1.5 Solution Sets of Linear Systems

Theory: Solution Sets of Linear Systems

Homogeneous vs. Nonhomogeneous Systems (Lay §1.5)

A linear system is

homogeneous if it can be written as $A\mathbf{x} = \mathbf{0}$ (every right-hand side is zero);
nonhomogeneous if it is $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} \ne \mathbf{0}$ .

A homogeneous system is always consistent: $\mathbf{x} = \mathbf{0}$ always works, because $A\mathbf{0} = \mathbf{0}$ . This zero solution is called the trivial solution. So for $A\mathbf{x} = \mathbf{0}$ the question is never “is there a solution?” but “is there a nontrivial solution, i.e. some $\mathbf{x} \ne \mathbf{0}$ ?”. Uniqueness depends on free variables:

no free variable → only the trivial solution (unique);
at least one free variable → infinitely many solutions (so nontrivial ones exist).

Fact (Lay §1.5): The homogeneous equation $A\mathbf{x} = \mathbf{0}$ has a nontrivial solution if and only if the equation has at least one free variable, i.e. if and only if some column of the coefficient matrix $A$ is not a pivot column.

How to recognise it fast: you only need an echelon form (REF), not the RREF. Circle the pivots of the coefficient part; any column without a pivot gives a free variable.

Shortcut — more unknowns than equations. If $A$ has fewer rows than columns (e.g. 2 equations, 3 unknowns), there are at most 2 pivots but 3 variables, so at least one variable must be free. Then $A\mathbf{x} = \mathbf{0}$ has a nontrivial solution — no row operations needed (Exercises 3 and 4 below).

Parametric Vector Form

Once the free variables are known, solve each basic variable in terms of the free ones and write the whole solution vector as a linear combination with the free variables as weights. For example, if $x_1 = 2x_2$ , $x_2$ free, $x_3 = 5$ :

\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 2x_2 \\ x_2 \\ 5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 5 \end{bmatrix} + \begin{bmatrix} 2x_2 \\ x_2 \\ 0 \end{bmatrix} = \underbrace{\begin{bmatrix} 0 \\ 0 \\ 5 \end{bmatrix}}_{\mathbf{p}} + x_2 \underbrace{\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}}_{\mathbf{v}}

The trick: split the vector into one piece per free variable plus a constant piece, then factor each free variable out.

Homogeneous: $A\mathbf{x} = \mathbf{0} \;\Rightarrow\; \mathbf{x} = x_n \mathbf{v}$ (or $s\mathbf{u} + t\mathbf{v} + \cdots$ ) — no constant vector, the solution set is $\operatorname{Span}\{\mathbf{v}, \dots\}$ , a line/plane through the origin.
Nonhomogeneous: $A\mathbf{x} = \mathbf{b} \;\Rightarrow\; \mathbf{x} = \mathbf{p} + x_n \mathbf{v}$ , where $\mathbf{p}$ is a constant vector and $\mathbf{v}$ is multiplied by a free variable.

Geometrically, one free variable gives a line, two free variables give a plane.

The 6-Step Solution-Set Recipe

Find REF of the augmented matrix and circle the pivots.
Check for free variables: if a coefficient column has no pivot, that variable is free and (for $A\mathbf{x} = \mathbf{0}$ ) a nontrivial solution exists. (For $A\mathbf{x} = \mathbf{b}$ , also check there is no row $[0 \ \cdots \ 0 \mid c]$ with $c \ne 0$ .)
Reduce to RREF.
Write the RREF as equations.
Express the basic variables in terms of the free variables.
Write the parametric vector form: split by free variable and factor.

Mini-example (Lay §1.5, Example 1): homogeneous

Determine if the system has a nontrivial solution and describe the solution set: $3x_1 + 5x_2 - 4x_3 = 0$ , $\ -3x_1 - 2x_2 + 4x_3 = 0$ , $\ 6x_1 + x_2 - 8x_3 = 0$ .

Step 1 ( $R_2 \leftarrow R_2 + R_1$ , $R_3 \leftarrow R_3 - 2R_1$ , then $R_3 \leftarrow R_3 + 3R_2$ ):

\left[\begin{array}{ccc|c} 3 & 5 & -4 & 0 \\ -3 & -2 & 4 & 0 \\ 6 & 1 & -8 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 3 & 5 & -4 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & -9 & 0 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} \boxed{3} & 5 & -4 & 0 \\ 0 & \boxed{3} & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

Step 2: column 3 has no pivot → $x_3$ is free → a nontrivial solution exists.

Step 3 ( $R_2 \leftarrow \tfrac13 R_2$ , $R_1 \leftarrow R_1 - 5R_2$ , $R_1 \leftarrow \tfrac13 R_1$ ): $\left[\begin{array}{ccc|c} 1 & 0 & -\frac43 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$

Step 4: $x_1 - \tfrac43 x_3 = 0$ , $\ x_2 = 0$ , $\ 0 = 0$ .

Step 5: $x_1 = \tfrac43 x_3$ , $x_2 = 0$ , $x_3$ free.

Step 6:

\mathbf{x} = \begin{bmatrix} \frac43 x_3 \\ 0 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} \frac43 \\ 0 \\ 1 \end{bmatrix} = x_3 \mathbf{v}

The solution set is $\operatorname{Span}\{\mathbf{v}\}$ : a line through the origin in $\mathbb{R}^3$ .

Mini-example (Lay §1.5, Example 3): nonhomogeneous

Same coefficient matrix, now with $\mathbf{b} = (7, -1, -4)$ . The same row operations give

\left[\begin{array}{ccc|c} 3 & 5 & -4 & 7 \\ -3 & -2 & 4 & -1 \\ 6 & 1 & -8 & -4 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & -\frac43 & -1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]

So $x_1 = -1 + \tfrac43 x_3$ , $x_2 = 2$ , $x_3$ free, and

\mathbf{x} = \begin{bmatrix} -1 + \frac43 x_3 \\ 2 \\ x_3 \end{bmatrix} = \underbrace{\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}}_{\mathbf{p}} + x_3 \underbrace{\begin{bmatrix} \frac43 \\ 0 \\ 1 \end{bmatrix}}_{\mathbf{v}}

Notice: the same $\mathbf{v}$ as in Example 1, just shifted by $\mathbf{p}$ . That is no accident:

Solutions of Nonhomogeneous Systems

Theorem 6 (Lay §1.5): Suppose $A\mathbf{x} = \mathbf{b}$ is consistent for some given $\mathbf{b}$ , and let $\mathbf{p}$ be a solution. Then the solution set of $A\mathbf{x} = \mathbf{b}$ is the set of all vectors of the form $\mathbf{w} = \mathbf{p} + \mathbf{v}_h,$ where $\mathbf{v}_h$ is any solution of the homogeneous equation $A\mathbf{x} = \mathbf{0}$ .

Why it works: if $A\mathbf{p} = \mathbf{b}$ and $A\mathbf{v}_h = \mathbf{0}$ , then $A(\mathbf{p} + \mathbf{v}_h) = \mathbf{b} + \mathbf{0} = \mathbf{b}$ . Conversely, if $A\mathbf{w} = \mathbf{b}$ then $A(\mathbf{w} - \mathbf{p}) = \mathbf{0}$ , so $\mathbf{w} - \mathbf{p}$ is a homogeneous solution.

Geometric picture: the solution set of $A\mathbf{x} = \mathbf{b}$ is the solution set of $A\mathbf{x} = \mathbf{0}$ translated by $\mathbf{p}$ . A line through the origin becomes a parallel line through $\mathbf{p}$ ; a plane through the origin becomes a parallel plane through $\mathbf{p}$ . Here $\mathbf{p}$ is called a particular solution.

Common mistakes:

Theorem 6 needs $A\mathbf{x} = \mathbf{b}$ to be consistent. If REF has a row $[0 \ \cdots \ 0 \mid b]$ with $b \ne 0$ , there is no $\mathbf{p}$ and the solution set is empty — it is not a shifted copy of anything.

A variable whose column is all zeros (e.g. $x_3$ in $[1 \ 3 \ 0 \ -4]$ ) is free, not zero. Don’t drop it from the parametric form.

Read free variables from the coefficient columns only — the augmented column is never a variable.

Solved Examples (Textbook Questions)

Exercise 1

(Adapted from Lay §1.5, Exercises 1–4)

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.

$2x_1 - 5x_2 + 8x_3 = 0$ , $\ -2x_1 - 7x_2 + x_3 = 0$ , $\ 4x_1 + 2x_2 + 7x_3 = 0$
$x_1 - 3x_2 + 7x_3 = 0$ , $\ -2x_1 + x_2 - 4x_3 = 0$ , $\ x_1 + 2x_2 + 9x_3 = 0$
$-3x_1 + 5x_2 - 7x_3 = 0$ , $\ -6x_1 + 7x_2 + x_3 = 0$
$-5x_1 + 7x_2 + 9x_3 = 0$ , $\ x_1 - 2x_2 + 6x_3 = 0$

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Solution

Strategy: reduce to echelon form only and look for a coefficient column with no pivot.

1. $R_2 \leftarrow R_2 + R_1$ , $R_3 \leftarrow R_3 - 2R_1$ , then $R_3 \leftarrow R_3 + R_2$ :

\left[\begin{array}{ccc|c} 2 & -5 & 8 & 0 \\ -2 & -7 & 1 & 0 \\ 4 & 2 & 7 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 2 & -5 & 8 & 0 \\ 0 & -12 & 9 & 0 \\ 0 & 12 & -9 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} \boxed{2} & -5 & 8 & 0 \\ 0 & \boxed{-12} & 9 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

Column 3 has no pivot → $x_3$ is free → nontrivial solution exists.

2. $R_2 \leftarrow R_2 + 2R_1$ , $R_3 \leftarrow R_3 - R_1$ , then $R_3 \leftarrow R_3 + R_2$ :

\left[\begin{array}{ccc|c} 1 & -3 & 7 & 0 \\ -2 & 1 & -4 & 0 \\ 1 & 2 & 9 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -3 & 7 & 0 \\ 0 & -5 & 10 & 0 \\ 0 & 5 & 2 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} \boxed{1} & -3 & 7 & 0 \\ 0 & \boxed{-5} & 10 & 0 \\ 0 & 0 & \boxed{12} & 0 \end{array}\right]

Every column is a pivot column → no free variable → only the trivial solution.

3. $R_2 \leftarrow R_2 - 2R_1$ :

\left[\begin{array}{ccc|c} -3 & 5 & -7 & 0 \\ -6 & 7 & 1 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} \boxed{-3} & 5 & -7 & 0 \\ 0 & \boxed{-3} & 15 & 0 \end{array}\right]

$x_3$ is free → nontrivial solutions exist.

4. Swap $R_1 \leftrightarrow R_2$ , then $R_2 \leftarrow R_2 + 5R_1$ :

\left[\begin{array}{ccc|c} 1 & -2 & 6 & 0 \\ -5 & 7 & 9 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} \boxed{1} & -2 & 6 & 0 \\ 0 & \boxed{-3} & 39 & 0 \end{array}\right]

$x_3$ is free → nontrivial solutions exist.

Alert-student shortcut for 3 and 4: with only 2 equations there can be at most 2 pivots, but there are 3 variables, so at least one variable must be free. Row operations are unnecessary.

Exercise 2

(Adapted from Lay §1.5, Exercises 5–6)

Write the solution set of each homogeneous system in parametric vector form.

$x_1 + 3x_2 + x_3 = 0$ , $\ -4x_1 - 9x_2 + 2x_3 = 0$ , $\ -3x_2 - 6x_3 = 0$
$x_1 + 3x_2 - 5x_3 = 0$ , $\ x_1 + 4x_2 - 8x_3 = 0$ , $\ -3x_1 - 7x_2 + 9x_3 = 0$

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Solution

5. $R_2 \leftarrow R_2 + 4R_1$ ; $R_3 \leftarrow R_3 + R_2$ ; $R_2 \leftarrow \tfrac13 R_2$ ; $R_1 \leftarrow R_1 - 3R_2$ :

\left[\begin{array}{ccc|c} 1 & 3 & 1 & 0 \\ -4 & -9 & 2 & 0 \\ 0 & -3 & -6 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 3 & 1 & 0 \\ 0 & 3 & 6 & 0 \\ 0 & -3 & -6 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 3 & 1 & 0 \\ 0 & 3 & 6 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & -5 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

Equations: $x_1 - 5x_3 = 0$ , $x_2 + 2x_3 = 0$ . So $x_1 = 5x_3$ , $x_2 = -2x_3$ , $x_3$ free:

\mathbf{x} = \begin{bmatrix} 5x_3 \\ -2x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} 5 \\ -2 \\ 1 \end{bmatrix}

6. $R_2 \leftarrow R_2 - R_1$ , $R_3 \leftarrow R_3 + 3R_1$ ; $R_3 \leftarrow R_3 - 2R_2$ ; $R_1 \leftarrow R_1 - 3R_2$ :

\left[\begin{array}{ccc|c} 1 & 3 & -5 & 0 \\ 1 & 4 & -8 & 0 \\ -3 & -7 & 9 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 3 & -5 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 2 & -6 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 3 & -5 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 4 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

Equations: $x_1 + 4x_3 = 0$ , $x_2 - 3x_3 = 0$ . So $x_1 = -4x_3$ , $x_2 = 3x_3$ , $x_3$ free:

\mathbf{x} = \begin{bmatrix} -4x_3 \\ 3x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} -4 \\ 3 \\ 1 \end{bmatrix}

Both solution sets are lines through the origin.

Exercise 3

(Adapted from Lay §1.5, Exercises 7–10)

Describe all solutions of $A\mathbf{x} = \mathbf{0}$ in parametric vector form, where $A$ is row equivalent to the given matrix.

\textbf{7. } \begin{bmatrix} 1 & 3 & -3 & 7 \\ 0 & 1 & -4 & 5 \end{bmatrix} \qquad \textbf{8. } \begin{bmatrix} 1 & -2 & -9 & 5 \\ 0 & 1 & 2 & -6 \end{bmatrix}

\textbf{9. } \begin{bmatrix} 3 & -9 & 6 \\ -1 & 3 & -2 \end{bmatrix} \qquad \textbf{10. } \begin{bmatrix} 1 & 3 & 0 & -4 \\ 2 & 6 & 0 & -8 \end{bmatrix}

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Solution

These are coefficient matrices, so append a zero column mentally (it never changes under row operations).

7. $R_1 \leftarrow R_1 - 3R_2$ :

\begin{bmatrix} 1 & 3 & -3 & 7 \\ 0 & 1 & -4 & 5 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 9 & -8 \\ 0 & 1 & -4 & 5 \end{bmatrix}

$x_1 + 9x_3 - 8x_4 = 0$ , $x_2 - 4x_3 + 5x_4 = 0$ . Basic: $x_1, x_2$ ; free: $x_3, x_4$ . So $x_1 = -9x_3 + 8x_4$ , $x_2 = 4x_3 - 5x_4$ :

\mathbf{x} = \begin{bmatrix} -9x_3 + 8x_4 \\ 4x_3 - 5x_4 \\ x_3 \\ x_4 \end{bmatrix} = x_3 \begin{bmatrix} -9 \\ 4 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 8 \\ -5 \\ 0 \\ 1 \end{bmatrix}

8. $R_1 \leftarrow R_1 + 2R_2$ :

\begin{bmatrix} 1 & -2 & -9 & 5 \\ 0 & 1 & 2 & -6 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & -5 & -7 \\ 0 & 1 & 2 & -6 \end{bmatrix}

$x_1 = 5x_3 + 7x_4$ , $x_2 = -2x_3 + 6x_4$ , $x_3, x_4$ free:

\mathbf{x} = x_3 \begin{bmatrix} 5 \\ -2 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 7 \\ 6 \\ 0 \\ 1 \end{bmatrix}

9. $R_1 \leftarrow \tfrac13 R_1$ , then $R_2 \leftarrow R_2 + R_1$ :

\begin{bmatrix} 3 & -9 & 6 \\ -1 & 3 & -2 \end{bmatrix} \sim \begin{bmatrix} 1 & -3 & 2 \\ 0 & 0 & 0 \end{bmatrix}

$x_1 - 3x_2 + 2x_3 = 0$ , so $x_1 = 3x_2 - 2x_3$ with $x_2, x_3$ free:

\mathbf{x} = x_2 \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}

(a plane through the origin in $\mathbb{R}^3$ ).

10. $R_2 \leftarrow R_2 - 2R_1$ :

\begin{bmatrix} 1 & 3 & 0 & -4 \\ 2 & 6 & 0 & -8 \end{bmatrix} \sim \begin{bmatrix} 1 & 3 & 0 & -4 \\ 0 & 0 & 0 & 0 \end{bmatrix}

The only equation is $x_1 + 3x_2 - 4x_4 = 0$ . The only basic variable is $x_1$ ; $x_2, x_3, x_4$ are all free (note $x_3$ is free, not zero — its column is all zeros). So $x_1 = -3x_2 + 4x_4$ :

\mathbf{x} = \begin{bmatrix} -3x_2 + 4x_4 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = x_2 \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 4 \\ 0 \\ 0 \\ 1 \end{bmatrix}

Exercise 4

(Adapted from Lay §1.5, Exercises 11–12)

Describe all solutions of $A\mathbf{x} = \mathbf{0}$ in parametric vector form, where $A$ is row equivalent to

\textbf{11. } \begin{bmatrix} 1 & -4 & -2 & 0 & 3 & -5 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \qquad \textbf{12. } \begin{bmatrix} 1 & 5 & 2 & -6 & 9 & 0 \\ 0 & 0 & 1 & -7 & 4 & -8 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

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Solution

11. Already in echelon form, pivots in columns 1, 3, 5. Clear above the pivots: $R_1 \leftarrow R_1 - 3R_3$ , then $R_1 \leftarrow R_1 + 2R_2$ :

A \sim \begin{bmatrix} 1 & -4 & -2 & 0 & 0 & 7 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \sim \begin{bmatrix} \boxed{1} & -4 & 0 & 0 & 0 & 5 \\ 0 & 0 & \boxed{1} & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & \boxed{1} & -4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

Equations: $x_1 - 4x_2 + 5x_6 = 0$ , $x_3 - x_6 = 0$ , $x_5 - 4x_6 = 0$ . Basic: $x_1, x_3, x_5$ . Free: $x_2, x_4, x_6$ (in particular $x_4$ is free, not zero!). $x_1 = 4x_2 - 5x_6$ , $x_3 = x_6$ , $x_5 = 4x_6$ :

\mathbf{x} = \begin{bmatrix} 4x_2 - 5x_6 \\ x_2 \\ x_6 \\ x_4 \\ 4x_6 \\ x_6 \end{bmatrix} = x_2 \begin{bmatrix} 4 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_6 \begin{bmatrix} -5 \\ 0 \\ 1 \\ 0 \\ 4 \\ 1 \end{bmatrix} = x_2\mathbf{u} + x_4\mathbf{v} + x_6\mathbf{w}

12. Pivots in columns 1, 3, 6. $R_2 \leftarrow R_2 + 8R_3$ , then $R_1 \leftarrow R_1 - 2R_2$ :

A \sim \begin{bmatrix} 1 & 5 & 2 & -6 & 9 & 0 \\ 0 & 0 & 1 & -7 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \sim \begin{bmatrix} \boxed{1} & 5 & 0 & 8 & 1 & 0 \\ 0 & 0 & \boxed{1} & -7 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & \boxed{1} \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

Equations: $x_1 + 5x_2 + 8x_4 + x_5 = 0$ , $x_3 - 7x_4 + 4x_5 = 0$ , $x_6 = 0$ . Basic: $x_1, x_3, x_6$ . Free: $x_2, x_4, x_5$ . $x_1 = -5x_2 - 8x_4 - x_5$ , $x_3 = 7x_4 - 4x_5$ , $x_6 = 0$ :

\mathbf{x} = x_2 \begin{bmatrix} -5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -8 \\ 0 \\ 7 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} -1 \\ 0 \\ -4 \\ 0 \\ 1 \\ 0 \end{bmatrix}

Two classic mistakes on this type (mentioned by the Study Guide): (1) treating a variable with an all-zero column, like $x_4$ in Exercise 11, as zero instead of free; (2) forgetting a basic variable that equals $0$ , like $x_6$ in Exercise 12 — it is basic, so it gets no vector of its own, but its entry $0$ still appears in every vector.

Exercise 5

(Adapted from Lay §1.5, Exercises 17–18)

Describe and compare the solution sets of $x_1 + 9x_2 - 4x_3 = 0$ and $x_1 + 9x_2 - 4x_3 = -2$ .
Describe and compare the solution sets of $x_1 - 3x_2 + 5x_3 = 0$ and $x_1 - 3x_2 + 5x_3 = 4$ .

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Solution

A single equation in 3 unknowns: $x_1$ is basic, $x_2$ and $x_3$ are free.

17. Nonhomogeneous: $x_1 = -2 - 9x_2 + 4x_3$ :

\mathbf{x} = \begin{bmatrix} -2 - 9x_2 + 4x_3 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} -9 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 4 \\ 0 \\ 1 \end{bmatrix} = \mathbf{p} + x_2\mathbf{u} + x_3\mathbf{v}

Homogeneous: $x_1 = -9x_2 + 4x_3$ , so $\mathbf{x} = x_2\mathbf{u} + x_3\mathbf{v}$ with the same $\mathbf{u}, \mathbf{v}$ .

Comparison: the homogeneous solution set is the plane through the origin spanned by $\mathbf{u}$ and $\mathbf{v}$ . The nonhomogeneous solution set is the parallel plane through $\mathbf{p} = (-2, 0, 0)$ (Theorem 6).

18. Nonhomogeneous: $x_1 = 4 + 3x_2 - 5x_3$ :

\mathbf{x} = \begin{bmatrix} 4 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -5 \\ 0 \\ 1 \end{bmatrix} = \mathbf{p} + x_2\mathbf{u} + x_3\mathbf{v}

Homogeneous: $x_1 = 3x_2 - 5x_3$ , so $\mathbf{x} = x_2\mathbf{u} + x_3\mathbf{v}$ .

Comparison: the homogeneous solution set is the plane through the origin spanned by $\mathbf{u}, \mathbf{v}$ ; the nonhomogeneous one is the parallel plane through $\mathbf{p} = (4, 0, 0)$ .

Exercise 6

(Adapted from Lay §1.5, Exercises 34–36)

Given $A = \begin{bmatrix} 4 & -6 \\ -8 & 12 \\ 6 & -9 \end{bmatrix}$ , find one nontrivial solution of $A\mathbf{x} = \mathbf{0}$ by inspection.
Construct a $3 \times 3$ nonzero matrix $A$ such that $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ is a solution of $A\mathbf{x} = \mathbf{0}$ .
Construct a $3 \times 3$ nonzero matrix $A$ such that $\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$ is a solution of $A\mathbf{x} = \mathbf{0}$ .

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Solution

Key idea: $A\mathbf{x} = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots$ , so a solution of $A\mathbf{x} = \mathbf{0}$ is a list of weights that makes the columns add to $\mathbf{0}$ .

34. Compare the columns: $-6 = -\tfrac32 \cdot 4$ , $12 = -\tfrac32 \cdot (-8)$ , $-9 = -\tfrac32 \cdot 6$ . So $\mathbf{a}_2 = -\tfrac32 \mathbf{a}_1$ , i.e. $3\mathbf{a}_1 + 2\mathbf{a}_2 = \mathbf{0}$ . Therefore

\mathbf{x} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \text{ satisfies } A\mathbf{x} = \mathbf{0}. \quad \text{Check row 1: } 4(3) - 6(2) = 0 \ ✓

35. We need $1\cdot\mathbf{a}_1 + 1\cdot\mathbf{a}_2 + 1\cdot\mathbf{a}_3 = \mathbf{0}$ : each row sum must be zero. One example:

A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 0 & -2 \\ 0 & 3 & -3 \end{bmatrix}, \qquad A\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

36. We need $\mathbf{a}_1 - 2\mathbf{a}_2 + \mathbf{a}_3 = \mathbf{0}$ , i.e. in every row, (first entry) + (third entry) = 2 × (middle entry): the sum of the first and third columns is twice the second column. One example:

A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 2 \end{bmatrix}, \qquad A\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 - 2 + 1 \\ 2 - 2 + 0 \\ 0 - 2 + 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

(Many answers are possible — any nonzero matrix with this row property works.)

Exercise 7

(Adapted from Lay §1.5, Exercise 38)

Suppose $A$ is a $3 \times 3$ matrix and $\mathbf{y}$ is a vector in $\mathbb{R}^3$ such that the equation $A\mathbf{x} = \mathbf{y}$ does not have a solution. Does there exist a vector $\mathbf{z}$ in $\mathbb{R}^3$ such that $A\mathbf{x} = \mathbf{z}$ has a unique solution? Discuss.

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Solution

No.

If $A\mathbf{x} = \mathbf{y}$ has no solution, then $A$ does not have a pivot in every row (otherwise $A\mathbf{x} = \mathbf{b}$ would be consistent for every $\mathbf{b}$ , Lay §1.4 Theorem 4).
$A$ is $3 \times 3$ , so it has at most two pivot positions.
Hence among its 3 columns at least one is not a pivot column: any equation $A\mathbf{x} = \mathbf{z}$ has at most two basic variables and at least one free variable.
So for any $\mathbf{z}$ , the solution set of $A\mathbf{x} = \mathbf{z}$ is either empty (inconsistent) or infinite (consistent with a free variable) — never a single vector.

This matches Theorem 6: if a $\mathbf{p}$ exists, the solution set is $\mathbf{p}$ + (solutions of $A\mathbf{x} = \mathbf{0}$ ), and $A\mathbf{x} = \mathbf{0}$ has infinitely many solutions here.