The Timpson Lyceum

1. (2 pts) Row reduce the matrix to reduced echelon form and list its pivot positions and pivot columns:

$A = \begin{bmatrix} 1 & 2 & 1 & 4 \\ 2 & 4 & 3 & 9 \\ 3 & 6 & 4 & 13 \end{bmatrix}.$

Solution

1. Reduce:

$\begin{bmatrix} 1 & 2 & 1 & 4 \\ 2 & 4 & 3 & 9 \\ 3 & 6 & 4 & 13 \end{bmatrix} \xrightarrow[R_3 - 3R_1]{R_2 - 2R_1} \begin{bmatrix} 1 & 2 & 1 & 4 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} \xrightarrow{R_3 - R_2} \begin{bmatrix} 1 & 2 & 1 & 4 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_1 - R_2} \begin{bmatrix} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$

The pivot positions are $(1,1)$ and $(2,3)$ ; the pivot columns are 1 and 3. Columns 2 and 4 are not pivot columns.

2. (3 pts) Solve the following system, giving a parametric description of its general solution (solve for the basic variables in terms of the free variable):

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

Solution

2. Augmented matrix:

$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & 5 & -1 & 7 \end{array}\right] \xrightarrow{R_2 - 2R_1} \left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & 1 & 1 \end{array}\right] \xrightarrow{R_1 - 2R_2} \left[\begin{array}{ccc|c} 1 & 0 & -3 & 1 \\ 0 & 1 & 1 & 1 \end{array}\right].$

Columns 1 and 2 hold the pivots, so $x_1, x_2$ are basic and $x_3$ is free. Solving each pivot row for its basic variable gives the parametric description $x_1 = 1 + 3x_3$ , $x_2 = 1 - x_3$ , with $x_3$ free (the free variable $x_3$ is the parameter).

3. (3 pts) For the coefficient matrix $\begin{bmatrix} 1 & -2 \\ 2 & h \end{bmatrix}$ , find the value of $h$ for which it fails to have a pivot in every column. For that value, determine whether the system with augmented matrix $\left[\begin{array}{cc|c} 1 & -2 & 3 \\ 2 & h & 5 \end{array}\right]$ is consistent.

Solution

3. $\begin{bmatrix} 1 & -2 \\ 2 & h \end{bmatrix} \xrightarrow{R_2 - 2R_1} \begin{bmatrix} 1 & -2 \\ 0 & h+4 \end{bmatrix}$ . Column 2 lacks a pivot exactly when $h + 4 = 0$ , i.e. $h = -4$ . For $h = -4$ the augmented matrix reduces to $\left[\begin{array}{cc|c} 1 & -2 & 3 \\ 0 & 0 & -1 \end{array}\right]$ (applying $R_2 - 2R_1$ gives $[\,0\;\;0\mid -1\,]$ ), so the system is inconsistent.

4. (3 pts) Determine whether each matrix is in (i) echelon form, (ii) reduced echelon form. If it is not in echelon form, state which condition fails.

(a) $\begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 3 \\ 0 & 1 & 0 \end{bmatrix}$

(b) $\begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}$

(c) $\begin{bmatrix} 2 & 1 & 0 \\ 0 & 0 & 5 \\ 0 & 0 & 0 \end{bmatrix}$

Solution

4.

(a) Not echelon. The leading entry of row 3 is in column 2, which is to the left of the leading entry of row 2 (in column 3); leading entries must move strictly right as you go down. Consequently it is not reduced echelon either.

(b) Echelon and reduced echelon. Leading entries step right, all leading entries are $1$ , each is the only nonzero entry in its column, and the zero row is at the bottom.

(c) Echelon but not reduced. The staircase is valid, but the leading entries ( $2$ and $5$ ) are not $1$ , so it fails the reduced-form requirement.

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

(a) The reduced echelon form of a matrix is unique.

(b) Every matrix has a unique echelon form.

(c) If a coefficient matrix has a pivot position in every row, then the corresponding system is consistent for every right-hand side.

Solution

5.

(a) True. Each matrix is row equivalent to one and only one matrix in reduced echelon form (Theorem 1).

(b) False. A matrix can be reduced to many different echelon forms depending on the operations chosen; only the reduced echelon form is unique.

(c) True. If every row has a pivot, then no echelon form of the augmented matrix can have a row $[\,0\cdots0\mid b\neq0\,]$ (that row would have no pivot in the coefficient part). Hence the system is consistent for every right-hand side.

6. (3 pts) Explain why a system of three equations in four unknowns can never have a unique solution. What are the only possibilities for its solution set?

Solution

6. The coefficient matrix has three rows, so there are at most three pivots. With four columns (unknowns), at least one column has no pivot, giving at least one free variable. Therefore the system can only be inconsistent (no solutions) or consistent with a free variable (infinitely many solutions). A unique solution would require a pivot in every column, which is impossible here.

7. (3 pts) The augmented matrix of a linear system has been reduced to

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

Identify the basic and free variables, and give a parametric description of the general solution (each basic variable in terms of the free variables).

Solution

7. Columns 1 and 3 are pivot columns, so $x_1$ and $x_3$ are basic; columns 2 and 4 have no pivot, so $x_2$ and $x_4$ are free. Solving each pivot row for its basic variable gives the parametric description

$x_1 = 5 + 2x_2 - 3x_4, \qquad x_3 = 2 + x_4, \qquad x_2, x_4 \text{ free}.$

8. (4 pts ••) Prove that a linear system is consistent if and only if the rightmost column of its augmented matrix is not a pivot column — equivalently, no row of an echelon form of the augmented matrix has the form $[\,0 \;\; \cdots \;\; 0 \mid b\,]$ with $b \neq 0$ .

Solution

8. ( $\Leftarrow$ ) Suppose no row of an echelon form of the augmented matrix is $[\,0\cdots0\mid b\,]$ with $b \neq 0$ . Then every nonzero row has its leading (pivot) entry in one of the first $n$ columns (the coefficient columns), so the rightmost column is not a pivot column. Assign any values to the free variables and back-substitute up the pivots; each pivot equation determines its basic variable, producing a solution. Hence the system is consistent.

( $\Rightarrow$ ) Conversely, suppose the rightmost column is a pivot column. Then some row of the echelon form has its pivot there, i.e. the row is $[\,0\cdots0\mid b\,]$ with $b \neq 0$ . This row asserts $0 = b \neq 0$ , an impossibility, so no solution exists and the system is inconsistent. Contrapositively, a consistent system has a non-pivot rightmost column. $\blacksquare$