The Timpson Lyceum

1. (2 pts) Solve the system by elimination or substitution:

$\begin{aligned} x_1 + 2x_2 &= 4 \\ 3x_1 - x_2 &= 5. \end{aligned}$

Solution

1. From the first equation, $x_1 = 4 - 2x_2$ . Substitute into the second: $3(4 - 2x_2) - x_2 = 5$ , so $12 - 7x_2 = 5$ , giving $x_2 = 1$ and $x_1 = 4 - 2 = 2$ . The unique solution is $(x_1, x_2) = (2, 1)$ .

2. (3 pts) Write the augmented matrix of the following system and use elementary row operations to solve it:

$\begin{aligned} x_1 + x_2 + x_3 &= 6 \\ 2x_1 - x_2 + x_3 &= 3 \\ x_1 + 2x_2 - x_3 &= 2. \end{aligned}$

Solution

2. The augmented matrix and reduction:

$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 2 \end{array}\right] \xrightarrow[R_3 - R_1]{R_2 - 2R_1} \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & -3 & -1 & -9 \\ 0 & 1 & -2 & -4 \end{array}\right] \xrightarrow{R_2 + 3R_3} \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 0 & -7 & -21 \\ 0 & 1 & -2 & -4 \end{array}\right].$

From row 2, $-7x_3 = -21$ , so $x_3 = 3$ . From $x_2 - 2x_3 = -4$ , $x_2 = -4 + 6 = 2$ . From $x_1 + x_2 + x_3 = 6$ , $x_1 = 6 - 2 - 3 = 1$ . The solution is $(1, 2, 3)$ .

3. (3 pts) Consider the system

$\begin{aligned} x_1 - 3x_2 &= 1 \\ 2x_1 + h\,x_2 &= 4. \end{aligned}$

Find all values of $h$ for which the system has

(a) a unique solution,

(b) no solution,

(c) infinitely many solutions.

Solution

3. Reduce the augmented matrix:

$\left[\begin{array}{cc|c} 1 & -3 & 1 \\ 2 & h & 4 \end{array}\right] \xrightarrow{R_2 - 2R_1} \left[\begin{array}{cc|c} 1 & -3 & 1 \\ 0 & h+6 & 2 \end{array}\right].$

(a) If $h + 6 \neq 0$ , i.e. $h \neq -6$ , there is a pivot in each column and the system has a unique solution.

(b) If $h = -6$ , row 2 becomes $[\,0\;\;0\mid 2\,]$ , which is impossible, so there is no solution.

(c) Infinitely many solutions occur only when row 2 vanishes entirely; here the right-hand entry is $2 \neq 0$ , so this never happens.

4. (3 pts) Each matrix below is the augmented matrix of a linear system that has already been row reduced. In each case state whether the system is consistent, and if so how many solutions it has.

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

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

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

Solution

4.

(a) A pivot in each of the three columns and no impossible row: consistent, unique solution $(3, -1, 2)$ .

(b) No row of the form $[\,0\cdots0\mid b\neq0\,]$ , and column 2 has no pivot, so $x_2$ is a free variable: consistent, infinitely many solutions.

(c) Row 3 reads $0 = 6$ , which is impossible: inconsistent, no solutions.

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

(a) Every elementary row operation is reversible.

(b) A system of two equations in three unknowns can have exactly one solution.

(c) If a linear system is inconsistent, then the reduced form of its augmented matrix has a row of the form $[\,0 \;\; 0 \;\; \cdots \;\; 0 \mid b\,]$ with $b \neq 0$ .

Solution

5.

(a) True. Replacement is undone by the opposite replacement, an interchange by repeating it, and scaling by $c \neq 0$ by scaling by $1/c$ . Each operation has an inverse operation of the same type.

(b) False. Three unknowns give three columns, but a coefficient matrix with two rows has at most two pivots, so at least one variable is free. A consistent such system therefore has infinitely many solutions, never exactly one (and an inconsistent one has none).

(c) True. Inconsistency means no solution exists; in a row-reduced augmented matrix this can only be signaled by an equation $0 = b$ with $b \neq 0$ , i.e. a row $[\,0\cdots0\mid b\,]$ , $b \neq 0$ .

6. (3 pts) Explain why performing any one elementary row operation on the augmented matrix of a linear system does not change its solution set. Address all three types of operation (replacement, interchange, scaling).

Solution

6. A solution of the system is a tuple satisfying every equation simultaneously.

Replacement ( $R_i \to R_i + cR_j$ ): any tuple satisfying equations $i$ and $j$ also satisfies their combination, and since the operation is reversible, the new system has exactly the same simultaneous solutions.
Interchange: reordering the equations does not change which tuples satisfy all of them.
Scaling by $c \neq 0$ : $E_i = 0$ and $cE_i = 0$ have identical solution sets because $c \neq 0$ .

Since each operation preserves the set of tuples satisfying all equations, the solution set is unchanged.

7. (3 pts) Describe geometrically the solution set of each system in $\mathbb{R}^2$ , and state how many solutions it has.

(a) $\begin{aligned} x_1 - x_2 &= 1 \\ 2x_1 - 2x_2 &= 5 \end{aligned}$

(b) $\begin{aligned} x_1 - x_2 &= 1 \\ 2x_1 - 2x_2 &= 2 \end{aligned}$

Solution

7. Each equation is a line in $\mathbb{R}^2$ .

(a) The lines $x_1 - x_2 = 1$ and $2x_1 - 2x_2 = 5$ (i.e. $x_1 - x_2 = 2.5$ ) have the same slope but different intercepts: they are parallel and distinct, so they never meet — no solution.

(b) Here $2x_1 - 2x_2 = 2$ is $x_1 - x_2 = 1$ , the same line as the first equation. The two lines coincide, so every point on the line is a solution — infinitely many solutions.

8. (4 pts ••) Suppose $u = (u_1, \dots, u_n)$ and $v = (v_1, \dots, v_n)$ are two distinct solutions of a system of linear equations. For a real number $t$ , let $w$ be the tuple whose $k$ -th entry is $w_k = u_k + t(u_k - v_k)$ . Prove that $w$ is also a solution for every $t$ , and conclude that the system has infinitely many solutions.

Solution

8. Take any one equation of the system, $a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b$ . Since $u$ and $v$ are solutions,

$a_1 u_1 + \cdots + a_n u_n = b \qquad\text{and}\qquad a_1 v_1 + \cdots + a_n v_n = b.$

Substitute the entries $w_k = u_k + t(u_k - v_k)$ into the left side of the equation and group terms:

$\sum_{k=1}^n a_k\big(u_k + t(u_k - v_k)\big) = \sum_{k=1}^n a_k u_k + t\Big(\sum_{k=1}^n a_k u_k - \sum_{k=1}^n a_k v_k\Big) = b + t(b - b) = b.$

So $w$ satisfies that equation, and since the equation was arbitrary it satisfies every equation of the system — $w$ is a solution for each $t \in \mathbb{R}$ . Because $u \neq v$ , some entry has $u_j - v_j \neq 0$ , so different values of $t$ give tuples $w$ with different $j$ -th entry $u_j + t(u_j - v_j)$ ; the solutions are therefore distinct, and the system has infinitely many solutions. $\blacksquare$