $a$, which gives $x = \frac{c-by}a$. Thus, for any $y \in {\bf R}$, the pair $(x=\frac{c-by}a, y)$ is a solution. A similar analysis works if $b \ne 0$. It remains to consider the case in which $a=0$ and $b=0$. In this case the solution set of the equation depends on $c$:

If $c = 0$, then any pair $(x,y)$ is a solution. Indeed: $0 x + 0y= 0$ holds true then. Thus, if $a=b=c=0$, there are infinitely many solutions.
If $c \ne 0$, the equation $0x+0y = c$ has no solution, since the left hand side is always 0, while the right hand side is nonzero. So, in the case $a=b=0$ but $c \ne 0$, there is no solution.

Solution 2.7

(See Exercise 2.7.) We apply Method 2.31 to bring each augmented matrix (see Definition 2.23) to reduced row-echelon form (see Definition 2.27).

We consider the system from Example 2.8: $x+y=4$, $x-y=1$. We bring the augmented matrix to reduced row-echelon form:

\[ \begin{align*} \left ( \begin{array}{cc|c} 1 & 1 & 4 \\ 1 & -1 & 1 \end{array} \right ) %Subtract row~$1$ from row~$2$, then scale row~$2$ by $-\tfrac{1}{2}$: & \xrightarrow{R_2 - R_1 \to R_2} & \left ( \begin{array}{cc|c} 1 & 1 & 4 \\ 0 & -2 & -3 \end{array} \right ) \\ & \xrightarrow{-\frac{1}{2}R_2 \to R_2} & \left ( \begin{array}{cc|c} 1 & 1 & 4 \\ 0 & 1 & \tfrac{3}{2} \end{array} \right ) \\ %Eliminate $y$ from row~$1$: & \xrightarrow{R_1 - R_2 \to R_1} & \left ( \begin{array}{cc|c} \underline{1} & 0 & \tfrac{5}{2} \\[4pt] 0 & \underline{1} & \tfrac{3}{2} \end{array} \right ). \end{align*} \]

Both variables are pivot variables; the system has the unique solution $(x, y) = \bigl(\tfrac{5}{2}, \tfrac{3}{2}\bigr)$.

We now consider the system from Example 2.10: $x+y=4$, $x+y=1$. Again we bring the augmented matrix to reduced row-echelon form:

\[ \begin{align*} \left ( \begin{array}{cc|c} 1 & 1 & 4 \\ 1 & 1 & 1 \end{array} \right ) & %Subtract row~$1$ from row~$2$, then scale row~$2$ by $-\tfrac{1}{3}$: \xrightarrow{R_2 - R_1 \to R_2} & \left ( \begin{array}{cc|c} 1 & 1 & 4 \\ 0 & 0 & -3 \end{array} \right ) \\ & \xrightarrow{-\frac{1}{3}R_2 \to R_2} & \left ( \begin{array}{cc|c} \underline{1} & 1 & 4 \\ 0 & 0 & \underline{1} \end{array} \right ). \end{align*} \]

The last row encodes $0 = 1$, which is a contradiction (cf. Method 2.31, specifically 3. there). The system has no solution.

We now consider the system from Example 2.11: $x+y=4$, $-2x-2y=-8$. The augmented matrix is

\[ \left ( \begin{array}{cc|c} 1 & 1 & 4 \\ -2 & -2 & -8 \end{array} \right ) \xrightarrow{R_2 + 2R_1 \to R_2} \left ( \begin{array}{cc|c} \underline{1} & 1 & 4 \\ 0 & 0 & 0 \end{array} \right ). \]

This is already in reduced row-echelon form. The variable $x$ is a pivot variable; $y$ is free. Setting $y = t$ with $t \in \mathbf{R}$, the solution set is

\[ \{(4 - t,\, t) \mid t \in \mathbf{R}\}. \]

Solution 2.8

(See Exercise 2.8.) By Definition 2.23, the augmented matrix of the system is

\[ \left ( \begin{array}{cc|c} 1 & 1 & 1 \\ 1 & -1 & b \end{array} \right ). \]

We apply Gaussian elimination (Method 2.29) and perform elementary row operations to bring the matrix to reduced row echelon form:

\[ \xrightarrow{R_2 - R_1 \to R_2} \left ( \begin{array}{cc|c} \underline{1} & 1 & 1 \\ 0 & -2 & b-1 \end{array} \right ) %.$$ %Dividing row~$2$ by $-2$ gives a leading $1$: %$$ \xrightarrow{-\tfrac{1}{2}R_2 \to R_2} \left ( \begin{array}{cc|c} \underline{1} & 1 & 1 \\ 0 & \underline{1} & \tfrac{1-b}{2} \end{array} \right ) %.$$ %Subtracting row~$2$ from row~$1$ yields the reduced row-echelon form (see \Cref{def:row-echelon-form}): %$$ \xrightarrow{R_1 - R_2 \to R_1} \left ( \begin{array}{cc|c} \underline{1} & 0 & \tfrac{1+b}{2} \\ 0 & \underline{1} & \tfrac{1-b}{2} \end{array} \right ). \]

Both variables are pivot variables; there are no free variables. By Method 2.31, the system has a unique solution for every $b \in \mathbf{R}$:

\[ x = \tfrac{1+b}{2}, \qquad y = \tfrac{1-b}{2}. \]

Geometric illustration. Each equation represents a line in the $xy$-plane. The first equation $x+y=1$ always gives the same line (slope $-1$, passing through $(0,1)$ and $(1,0)$). The second equation $x-y=b$ gives a family of parallel lines (slope $1$); a change of the parameter $b$ amounts to a shift of this line. Since the two lines have different slopes ($-1$ and $1$), they always intersect in exactly one point, confirming the unique solution for all $b$.

For $b=0$: the unique solution is $\bigl(\tfrac{1}{2}, \tfrac{1}{2}\bigr)$ (intersection of $x+y=1$ and $x-y=0$). For $b=1$: the unique solution is $(1, 0)$ (intersection of $x+y=1$ and $x-y=1$).

Solution 2.9

(See Exercise 2.9.) By Definition 2.23, the augmented matrix of the system is

\[ \left ( \begin{array}{cc|c} a & b & 1 \\ 1 & -1 & 2 \end{array} \right ). \]

Our goal is to use elementary row operations to bring this matrix to reduced row-echelon form. We swap rows $1$ and $2$ (see Definition 2.28) to obtain a leading $1$ in the first row:

\[ \xrightarrow{R_1 \leftrightarrow R_2} \left ( \begin{array}{cc|c} \underline{1} & -1 & 2 \\ a & b & 1 \end{array} \right ). \]

Subtracting $a$ times row $1$ from row $2$ eliminates $x$ from row $2$:

\[ \xrightarrow{R_2 - aR_1 \to R_2} \left ( \begin{array}{cc|c} \underline{1} & -1 & 2 \\ 0 & a+b & 1-2a \end{array} \right ). \]

In order to arrive at a row echelon form, we need to have “1” or “0” in place of the entry “$a+b$”. So, we now distinguish cases depending on $a+b$.

This is the case $a + b \neq 0$. In this case we can divide row $2$ by $a+b$:

\[ \xrightarrow{\tfrac{1}{a+b}R_2 \to R_2} \left ( \begin{array}{cc|c} \underline{1} & -1 & 2 \\ 0 & \underline{1} & \tfrac{1-2a}{a+b} \end{array} \right ). \]

Add row $2$ to row $1$:

\[ \xrightarrow{R_1 + R_2 \to R_1} \left ( \begin{array}{cc|c} \underline{1} & 0 & \tfrac{2b+1}{a+b} \\ 0 & \underline{1} & \tfrac{1-2a}{a+b} \end{array} \right ). \]

This is the reduced row-echelon form (Definition 2.27). Both variables are pivot variables; by Method 2.31 the system has a unique solution:

\[ x = \frac{2b+1}{a+b}, \qquad y = \frac{1-2a}{a+b}. \]

This is the case $a + b = 0$, i.e., $b = -a$. The matrix reads

\[ \left ( \begin{array}{cc|c} 1 & -1 & 2 \\ 0 & 0 & 1-2a \end{array} \right ). \]

In order to decide whether 3. of Method 2.31 applies, we again distinguish two cases:

1.  This is the subcase $a \neq \tfrac{1}{2}$ (so $1 - 2a \neq 0$). The second row encodes $0 = 1-2a \neq 0$, a contradiction. The system has *no solution*.

2.  This is the subcase $a = \tfrac{1}{2}$, $b = -\tfrac{1}{2}$. The second row is $0 = 0$ and vanishes. The variable $y$ is free; set $y = t$ with $t \in \mathbf{R}$. Then $x = 2 + t$, giving *infinitely many solutions*:

\[ \{(2+t,\, t) \mid t \in \mathbf{R}\}. \]

Summary. The system has:

a unique solution if and only if $a + b \neq 0$;
no solution if and only if $a + b = 0$ and $a \neq \tfrac{1}{2}$;
infinitely many solutions if and only if $a = \tfrac{1}{2}$ and $b = -\tfrac{1}{2}$.

Geometric explanation. The second equation $x - y = 2$ always defines the same line $\ell_2$ (slope $1$, passing through $(2,0)$ and $(0,-2)$). The first equation $ax + by = 1$ defines a line $\ell_1$ provided $(a,b) \neq (0,0)$; if $a = b = 0$ it reads $0 = 1$, which has no solution (this falls under sub-case 2a since $a+b=0$ and $a \neq \frac12$).

The two lines $\ell_1$ and $\ell_2$ are parallel precisely when $(a,b)$ is a scalar multiple of $(1,-1)$, i.e., $b = -a$ (equivalently $a+b=0$; in terminology introduced later, in Definition 3.49 the vector $(a,b)$ is called linearly dependent of $(1,1)$):

If $a + b \neq 0$, the lines are not parallel and intersect in exactly one point: unique solution.
If $a + b = 0$ and $a \neq \tfrac{1}{2}$, the lines are parallel but distinct: no solution.
If $a = \tfrac{1}{2}$, $b = -\tfrac{1}{2}$, then $\ell_1$ is $\tfrac{1}{2}x - \tfrac{1}{2}y = 1$, i.e., $x - y = 2$, which is the same as $\ell_2$: infinitely many solutions.

Solution 2.10

(See Exercise 2.10.) The matrix associated to the system is as follows, and we bring it to reduced row echelon form:

\[ \begin{align*} \left ( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ -2 & -3 & 1 & 1 \\ 0 & 1 & -1 & 1 \end{array} \right ) \leadsto & \left ( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 1 & -1 & 1 \\ 0 & 1 & -1 & 1 \end{array} \right ) \\ \leadsto & \left ( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{ccc|c} \underline 1 & 0 & +1 & -2 \\ 0 & \underline 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right ) \end{align*} \]

We have two leading ones (underlined), so the third unknown $x_3$ is a free variable and $x_1$ and $x_2$ are non-free, and we have $x_2 = 1+x_3$ and $x_1 = -2 -x_3$. Thus, the solution set is

\[ \{(-2-x_3, 1+x_3, x_3) \ | \ x_3 \in {\bf R} \}. \]

Solution 2.11

(See Exercise 2.11.) We apply Method 2.31. We compute the reduced row-echelon form of the augmented matrix using Gaussian elimination (Method 2.29). We subtract the first row from the third and from the fourth:

\[ \left ( \begin{array}{cccc|c} 2 & -1 & 1 & -1 & 1 \\ 0 & 1 & -3 & 1 & 3 \\ 0 & 2 & -5 & 2 & 5 \\ 0 & 1 & -3 & 2 & 1 \\ \end{array} \right ). \]

We subtract twice the second row from the third, and subtract the second row from the fourth:

\[ \left ( \begin{array}{cccc|c} 2 & -1 & 1 & -1 & 1 \\ 0 & 1 & -3 & 1 & 3 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & -2 \\ \end{array} \right ). \]

This matrix is in row-echelon form (Definition 2.27). We now eliminate the remaining non-pivot entries. We subtract the third row from the first and add three times the third row to the second:

\[ \left ( \begin{array}{cccc|c} 2 & -1 & 0 & -1 & 2 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & -2 \\ \end{array} \right ). \]

We add the fourth row to the first and subtract the fourth row from the second:

\[ \left ( \begin{array}{cccc|c} 2 & -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & -2 \\ \end{array} \right ). \]

We add the second row to the first:

\[ \left ( \begin{array}{cccc|c} 2 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & -2 \\ \end{array} \right ). \]

Finally, we multiply the first row by $\tfrac{1}{2}$:

\[ \left ( \begin{array}{cccc|c} {\underline 1} & 0 & 0 & 0 & 1 \\ 0 & {\underline 1} & 0 & 0 & 2 \\ 0 & 0 & {\underline 1} & 0 & -1 \\ 0 & 0 & 0 & {\underline 1} & -2 \\ \end{array} \right ). \]

This is the reduced row-echelon form, with all four variables $x_1, x_2, x_3, x_4$ being pivot variables (there are no free variables). The unique solution is

\[ (x_1, x_2, x_3, x_4) = (1, 2, -1, -2). \]

Solution 2.12

(See Exercise 2.12.) We apply Method 2.31. The matrix associated to the system is

\[ \left ( \begin{array}{cccc|c} 1 & -1 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 & 1 \\ 1 & -1 & 0 & 1 & -3 \\ 1 & -1 & 3 & -2 & 0 \end{array} \right ). \]

We compute the reduced row-echelon form of that matrix using Gaussian eliminiation (Method 2.29): we subtract the first row from the third, which gives

\[ \left ( \begin{array}{cccc|c} 1 & -1 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 & 1 \\ 0 & 0 & -1 & 1 & -1 \\ 1 & -1 & 3 & -2 & 0 \end{array} \right ). \]

We then subtract the first row from the fourth:

\[ \left ( \begin{array}{cccc|c} 1 & -1 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 & 1 \\ 0 & 0 & -1 & 1 & -1 \\ 0 & 0 & 2 & -2 & 2 \end{array} \right ). \]

We add the second line to the third:

\[ \left ( \begin{array}{cccc|c} 1 & -1 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & -2 & 2 \end{array} \right ). \]

We then add $(-2)$ times the second line to the fourth (equivalently, subtract $2$ times the second line from the fourth):

\[ \left ( \begin{array}{cccc|c} {\underline 1} & -1 & 1 & 0 & -2 \\ 0 & 0 & {\underline 1} & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

This matrix is in row-echelon form, with the leading 1’s being underlined above. We finally bring it into reduced row-echelon form by subtracting the second from the first line, which gives

\[ \left ( \begin{array}{cccc|c} {\underline 1} & -1 & 0 & 1 & -3 \\ 0 & 0 & {\underline 1} & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

The matrix has no entry of the form $0 \ \dots \ 0 \ 1$, so the system does have a solution. The first column of the matrix corresponds to the variable $x_1$ etc., so that the free variables are $x_2$ and $x_4$. We let $x_2 = \alpha$, $x_4 = \beta$, where $\alpha$ and $\beta$ are arbitrary real numbers. The non-free variables $x_1$ and $x_3$ are uniquely determined by $\alpha$ and $\beta$. To compute them, we use the equations obtained by the matrix

\[ \begin{align*} x_3 - \beta & = 1 \\ x_1 -\alpha + \beta & = -3 \end{align*} \]

which we solve as $x_3 = 1 + \beta$ and $x_1 = \alpha-\beta-3$. Thus, the solution set is

\[ \{(\alpha - \beta - 3, \alpha, 1+\beta, \beta) \ | \ \alpha, \beta \in {\bf R}\}. \]

Solution 2.13

(See Exercise 2.13.) By Definition 2.23, the augmented matrix of the system is

\[ \left ( \begin{array}{cc|c} 1 & h & 4 \\ 3 & 6 & 8 \end{array} \right ). \]

We apply Method 2.29. Subtracting $3$ times row $1$ from row $2$ eliminates $x$:

\[ \xrightarrow{R_2 - 3R_1 \to R_2} \left ( \begin{array}{cc|c} 1 & h & 4 \\ 0 & 6-3h & -4 \end{array} \right ). \]

We now distinguish cases depending on the value of $6 - 3h = 3(2-h)$.

We consider the case $6 - 3h \neq 0$, i.e., $h \neq 2$. In this case we can divide row $2$ by $3(2-h)$:

\[ \xrightarrow{\tfrac{1}{3(2-h)}R_2 \to R_2} \left ( \begin{array}{cc|c} \underline{1} & h & 4 \\ 0 & \underline{1} & \tfrac{-4}{3(2-h)} \end{array} \right ) %.$$ %Subtract $h$ times row~$2$ from row~$1$: %$$ \xrightarrow{R_1 - hR_2 \to R_1} \left ( \begin{array}{cc|c} \underline{1} & 0 & \tfrac{8(3-h)}{3(2-h)} \\ 0 & \underline{1} & \tfrac{-4}{3(2-h)} \end{array} \right ). \]

This is the reduced row-echelon form (Definition 2.27). Both variables are pivot variables; by Method 2.31 the system has the unique solution

\[ x = \frac{8(3-h)}{3(2-h)}, \qquad y = \frac{-4}{3(2-h)}. \]

We now consider the case $h=2$. The matrix becomes

\[ \left ( \begin{array}{cc|c} 1 & 2 & 4 \\ 0 & 0 & -4 \end{array} \right ). \]

The second row encodes $0 = -4$, a contradiction. The system has no solution.

Thus, the system has a unique solution if $h \neq 2$, and no solution if $h = 2$.

Geometric illustration. The second equation $3x + 6y = 8$ always defines the fixed line $\ell_2\colon y = \tfrac{4}{3} - \tfrac{x}{2}$ (slope $-\tfrac{1}{2}$). The first equation $x + hy = 4$ defines a line $\ell_1$ with slope $-\tfrac{1}{h}$ (for $h \neq 0$); for $h = 0$ it is the vertical line $x = 4$.

For $h \neq 2$, the lines $\ell_1$ and $\ell_2$ are not parallel (their slopes differ), so they intersect in exactly one point: unique solution. For $h = 2$, both lines have slope $-\tfrac{1}{2}$; the first is $x + 2y = 4$ and the second is $x + 2y = \tfrac{8}{3}$. Since $4 \neq \tfrac{8}{3}$, the lines are parallel and distinct: no solution.

For $h = 3$: the unique solution is $\bigl(0,\,\tfrac{4}{3}\bigr)$ (intersection of $\ell_1\colon x+3y=4$ and $\ell_2\colon 3x+6y=8$). For $h = 2$: the lines $x+2y=4$ and $3x+6y=8$ are parallel; no solution.

Solution 2.14

(See Exercise 2.14.) By Definition 2.13 this is a homogeneous system; in particular the trivial solution $(x_1,x_2,x_3,x_4)=(0,0,0,0)$ always exists (Remark 2.14). (Later on, in Proposition 3.18, we will observe that the solution set of a homogeneous system in four variables is a sub-vector space of $\mathbf{R}^4$).

For $t=0$, we perform Gaussian elimination similarly as in the exercises above. One has

[ \left ( \begin{array}{cccc|c} 2 & 0 & 1 & 0 & 0 \ 1 & -2 & 0 & 3 & 0 \ 4 & -4 & 0 & 5 & 0 \end{array} \right ) \leadsto

\left ( \begin{array}{cccc|c} {\underline 1} & 0 & 0 & -\tfrac{1}{2} & 0 \ 0 & {\underline 1} & 0 & -\tfrac{7}{4} & 0 \ 0 & 0 & {\underline 1} & 1 & 0 \end{array} \right ). ]

This is the reduced row-echelon form of the above matrix. The pivot variables are $x_1, x_2, x_3$; the free variable is $x_4$. Setting $x_4 = s$ and applying Method 2.31:

\[ x_1 = \tfrac{1}{2}s, \quad x_2 = \tfrac{7}{4}s, \quad x_3 = -s. \]

The solution set is $\bigl\{s\cdot\bigl(\tfrac{1}{2},\, \tfrac{7}{4},\, -1,\, 1\bigr) \mid s \in \mathbf{R}\bigr\}$.

We now consider general $t$. We swap row $1$ and row $2$ and eliminate the first column:

\[ \left ( \begin{array}{cccc|c} 2 & 0 & 1 & -t & 0 \\ 1 & -2 & 0 & 3 & 0 \\ 4 & -4 & t & 5 & 0 \end{array} \right ) \xrightarrow{R_1 \leftrightarrow R_2} \left ( \begin{array}{cccc|c} 1 & -2 & 0 & 3 & 0 \\ 2 & 0 & 1 & -t & 0 \\ 4 & -4 & t & 5 & 0 \end{array} \right ) \]

\[ \xrightarrow{\substack{R_2 - 2R_1 \to R_2 \\ R_3 - 4R_1 \to R_3}} \left ( \begin{array}{cccc|c} 1 & -2 & 0 & 3 & 0 \\ 0 & 4 & 1 & -t-6 & 0 \\ 0 & 4 & t & -7 & 0 \end{array} \right ) \xrightarrow{R_3 - R_2 \to R_3} \left ( \begin{array}{cccc|c} 1 & -2 & 0 & 3 & 0 \\ 0 & 4 & 1 & -t-6 & 0 \\ 0 & 0 & t-1 & t-1 & 0 \end{array} \right ). \]

We distinguish cases depending on $t-1$.

Case 1: $t = 1$. Row $3$ becomes all zeros. Scaling row $2$ by $\tfrac{1}{4}$ and then eliminating:

\[ \xrightarrow{\frac{1}{4}R_2 \to R_2} \left ( \begin{array}{cccc|c} 1 & -2 & 0 & 3 & 0 \\ 0 & 1 & \tfrac{1}{4} & -\tfrac{7}{4} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ) \xrightarrow{R_1 + 2R_2 \to R_1} \left ( \begin{array}{cccc|c} {\underline 1} & 0 & \tfrac{1}{2} & -\tfrac{1}{2} & 0 \\ 0 & {\underline 1} & \tfrac{1}{4} & -\tfrac{7}{4} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

Pivot variables are $x_1, x_2$; free variables are $x_3 = s$ and $x_4 = r$. By Method 2.31: $x_1 = -\tfrac{1}{2}s + \tfrac{1}{2}r$ and $x_2 = -\tfrac{1}{4}s + \tfrac{7}{4}r$. The solution set is

\[ \Bigl\{s\cdot\Bigl(-\tfrac{1}{2},\,-\tfrac{1}{4},\,1,\,0\Bigr) + r\cdot\Bigl(\tfrac{1}{2},\,\tfrac{7}{4},\,0,\,1\Bigr) \;\Big|\; s,r\in\mathbf{R}\Bigr\}. \]

We observe that we have two free variables; in terminology introduced later on (Definition 3.66) we will say that the solution set is a two-dimensional subspace of $\mathbf{R}^4$.

Case 2: $t \neq 1$. Divide row $3$ by $t-1$, then eliminate:

\[ \xrightarrow{\frac{1}{t-1}R_3 \to R_3} \left ( \begin{array}{cccc|c} 1 & -2 & 0 & 3 & 0 \\ 0 & 4 & 1 & -t-6 & 0 \\ 0 & 0 & 1 & 1 & 0 \end{array} \right ) \xrightarrow{R_2 - R_3 \to R_2} \left ( \begin{array}{cccc|c} 1 & -2 & 0 & 3 & 0 \\ 0 & 4 & 0 & -t-7 & 0 \\ 0 & 0 & 1 & 1 & 0 \end{array} \right ). \]

Scale row $2$ by $\tfrac{1}{4}$ and eliminate column $2$:

\[ \xrightarrow{\frac{1}{4}R_2 \to R_2} \left ( \begin{array}{cccc|c} 1 & -2 & 0 & 3 & 0 \\ 0 & 1 & 0 & \frac{-t-7}{4} & 0 \\ 0 & 0 & 1 & 1 & 0 \end{array} \right ) \xrightarrow{R_1 + 2R_2 \to R_1} \left ( \begin{array}{cccc|c} {\underline 1} & 0 & 0 & \frac{-1-t}{2} & 0 \\ 0 & {\underline 1} & 0 & \frac{-t-7}{4} & 0 \\ 0 & 0 & {\underline 1} & 1 & 0 \end{array} \right ). \]

Pivot variables are $x_1, x_2, x_3$; free variable is $x_4 = s$. By Method 2.31: $x_1 = \tfrac{1+t}{2}s$, $x_2 = \tfrac{t+7}{4}s$, $x_3 = -s$. The solution set is

\[ \Bigl\{s\cdot\Bigl(\tfrac{1+t}{2},\,\tfrac{t+7}{4},\,-1,\,1\Bigr) \;\Big|\; s\in\mathbf{R}\Bigr\}. \]

Unlike before, we have one free variable only, so the solution set is a one-dimensional subspace of $\mathbf{R}^4$.

Summary. For $t\neq 1$ the solution set is the span of a single vector (one-dimensional), while for $t=1$ it is the span of two linearly independent vectors (two-dimensional). In both cases the trivial solution, i.e., the zero vector $0$ is included.

Solution 2.15

(See Exercise 2.15.) By Definition 2.13 this is a homogeneous system, so the trivial solution always exists (Remark 2.14). We form the augmented matrix and apply Gaussian elimination (Method 2.29):

\[ \left ( \begin{array}{cccc|c} 1 & 0 & -1 & 2 & 0 \\ 0 & 1 & 2 & -2 & 0 \\ 1 & 1 & 1 & 0 & 0 \end{array} \right ). \]

Subtract row $1$ from row $3$:

\[ \xrightarrow{R_3 - R_1 \to R_3} \left ( \begin{array}{cccc|c} 1 & 0 & -1 & 2 & 0 \\ 0 & 1 & 2 & -2 & 0 \\ 0 & 1 & 2 & -2 & 0 \end{array} \right ). \]

Subtract row $2$ from row $3$:

\[ \xrightarrow{R_3 - R_2 \to R_3} \left ( \begin{array}{cccc|c} {\underline 1} & 0 & -1 & 2 & 0 \\ 0 & {\underline 1} & 2 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

This is already the reduced row-echelon form (Definition 2.27). The pivot variables are $x_1$ and $x_2$; the free variables are $x_3$ and $x_4$.

Setting $x_3 = s$ and $x_4 = t$ with $s, t \in \mathbf{R}$, Method 2.31 gives

\[ x_1 = x_3 - 2x_4 = s - 2t, \qquad x_2 = -2x_3 + 2x_4 = -2s + 2t. \]

The solution set is

\[ \bigl\{(s-2t,\,-2s+2t,\,s,\,t) \mid s,t\in\mathbf{R}\bigr\} = \bigl\{s(1,-2,1,0)+t(-2,2,0,1) \mid s,t\in\mathbf{R}\bigr\}. \]

Solution 2.16

(See Exercise 2.16.) Suppose $x_1 = 1-t$, $x_2 = 2+3t$ and $x_3 = 4t$. We have to determine whether there is some $t\in {\bf R}$ such that for these choices of $x_1, x_2, x_3$, we have a solution of the given system, i.e., whether

\[ \begin{align*} x_1 + x_2 + x_3 & = (1-t)+(2+3t)+4t & = 1 \\ x_1 - x_3 & = 1-t-4t & = 0. \end{align*} \]

Simplifying these equations gives the system

\[ \begin{align*} 6t + 3 & = 0 \\ 1-5t & = 0. \end{align*} \]

This system has no solutions, so there is no $t \in {\bf R}$ such that the vector $(1-t, 2+3t, 4t)$ is a solution to the original system.

Solution 2.17

(See Exercise 2.17.) We substitute $x_1 = 3+t$, $x_2 = 2+t$, $x_3 = \tfrac{2}{3}+t$ into both equations and determine for which $t \in \mathbf{R}$ both are satisfied simultaneously. We begin by considering the first equation:

\[ \begin{align*} x_1 - x_2 + 3x_3 &= (3+t)-(2+t)+3\!\left(\tfrac{2}{3}+t\right) \\ &= 1 + 2 + 3t \\ &= 3 + 3t. \end{align*} \]

Setting this equal to $0$ gives $3+3t = 0$, i.e., $t = -1$. The second equation reads

\[ x_1 - x_2 = (3+t)-(2+t) = 1. \]

This equals $1$ for every $t \in \mathbf{R}$, so the second equation imposes no restriction on $t$.

The first equation uniquely determines $t = -1$, while the second is satisfied for all $t$. Hence there is exactly one value $t = -1$ such that $(3+t, 2+t, \tfrac{2}{3}+t)$ is a solution of the system. For this value the solution vector is

\[ (x_1, x_2, x_3) = \bigl(2,\; 1,\; -\tfrac{1}{3}\bigr). \]

Solution 2.18

(See Exercise 2.18.) We substitute $x_1 = 1+t$, $x_2 = t+q$ and $x_3 = -t+2q+1$ into the given equation and get the equation

\[ 3(1+t)+2(t+q)-(-t+2q+1)=5. \]

This simplifies to

\[ 6t+2=5 \]

which has the solution $t = -\frac 12$. Since the variable $q$ does not appear in that equation it is a free variable. Thus, for all $q \in {\bf R}$, the vector

\[ (x_1 = 1-\frac 12, x_2 = -\frac 12 + q, x_3 = \frac 12 + 2q+1) = (\frac 12, -\frac 12+q, \frac 32 + 2q) \]

satisfies the requested conditions. (Note that these are infinitely many solutions.)

Solution 2.19

(See Exercise 2.19.) We have to find $a_0, \dots, a_3$, so these are the unknowns. The conditions amount to the linear (!) system

\[ \begin{align*} p(1) & = a_0 + a_1 + a_2 + a_3 & = 0 \\ p(2) & = a_0 + a_1 \cdot 2+ a_2 \cdot 2^2 + a_3 \cdot 2^3 & = 3. \end{align*} \]

This can be rewritten as

\[ \begin{align*} a_0 + a_1 + a_2 + a_3 & = 0 \\ a_0 + 2 a_1 + 4 a_2 + 8 a_3 & = 3. \end{align*} \]

Using Gaussian elimination to solve this: the associated matrix is

\[ \left ( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 0 \\ 1 & 2 & 4 & 8 & 3 \end{array} \right ). \]

Subtracting the first from the second row gives

\[ \left ( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 3 & 7 & 3 \end{array} \right ). \]

Subtracting the second from the first yields a reduced row echelon matrix:

\[ \left ( \begin{array}{cccc|c} 1 & 0 & -2 & -6 & -3 \\ 0 & 1 & 3 & 7 & 3 \end{array} \right ). \]

The variables $a_0$ and $a_1$ correspond to the leading 1’s, the variables $a_2$ and $a_3$ are therefore free variables. Thus, there are infintely many solutions. One solution, for $a_2 = a_3 = 0$ is

\[ a_0 = -3, \ a_1 = 3, \]

so that

\[ p(x) = -3 + 3x \]

is a solution to the problem. Another solution would be $a_2 = a_3 = 1$, which gives $a_1 = -7$ and $a_0 = 5$, i.e.,

\[ p(x) = 5 - 7 x + x^2 + x^3. \]

Solution 2.20

(See Exercise 2.20.) We apply Method 2.29 to the given augmented matrix. Subtracting twice the first row from the second, and subtracting the first row from the third, gives

\[ \begin{align*} \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & 1 \\ 2 & 0 & 1 & 2 & 1 \\ 1 & 3 & 5 & 7 & 2 \end{array} \right ) & \xrightarrow{R_2 \leftarrow R_2-2R_1,\; R_3 \leftarrow R_3-R_1} \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & 1 \\ 0 & -2 & -3 & -4 & -1 \\ 0 & 2 & 3 & 4 & 1 \end{array} \right ) \\ %Adding the second row to the third yields a zero row: %$$ & \xrightarrow{R_3 \leftarrow R_3+R_2} \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & 1 \\ 0 & -2 & -3 & -4 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ) \\ %Multiplying the second row by $-\tfrac{1}{2}$ gives the row-echelon form (see \Cref{def:row-echelon-form}): % $$ \xrightarrow{R_2 \leftarrow -\frac{1}{2}R_2} \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & 1 \\ 0 & 1 & \tfrac{3}{2} & 2 & \tfrac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \end{align*} \]

This matrix is in row-echelon form (see Definition 2.27). In order to get to reduced row-echelon form we continue like so:

\[ \xrightarrow{R_1 \leftarrow R_1-R_2} \left ( \begin{array}{cccc|c} 1 & 0 & \tfrac{1}{2} & 1 & \tfrac{1}{2} \\ 0 & 1 & \tfrac{3}{2} & 2 & \tfrac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

By Method 2.31, the pivot columns are $1$ and $2$, so $x_1$ and $x_2$ are the leading variables, while $x_3$ and $x_4$ are free variables. Setting $x_3 = s$ and $x_4 = t$ with $s, t \in {\bf R}$, the two non-trivial rows give

\[ \begin{align*} x_1 &= \tfrac{1}{2} - \tfrac{1}{2}s - t, \\ x_2 &= \tfrac{1}{2} - \tfrac{3}{2}s - 2t. \end{align*} \]

The solution set is therefore

\[ \left\{ \begin{pmatrix} \tfrac{1}{2} \\ \tfrac{1}{2} \\ 0 \\ 0 \end{pmatrix} + s \begin{pmatrix} -\tfrac{1}{2} \\ -\tfrac{3}{2} \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1 \\ -2 \\ 0 \\ 1 \end{pmatrix} : s, t \in {\bf R} \right\}. \]

Solution 2.21

(See Exercise 2.21.) We apply Method 2.29 to the given augmented matrix. Subtracting twice the first row from the second, and subtracting the first row from the third, gives

\[ \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & -1 \\ 2 & 0 & 1 & 2 & \alpha \\ 1 & 3 & 5 & 7 & 0 \end{array} \right ) \xrightarrow{R_2 \leftarrow R_2-2R_1,\; R_3 \leftarrow R_3-R_1} \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & -1 \\ 0 & -2 & -3 & -4 & \alpha+2 \\ 0 & 2 & 3 & 4 & 1 \end{array} \right ). \]

Adding the second row to the third:

\[ \xrightarrow{R_3 \leftarrow R_3+R_2} \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & -1 \\ 0 & -2 & -3 & -4 & \alpha+2 \\ 0 & 0 & 0 & 0 & \alpha+3 \end{array} \right ). \]

The last row encodes the equation $0 = \alpha+3$, which requires $\alpha = -3$ for any solution to exist (compare Method 2.31, 3. there).

Case $\alpha \neq -3$. The last row gives the contradiction $0 = \alpha+3 \neq 0$, so the system has no solution.
Case $\alpha = -3$. The last row vanishes. With $\alpha+2 = -1$, the matrix becomes

\[ \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & -1 \\ 0 & -2 & -3 & -4 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

Multiplying the second row by $-\tfrac{1}{2}$ and then subtracting it from the first produces the reduced row-echelon form (see Definition 2.27):

\[ \xrightarrow{R_2 \leftarrow -\frac{1}{2}R_2} \left ( \begin{array}{cccc|c} 1 & 1 & 2 & 3 & -1 \\ 0 & 1 & \tfrac{3}{2} & 2 & \tfrac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ) \xrightarrow{R_1 \leftarrow R_1-R_2} \left ( \begin{array}{cccc|c} 1 & 0 & \tfrac{1}{2} & 1 & -\tfrac{3}{2} \\ 0 & 1 & \tfrac{3}{2} & 2 & \tfrac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

By Method 2.31, $x_1$ and $x_2$ are leading variables while $x_3 = s$ and $x_4 = t$ are free. The two non-trivial rows give

\[ \begin{align*} x_1 &= -\tfrac{3}{2} - \tfrac{1}{2}s - t, \\ x_2 &= \tfrac{1}{2} - \tfrac{3}{2}s - 2t. \end{align*} \]

The solution set is therefore

\[ \left\{ \begin{pmatrix} -\tfrac{3}{2} \\ \tfrac{1}{2} \\ 0 \\ 0 \end{pmatrix} + s \begin{pmatrix} -\tfrac{1}{2} \\ -\tfrac{3}{2} \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1 \\ -2 \\ 0 \\ 1 \end{pmatrix} : s, t \in {\bf R} \right\}. \]

Solution 2.22

(See Exercise 2.22.) We apply Method 2.29 to the augmented matrix of the system. Dividing the first row by $2$ and then subtracting $(\alpha+2)$ times the first row from the second gives

\[ \begin{align*} \left ( \begin{array}{ccc|c} 2 & -1 & 1 & 1 \\ \alpha+2 & -2 & \alpha & -\alpha \end{array} \right ) & \xrightarrow{R_1 \leftarrow \frac{1}{2}R_1} \left ( \begin{array}{ccc|c} 1 & -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \\ \alpha+2 & -2 & \alpha & -\alpha \end{array} \right ) \\ & \xrightarrow{R_2 \leftarrow R_2 - (\alpha+2)R_1} \left ( \begin{array}{ccc|c} 1 & -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \\ 0 & \tfrac{\alpha-2}{2} & \tfrac{\alpha-2}{2} & \tfrac{-3\alpha-2}{2} \end{array} \right ). \end{align*} \]

The second row encodes the equation $\tfrac{\alpha-2}{2}(y + z) = \tfrac{-3\alpha-2}{2}$, which forces a case split on $\alpha = 2$:

Case $\alpha = 2$. The second row becomes $0 \cdot y + 0 \cdot z = -4$, i.e., $0 = -4$, a contradiction. The system has no solution.
Case $\alpha \neq 2$. We may divide the second row by $\tfrac{\alpha-2}{2}$ and then eliminate the $-\tfrac{1}{2}$ entry in the first row (Definition 2.28):

\[ \xrightarrow{R_2 \leftarrow \frac{2}{\alpha-2}R_2} \left ( \begin{array}{ccc|c} 1 & -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \\ 0 & 1 & 1 & \tfrac{-3\alpha-2}{\alpha-2} \end{array} \right ) \xrightarrow{R_1 \leftarrow R_1 + \frac{1}{2}R_2} \left ( \begin{array}{ccc|c} 1 & 0 & 1 & \tfrac{-(\alpha+2)}{\alpha-2} \\ 0 & 1 & 1 & \tfrac{-3\alpha-2}{\alpha-2} \end{array} \right ). \]

By Method 2.31, $z$ is free; setting $z = t$ gives

\[ \begin{align*} x &= \tfrac{-(\alpha+2)}{\alpha-2} - t, \\ y &= \tfrac{-3\alpha-2}{\alpha-2} - t. \end{align*} \]

The system has infinitely many solutions, forming the line

\[ \left\{ \begin{pmatrix} \tfrac{-(\alpha+2)}{\alpha-2} \\ \tfrac{-3\alpha-2}{\alpha-2} \\ 0 \end{pmatrix} + t \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} : t \in {\bf R} \right\} \subset {\bf R}^3. \]

Solution 2.23

(See Exercise 2.23.)

Setting up the system. Let $U, V, W, X, Y, Z \geq 0$ denote the number of cars per hour flowing through the respective streets. Since each intersection neither creates nor destroys cars, inflow must equal outflow at each node:

\[ \begin{align*} A\colon &\quad U + V + X = 500, \\ B\colon &\quad U + W + Z = 400, \\ C\colon &\quad X + Y = Z + 100, \\ D\colon &\quad V = W + Y. \end{align*} \]

Rewriting with all unknowns on the left and forming the augmented matrix (columns ordered $U, V, W, X, Y, Z$):

\[ \left ( \begin{array}{cccccc|c} 1 & 1 & 0 & 1 & 0 & 0 & 500 \\ 1 & 0 & 1 & 0 & 0 & 1 & 400 \\ 0 & 0 & 0 & 1 & 1 & -1 & 100 \\ 0 & 1 & -1 & 0 & -1 & 0 & 0 \end{array} \right ). \]

Gaussian elimination. We apply Method 2.29. Subtracting the first row from the second eliminates $U$ from row $2$; then adding the (updated) second row to the fourth eliminates $V$ from row $4$:

\[ \begin{align*} \xrightarrow{R_2 \leftarrow R_2 - R_1} & \left ( \begin{array}{cccccc|c} 1 & 1 & 0 & 1 & 0 & 0 & 500 \\ 0 & -1 & 1 & -1 & 0 & 1 & -100 \\ 0 & 0 & 0 & 1 & 1 & -1 & 100 \\ 0 & 1 & -1 & 0 & -1 & 0 & 0 \end{array} \right ) \\ \xrightarrow{R_4 \leftarrow R_4 + R_2} & \left ( \begin{array}{cccccc|c} 1 & 1 & 0 & 1 & 0 & 0 & 500 \\ 0 & -1 & 1 & -1 & 0 & 1 & -100 \\ 0 & 0 & 0 & 1 & 1 & -1 & 100 \\ 0 & 0 & 0 & -1 & -1 & 1 & -100 \end{array} \right ) % Adding the third row to the fourth yields a zero row: %$$ \xrightarrow{R_4 \leftarrow R_4 + R_3} & \left ( \begin{array}{cccccc|c} 1 & 1 & 0 & 1 & 0 & 0 & 500 \\ 0 & -1 & 1 & -1 & 0 & 1 & -100 \\ 0 & 0 & 0 & 1 & 1 & -1 & 100 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right ). \end{align*} \]

This is in row-echelon form (Definition 2.27).

Reading off the solution. By Method 2.31, the pivot columns are $1$, $2$, $4$, so $U$, $V$, $X$ are the non-free variables, while $W = s$, $Y = t$, $Z = r$ are free ($s, t, r \in {\bf R}$). The three non-trivial rows give

\[ \begin{align*} X &= 100 - t + r. V &= 100 + s - X + Z = s + t, \\ U &= 500 - V - X = 400 - s - r. \\ \end{align*} \]

The solution set is

\[ \left\{ \begin{pmatrix} 400 \\ 0 \\ 0 \\ 100 \\ 0 \\ 0 \end{pmatrix} + s \begin{pmatrix} -1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \\ 1 \\ 0 \end{pmatrix} + r \begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 1 \end{pmatrix} : s, t, r \in {\bf R} \right\} \subset {\bf R}^6, \]

where the entries correspond to $(U, V, W, X, Y, Z)$. The physically admissible scenarios are those in which all six flow values are non-negative; these are the parameter triples $(s, t, r)$ additionally satisfying $s, t, r \geq 0$, $s + r \leq 400$, and $r - t \leq 100$.

Solution 2.24

(See Exercise 2.24.) We solve, for each $\lambda\in\mathbf R$, the system by Gaussian elimination (Method 2.31):

\[ \left ( \begin{array}{ccc|c} \lambda & 0 & 0 & 0\\ 0 & \lambda & 1+\lambda & 1\\ \lambda & 1 & 2 & 3 \end{array} \right ) \xrightarrow{R_3\leftarrow R_3-R_1} \left ( \begin{array}{ccc|c} \lambda & 0 & 0 & 0\\ 0 & \lambda & 1+\lambda & 1\\ 0 & 1 & 2 & 3 \end{array} \right ). \]

We would like to divide the first row by $\lambda$, which however is only a legitimate row operation (Definition 2.28) if $\lambda \ne 0$. So we distinguish two cases:

Case $\lambda=0$. The matrix becomes

\[ \left ( \begin{array}{ccc|c} 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 1 & 2 & 3 \end{array} \right ), \]

so $x_3=1$, $x_2=1$, and $x_1$ is free. Hence the solution set is $\{(t, 1, 1) \mid t \in \mathbf R\}$.

Case $\lambda\neq0$. Divide row 2 by $\lambda$ and eliminate the $1$ in row 3, column 2:

\[ \left ( \begin{array}{ccc|c} \lambda & 0 & 0 & 0\\ 0 & 1 & \frac{1+\lambda}{\lambda} & \frac1\lambda\\ 0 & 1 & 2 & 3 \end{array} \right ) \xrightarrow{R_3\leftarrow R_3-R_2} \left ( \begin{array}{ccc|c} \lambda & 0 & 0 & 0\\ 0 & 1 & \frac{1+\lambda}{\lambda} & \frac1\lambda\\ 0 & 0 & \frac{\lambda-1}{\lambda} & \frac{3\lambda-1}{\lambda} \end{array} \right ). \]

The next possible division is by $\lambda-1$, so we split again:

If $\lambda=1$, the last row is $0\;0\;0\mid2$, so there is no solution.
If $\lambda\neq1$, then

\[ x_1=0,\qquad x_3=\frac{3\lambda-1}{\lambda-1}=\frac{1-3\lambda}{1-\lambda},\qquad x_2=3-2x_3. \]

So the solution is unique.

Summary:

If $\lambda=0$: infinitely many solutions, namely $(t,1,1)$ with $t\in\mathbf R$.
If $\lambda=1$: no solution.
If $\lambda\neq0,1$: unique solution $\left(0,\,3-2\frac{1-3\lambda}{1-\lambda},\,\frac{1-3\lambda}{1-\lambda}\right)$.