The Timpson Lyceum

1. (2 pts) Solve the homogeneous system and write its solution set in parametric vector form:

$\begin{aligned} x_1 - 2x_2 + 3x_3 &= 0 \\ 2x_1 - 3x_2 + 4x_3 &= 0. \end{aligned}$

Solution

1. Reduce the coefficient matrix:

$\begin{bmatrix} 1 & -2 & 3 \\ 2 & -3 & 4 \end{bmatrix} \xrightarrow{R_2 - 2R_1} \begin{bmatrix} 1 & -2 & 3 \\ 0 & 1 & -2 \end{bmatrix} \xrightarrow{R_1 + 2R_2} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \end{bmatrix}.$

So $x_1 = x_3$ , $x_2 = 2x_3$ , with $x_3$ free:

$\mathbf{x} = x_3 \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}.$

2. (3 pts) Find the solution set of $A\mathbf{x} = \mathbf{0}$ in parametric vector form, where

$A = \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & -1 & 8 \\ 1 & 2 & 0 & 5 \end{bmatrix}.$

Solution

2. Reduce $A$ :

$\begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & -1 & 8 \\ 1 & 2 & 0 & 5 \end{bmatrix} \xrightarrow[R_3 - R_1]{R_2 - 2R_1} \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 2 \end{bmatrix} \xrightarrow{R_3 - R_2} \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_1 + R_2} \begin{bmatrix} 1 & 2 & 0 & 5 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$

Pivots in columns 1 and 3, so $x_2, x_4$ are free: $x_1 = -2x_2 - 5x_4$ , $x_3 = -2x_4$ . Thus

$\mathbf{x} = x_2 \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -5 \\ 0 \\ -2 \\ 1 \end{bmatrix}.$

3. (3 pts) Solve the system below and express its solution in the form $\mathbf{x} = \mathbf{p} + x_2\mathbf{v}$ :

$\begin{aligned} x_1 + 2x_2 + x_3 &= 4 \\ 2x_1 + 4x_2 + x_3 &= 5. \end{aligned}$

Solution

3. Augmented reduction:

$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 2 & 4 & 1 & 5 \end{array}\right] \xrightarrow{R_2 - 2R_1} \left[\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 0 & 0 & -1 & -3 \end{array}\right].$

Row 2 gives $x_3 = 3$ ; then $x_1 = 4 - 2x_2 - 3 = 1 - 2x_2$ , with $x_2$ free. So

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

4. (3 pts) For the system in Problem 3, write the solution set of the associated homogeneous system $A\mathbf{x} = \mathbf{0}$ , and describe geometrically how the two solution sets are related.

Solution

4. Dropping the constants (setting the right sides to $0$ ) gives $x_1 = -2x_2$ , $x_3 = 0$ , so the homogeneous solution set is $\left\{\, x_2\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} \,\right\}$ — a line through the origin in the direction $(-2, 1, 0)$ . The solution set of Problem 3 is that same line translated by $\mathbf{p} = (1, 0, 3)$ : a parallel line through $(1,0,3)$ that does not pass through the origin.

5. (3 pts) State whether each of the following is true or false, and justify each answer.

(a) A homogeneous system $A\mathbf{x} = \mathbf{0}$ is always consistent.

(b) If $A\mathbf{x} = \mathbf{0}$ has only the trivial solution, then $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$ .

(c) If $A\mathbf{x} = \mathbf{b}$ is consistent, then its solution set is a translate of the solution set of $A\mathbf{x} = \mathbf{0}$ .

Solution

5.

(a) True. Setting all variables to $0$ (the trivial solution) always satisfies $A\mathbf{x} = \mathbf{0}$ , so the system is consistent.

(b) False. "Only the trivial solution" means the columns are independent, so when $A\mathbf{x}=\mathbf{b}$ is consistent the solution is unique — but for a non-square $A$ the equation may be inconsistent for some $\mathbf{b}$ . Uniqueness for every $\mathbf{b}$ requires a pivot in every row as well.

(c) True. This is Theorem 6: for a consistent system with particular solution $\mathbf{p}$ , the solution set is $\{\mathbf{p} + \mathbf{v}_h : A\mathbf{v}_h = \mathbf{0}\}$ , i.e. a translate by $\mathbf{p}$ of the solution set of $A\mathbf{x} = \mathbf{0}$ .

6. (3 pts) Describe geometrically, in $\mathbb{R}^3$ , the two solution sets, and state how they differ:

(a) $\mathbf{x} = t\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$

(b) $\mathbf{x} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + t\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$

Solution

6.

(a) $\mathbf{x} = t(2,1,-1)$ traces all scalar multiples of $(2,1,-1)$ : a line through the origin with direction $(2,1,-1)$ .

(b) $\mathbf{x} = (1,0,2) + t(2,1,-1)$ is the same line translated so it passes through $(1,0,2)$ ; it is parallel to the line in (a) but does not contain the origin (since $(1,0,2)$ is not a multiple of $(2,1,-1)$ ).

7. (3 pts) Prove that every homogeneous linear system with more unknowns than equations has a nontrivial solution.

Solution

7. Let the system be $A\mathbf{x} = \mathbf{0}$ with $A$ of size $m \times n$ where $n > m$ (more unknowns than equations). The number of pivots is at most $m < n$ , so at least one of the $n$ columns has no pivot, giving at least one free variable. A homogeneous system is always consistent, and a free variable may be assigned any nonzero value to produce a solution $\mathbf{x} \neq \mathbf{0}$ . Hence a nontrivial solution exists. $\blacksquare$

8. (4 pts ••) Suppose $A\mathbf{x} = \mathbf{b}$ is consistent, and let $\mathbf{p}$ be one solution. Prove that the solution set of $A\mathbf{x} = \mathbf{b}$ equals $\{\mathbf{p} + \mathbf{v}_h : A\mathbf{v}_h = \mathbf{0}\}$ .

Solution

8. Let $S$ be the solution set of $A\mathbf{x} = \mathbf{b}$ and let $H = \{\mathbf{v}_h : A\mathbf{v}_h = \mathbf{0}\}$ .

( $S \subseteq \mathbf{p} + H$ ) Let $\mathbf{w} \in S$ , so $A\mathbf{w} = \mathbf{b}$ . Put $\mathbf{v}_h = \mathbf{w} - \mathbf{p}$ . Then $A\mathbf{v}_h = A\mathbf{w} - A\mathbf{p} = \mathbf{b} - \mathbf{b} = \mathbf{0}$ , so $\mathbf{v}_h \in H$ and $\mathbf{w} = \mathbf{p} + \mathbf{v}_h$ .

( $\mathbf{p} + H \subseteq S$ ) Let $\mathbf{v}_h \in H$ . Then $A(\mathbf{p} + \mathbf{v}_h) = A\mathbf{p} + A\mathbf{v}_h = \mathbf{b} + \mathbf{0} = \mathbf{b}$ , so $\mathbf{p} + \mathbf{v}_h \in S$ .

Both inclusions hold, so $S = \{\mathbf{p} + \mathbf{v}_h : A\mathbf{v}_h = \mathbf{0}\}$ . $\blacksquare$