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Solving a system by row reduction: one, none or infinitely many solutions

Row reduction always ends in a staircase, but not every staircase names one point. How do the last rows say whether a system has one solution, none, or infinitely many, and how is an infinite family of solutions written down?

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Echelon forms and Gaussian eliminationPrecalculus
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When you reach for it

Any system of linear equations, in any number of variables, can be written as an augmented matrix and reduced to a row echelon form. That lesson kept to systems whose staircase has a pivot in every variable column, which gives exactly one solution. Most systems met in practice are not so tidy: an equation may contradict the others, or repeat what they already say. Row reduction handles these too, and the evidence is in the last rows of the echelon form, even when a coefficient is a letter whose value decides the case.

A system is consistent when it has at least one solution and inconsistent when it has none. Every linear system is in exactly one of three cases: one solution, no solution, or infinitely many.

The procedure

Day 1: reading the final rows

Reduce the augmented matrix to echelon form with row operations, naming each one (R_i \leftrightarrow R_j, kR_i \to R_i, R_i + cR_j \to R_i). Row operations never change the solution set, so whatever the final matrix says about solutions is true of the original system. Then read the rows from the bottom up.

A row \left[\begin{array}{ccc|c} 0 & 0 & 0 & c \end{array}\right] with c \ne 0. It is the equation 0x + 0y + 0z = c, that is 0 = c. No triple of numbers makes it true, so no triple solves the reduced system, and so none solves the original. The system is inconsistent; stop. In the language of pivots, a leading entry has appeared in the constants column, to the right of the bar.

A row of all zeros, \left[\begin{array}{ccc|c} 0 & 0 & 0 & 0 \end{array}\right]. It says 0 = 0, true for every triple. It adds no condition and can be ignored. It appears when one equation of the system was a combination of the others.

Otherwise the system is consistent, and the count of solutions depends on the variable columns.

  • A pivot in every variable column: exactly one solution, found by back-substitution, just as in Echelon forms and Gaussian elimination.
  • A variable column with no pivot: that variable is free, and there are infinitely many solutions. The variables whose columns hold pivots are the leading variables.
■ ∗ ∗ ∗ 0 ■ ∗ ∗ 0 0 ■ ∗ x y z one solution pivot in x, y, z ■ ∗ ∗ ∗ 0 ■ ∗ ∗ 0 0 0 ■ x y z no solution last row: 0 = c ■ ∗ ∗ ∗ 0 ■ ∗ ∗ 0 0 0 0 x y z infinitely many no pivot: z free
Figure 1The three ways an echelon form can end. A shaded square is a leading entry (never 0), a star is any number, and the bar separates the variable columns from the constants.

Why a free variable means infinitely many solutions: each nonzero row of the echelon form begins with its own leading variable, and the row below it involves only variables further right. Give the free variable any value at all. The bottom nonzero row then fixes its leading variable, since its pivot is not 0 and can be divided by; the row above fixes the next, and so on up. Every value of the free variable completes to exactly one solution, and different values give different solutions, so there are infinitely many.

Why the count is never two. If p and q are two different solutions, every point p + t(q - p) is one too. In an equation a_1x + a_2y + a_3z = b, write a \cdot v for the left side at the triple v. The left side is linear, so

\begin{aligned} a \cdot \big(p + t(q - p)\big) &= a \cdot p + t\,(a \cdot q - a \cdot p) \\ &= b + t(b - b) = b. \end{aligned}

Each real t gives a different point, so two solutions force infinitely many. In space each equation in three variables is a plane, and the three cases are the planes meeting in one point, having no point common to all three, or sharing a whole line or plane.

Day 2: writing the solution set

When a column has no pivot, keep going to reduced row echelon form (Gauss-Jordan elimination: make each pivot 1 and clear above it too). Each nonzero row now reads “leading variable plus some multiple of the free variables equals a number”. Name each free variable by a parameter, a letter such as t that may be any real number, and move the free terms to the right side. The result lists every solution at once, for example

(x, y, z) = (2 - t,\ 1 + 3t,\ t).

Each real t gives one solution; t = 0 gives (2, 1, 0). Reduced form is used because each leading variable then appears in only one row, so no back-substitution is needed.

The count of free variables is the count of variable columns minus the count of pivots. Two consequences follow. A consistent system with more variables than equations always has infinitely many solutions, since there are not enough rows to hold a pivot in every column. And a system with a letter k among its coefficients is settled by reducing with k carried along: the final rows are expressions in k, and each value of k that makes a would-be pivot 0 is a separate case.

Worked examples

Example 1

Day 1: reading three final forms

Each matrix is an echelon form of a system in x, y, z. Say how many solutions each system has.

A = \left[\begin{array}{ccc|c} 1 & 3 & -1 & 4 \\ 0 & 1 & 2 & -2 \\ 0 & 0 & 0 & 5 \end{array}\right]
B = \left[\begin{array}{ccc|c} 1 & -3 & 2 & 1 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right]
C = \left[\begin{array}{ccc|c} 2 & 1 & 0 & 3 \\ 0 & -1 & 4 & 2 \\ 0 & 0 & 3 & -6 \end{array}\right]
Solution

A: no solution. The last row says 0 = 5, false for every triple, whatever the first two rows say.

B: infinitely many solutions. The zero row says nothing. The pivots sit in the x and z columns, so the system is consistent and y is free; the free variable need not be the last one.

C: one solution. There are pivots in all three variable columns. Back-substitution gives z = -2, then -y - 8 = 2, so y = -10, then 2x - 10 = 3, so \displaystyle x = \frac{13}{2}. The solution is \displaystyle \left(\frac{13}{2}, -10, -2\right).

Example 2

Day 1: a contradiction hidden in the system

Solve x + y + 3z = 2, 2x + 3y + z = 5, 3x + 4y + 4z = 9.

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

Clear below the first pivot with R_2 - 2R_1 \to R_2 and R_3 - 3R_1 \to R_3. Row 2 becomes 0 \;\; 1 \;\; -5 \mid 1 and row 3 becomes 0 \;\; 1 \;\; -5 \mid 3. Then R_3 - R_2 \to R_3:

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

The last row reads 0 = 2. The system is inconsistent: no solution.

The reason: adding the first two equations gives 3x + 4y + 4z = 7, while the third says the same left side equals 9. As planes, no two are parallel, yet no point lies on all three.

Example 3

Day 1: four equations, three unknowns, one solution

Solve x - y + z = 2, 2x + y - z = 1, x + 2y + z = 8, x + 4y - z = 6.

Solution
\left[\begin{array}{ccc|c} 1 & -1 & 1 & 2 \\ 2 & 1 & -1 & 1 \\ 1 & 2 & 1 & 8 \\ 1 & 4 & -1 & 6 \end{array}\right]

R_2 - 2R_1 \to R_2, R_3 - R_1 \to R_3 and R_4 - R_1 \to R_4 clear the first column, giving rows 0 \;\; 3 \;\; -3 \mid -3, 0 \;\; 3 \;\; 0 \mid 6 and 0 \;\; 5 \;\; -2 \mid 4. Make the second pivot 1 with \displaystyle \frac{1}{3}R_2 \to R_2, giving 0 \;\; 1 \;\; -1 \mid -1, then clear below it: R_3 - 3R_2 \to R_3 and R_4 - 5R_2 \to R_4 both give 0 \;\; 0 \;\; 3 \mid 9. Finally R_4 - R_3 \to R_4:

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

The zero row says 0 = 0 and is dropped. No row is 0 = c with c \ne 0, and every variable column has a pivot, so there is one solution. Back-substitution: 3z = 9 gives z = 3; y - 3 = -1 gives y = 2; x - 2 + 3 = 2 gives x = 1. The solution is (1, 2, 3). Check in the fourth equation: 1 + 8 - 3 = 6. The fourth equation was a combination of the others.

Example 4

Day 2: one free variable

Solve x + 3y - 2z = 1, 2x + 7y - z = 5, x + 4y + z = 4, and write the solution set.

Solution
\left[\begin{array}{ccc|c} 1 & 3 & -2 & 1 \\ 2 & 7 & -1 & 5 \\ 1 & 4 & 1 & 4 \end{array}\right]

R_2 - 2R_1 \to R_2 and R_3 - R_1 \to R_3 both give 0 \;\; 1 \;\; 3 \mid 3, and R_3 - R_2 \to R_3 turns row 3 into zeros. The system is consistent, the pivots are in the x and y columns, and z is free. Go on to reduced form with R_1 - 3R_2 \to R_1:

\left[\begin{array}{ccc|c} 1 & 0 & -11 & -8 \\ 0 & 1 & 3 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]

The rows read x - 11z = -8 and y + 3z = 3. Let z = t. Then

(x, y, z) = (-8 + 11t,\ 3 - 3t,\ t)

for every real t. Check the whole family in the second equation: 2(-8 + 11t) + 7(3 - 3t) - t is -16 + 21 + 22t - 21t - t = 5. The value t = 1 gives the single solution (3, 0, 1), and 3 + 0 - 2 = 1 confirms it in the first equation. The third equation was the second minus the first, so the three planes share a line.

Example 5

Day 2: the free variable in the middle

Solve x - 2y + z = 4, 2x - 4y + 5z = 17.

Solution
\left[\begin{array}{ccc|c} 1 & -2 & 1 & 4 \\ 2 & -4 & 5 & 17 \end{array}\right]

R_2 - 2R_1 \to R_2 gives 0 \;\; 0 \;\; 3 \mid 9. The second pivot is in the z column, not the y column, because the y entry became 0 along with the x entry. So y is free. Reduce: \displaystyle \frac{1}{3}R_2 \to R_2 gives 0 \;\; 0 \;\; 1 \mid 3, and R_1 - R_2 \to R_1 gives 1 \;\; -2 \;\; 0 \mid 1. The rows read x - 2y = 1 and z = 3. With y = s,

(x, y, z) = (1 + 2s,\ s,\ 3)

for every real s. Infinitely many solutions, yet z is 3 in every one of them: a free variable lets the leading variables that depend on it move, not all of them. Check: (1 + 2s) - 2s + 3 = 4 and 2 + 4s - 4s + 15 = 17.

Example 6

Day 2: choosing the value of k

For which values of k does the system x + y + z = 2, x + 2y + 3z = 5, 2x + 3y + k^2 z = k + 5 have one solution, no solution, or infinitely many?

Solution
\left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 1 & 2 & 3 & 5 \\ 2 & 3 & k^2 & k + 5 \end{array}\right]

R_2 - R_1 \to R_2 gives 0 \;\; 1 \;\; 2 \mid 3, and R_3 - 2R_1 \to R_3 gives 0 \;\; 1 \;\; k^2 - 2 \mid k + 1. Then R_3 - R_2 \to R_3 leaves the last row

\left[\begin{array}{ccc|c} 0 & 0 & k^2 - 4 & k - 2 \end{array}\right]

None of these operations divided by an expression in k, so they are valid for every k. The entry k^2 - 4 = (k - 2)(k + 2) decides the case.

  • k \ne 2 and k \ne -2: the entry is a pivot, every variable column has one, and there is one solution, with \displaystyle z = \frac{k - 2}{k^2 - 4} = \frac{1}{k + 2}. For k = 0: \displaystyle z = \frac{1}{2}, y = 3 - 1 = 2, \displaystyle x = 2 - 2 - \frac{1}{2} = -\frac{1}{2}.
  • k = 2: the row is 0 \;\; 0 \;\; 0 \mid 0, z is free, and there are infinitely many solutions. With z = t, y = 3 - 2t and x = 2 - y - z = -1 + t, so (x, y, z) = (-1 + t,\ 3 - 2t,\ t).
  • k = -2: the row is 0 \;\; 0 \;\; 0 \mid -4, which says 0 = -4: no solution.

Dividing by k^2 - 4 at the start would have hidden both special cases, because division by 0 is undefined. Check the k = 2 family in the third equation, 2x + 3y + 4z = 7: -2 + 2t + 9 - 6t + 4t = 7.

Example 7

Day 2: a livestock mix with room to choose

Kuanysh mixes 100 kilograms of livestock ration from alfalfa (20\% protein, 0.30 dollars a kilogram), bran (15\% protein, 0.25 dollars) and corn (10\% protein, 0.20 dollars). The mix must hold 14 kilograms of protein and cost 24 dollars. Find every possible mix.

Solution

Let a, b, c be the kilograms of alfalfa, bran and corn. The conditions are a + b + c = 100, 0.20a + 0.15b + 0.10c = 14 and 0.30a + 0.25b + 0.20c = 24. Multiplying the last two by 20 clears the decimals:

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 100 \\ 4 & 3 & 2 & 280 \\ 6 & 5 & 4 & 480 \end{array}\right]

R_2 - 4R_1 \to R_2 and R_3 - 6R_1 \to R_3 both give 0 \;\; -1 \;\; -2 \mid -120, and R_3 - R_2 \to R_3 makes row 3 zero. The cost condition was already implied by the other two, so c is free. -R_2 \to R_2 gives 0 \;\; 1 \;\; 2 \mid 120, and R_1 - R_2 \to R_1 gives 1 \;\; 0 \;\; -1 \mid -20. With c = t:

(a, b, c) = (-20 + t,\ 120 - 2t,\ t).

Amounts cannot be negative. a \ge 0 needs t \ge 20 and b \ge 0 needs t \le 60, so every mix with between 20 and 60 kilograms of corn works, and only those. With t = 40 the mix is 20, 40 and 40 kilograms: protein 4 + 6 + 4 = 14, cost 6 + 10 + 8 = 24 dollars.

Had the budget been 25 dollars, the third row would end in 500, and the same operations would leave 0 \;\; 0 \;\; 0 \mid 20: no mix at all.

Boundaries

  • Linear equations only. The three-case count and the reading of final rows rely on every equation being linear. A system such as x^2 + y^2 = 25, y = x + 1 has exactly two solutions.
  • Read only after reaching echelon form. A row of zeros left of the bar means nothing until the matrix is in echelon form, since another row may still need to be cleared against it.
  • Never divide by an expression that may be 0. With a letter in the matrix, a scaling \displaystyle \frac{1}{k - 2}R_i \to R_i is a row operation only when k \ne 2; the value k = 2 must be treated as its own case.
  • The parameter is a choice. Which variables are free depends on the order of the columns, and the same set of solutions can be written with different parameters. Two descriptions agree when each value of one parameter matches exactly one value of the other.
  • Exact arithmetic. Rounded decimals can leave a tiny entry where a 0 belongs, faking a contradiction or a pivot.

Where this goes next

Solving a system by row reduction: one, none or infinitely many solutions studies families of solutions with several parameters, and Homogeneous systems and solutions in parametric vector form shows every solution as one particular solution plus a solution of the system with all constants 0. The number of pivots is the rank, which decides the case in one count. The same reduction balances chemical equations and fits curves through points in Linear systems in use: networks, chemical equations, fitting curves, and the case analysis repeats, by hand, what Systems in three variables and No solution, one solution, or every number found by elimination.

Exercises

Day 1: reading the final rows

For exercises 1 to 4, the matrix is an echelon form of a system. Say whether the system has no solution, one solution or infinitely many, and give the solution when there is exactly one.

  1. \left[\begin{array}{ccc|c} 1 & 4 & -3 & 2 \\ 0 & 1 & 5 & -1 \\ 0 & 0 & 0 & -3 \end{array}\right]
  2. \left[\begin{array}{ccc|c} 1 & -1 & 3 & 2 \\ 0 & 2 & -4 & 6 \\ 0 & 0 & 5 & 10 \end{array}\right]
  3. \left[\begin{array}{ccc|c} 3 & 1 & -1 & 4 \\ 0 & 0 & 2 & 6 \\ 0 & 0 & 0 & 0 \end{array}\right]
  4. \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & -3 & 6 \\ 0 & 0 & 0 \end{array}\right]

Day 1: solve, or show there is no solution

  1. 3x - 6y = 4, -x + 2y = 1
  2. x - y + 2z = 2, 2x + y + z = 1, 3x + 3z = 5
  3. x - 3y + z = 6, 2x - 6y + 3z = 14, -x + 4y + 2z = -1
  4. x + y = 3, 2x - y = 0, x - y = 2

Day 2: writing the solution set

Solve each system and write the solution set with parameters.

  1. x - 3y = 5, -2x + 6y = -10
  2. \left[\begin{array}{ccc|c} 1 & 0 & 2 & 5 \\ 0 & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]
  3. x + y + 3z = 4, 2x + 3y + 7z = 10, x + 2y + 4z = 6
  4. The single equation x + 2y - 3z = 6, in x, y, z
  5. 2x + 4y - z = 1, x + 2y + z = 5

Day 2: choosing k

For each system, find the values of the letters that give one solution, no solution and infinitely many solutions.

  1. x + 2y = 3, 2x + ky = 6
  2. x - y = 2, 3x - 3y = k
  3. x + y - z = 1, 2x + 3y + z = 4, 3x + 4y + kz = m
  4. x + ky = 2, kx + 4y = 4

Day 2: in context

  1. Lazzat mixes 60 kilograms of ration from the same three ingredients as Kuanysh: alfalfa (20\% protein), bran (15\%) and corn (10\%). It must hold 9 kilograms of protein. Find every possible mix (a, b, c), including the limits on each amount.
  2. Mukhtar sells Lazzat the alfalfa at 0.40 dollars a kilogram, the bran at 0.25 dollars and the corn at 0.20 dollars, and she wants her 60 kilograms with 9 kilograms of protein to cost exactly 16 dollars. Write the three equations without decimals, reduce, and find the mix.

Reasoning

  1. Explain why a final row 0 \;\; 0 \;\; 0 \mid 0 can be deleted without changing the solutions, while a single row 0 \;\; 0 \;\; 0 \mid 3 decides the whole system.
  2. A system has two equations in three variables. Explain why it cannot have exactly one solution, and give an example of such a system with no solution.
  3. Kuanysh finds that (1, 2, 0) and (3, 1, 1) both solve the same system of linear equations. Show that (5, 0, 2) solves it too, and describe infinitely many more solutions.
  4. Mukhtar writes the solutions of exercise 11 as (-2 + 2u,\ u,\ 2 - u). Is he right? Explain.
Answers
  1. No solution. The last row says 0 = -3.
  2. One solution, (3, 7, 2). 5z = 10 gives z = 2; 2y - 8 = 6 gives y = 7; x - 7 + 6 = 2 gives x = 3.
  3. Infinitely many solutions. Consistent, with pivots in the x and z columns, so y is free.
  4. One solution, (9, -2). The zero row says nothing; -3y = 6 gives y = -2, and x - 4 = 5 gives x = 9.
  5. No solution. R_1 \leftrightarrow R_2, then R_2 + 3R_1 \to R_2 gives 0 \;\; 0 \mid 7, that is 0 = 7.
  6. No solution. R_2 - 2R_1 \to R_2 gives 0 \;\; 3 \;\; -3 \mid -3, R_3 - 3R_1 \to R_3 gives 0 \;\; 3 \;\; -3 \mid -1, and R_3 - R_2 \to R_3 gives 0 \;\; 0 \;\; 0 \mid 2.
  7. One solution, (1, -1, 2). R_2 - 2R_1 \to R_2 gives 0 \;\; 0 \;\; 1 \mid 2 and R_3 + R_1 \to R_3 gives 0 \;\; 1 \;\; 3 \mid 5; R_2 \leftrightarrow R_3 finishes the staircase. So z = 2, y + 6 = 5 gives y = -1, and x + 3 + 2 = 6 gives x = 1.
  8. No solution. R_2 - 2R_1 \to R_2 and R_3 - R_1 \to R_3 give 0 \;\; -3 \mid -6 and 0 \;\; -2 \mid -1; \displaystyle -\frac{1}{3}R_2 \to R_2 gives 0 \;\; 1 \mid 2, and R_3 + 2R_2 \to R_3 gives 0 \;\; 0 \mid 3. The first two lines meet at (1, 2), which is not on the third.
  9. (x, y) = (5 + 3t,\ t). R_2 + 2R_1 \to R_2 gives a zero row, so y is free and x = 5 + 3y.
  10. (x, y, z) = (5 - 2t,\ 3 + t,\ t). Already reduced: x + 2z = 5, y - z = 3, z free.
  11. (x, y, z) = (2 - 2t,\ 2 - t,\ t). R_2 - 2R_1 \to R_2 and R_3 - R_1 \to R_3 both give 0 \;\; 1 \;\; 1 \mid 2; R_3 - R_2 \to R_3 gives a zero row, and R_1 - R_2 \to R_1 gives 1 \;\; 0 \;\; 2 \mid 2.
  12. (x, y, z) = (6 - 2s + 3t,\ s,\ t). One pivot, in the x column, so y = s and z = t are both free.
  13. (x, y, z) = (2 - 2s,\ s,\ 3). R_1 \leftrightarrow R_2, then R_2 - 2R_1 \to R_2 gives 0 \;\; 0 \;\; -3 \mid -9, so z = 3 and y is free; x + 2y + 3 = 5 gives x = 2 - 2y.
  14. k \ne 4: one solution, (3, 0). k = 4: infinitely many, (3 - 2t,\ t). Never none. R_2 - 2R_1 \to R_2 gives 0 \;\; k - 4 \mid 0.
  15. k = 6: infinitely many, (2 + t,\ t). k \ne 6: no solution. Never exactly one. R_2 - 3R_1 \to R_2 gives 0 \;\; 0 \mid k - 6.
  16. k \ne 0: one solution, for every m. k = 0 and m = 5: infinitely many. k = 0 and m \ne 5: no solution. R_2 - 2R_1 \to R_2 gives 0 \;\; 1 \;\; 3 \mid 2, R_3 - 3R_1 \to R_3 gives 0 \;\; 1 \;\; k + 3 \mid m - 3, and R_3 - R_2 \to R_3 gives 0 \;\; 0 \;\; k \mid m - 5.
  17. k \ne 2 and k \ne -2: one solution, \displaystyle \left(\frac{4}{k + 2}, \frac{2}{k + 2}\right). k = 2: infinitely many, (2 - 2t,\ t). k = -2: no solution. R_2 - kR_1 \to R_2 gives 0 \;\; 4 - k^2 \mid 4 - 2k, and 4 - k^2 = (2 - k)(2 + k).
  18. (a, b, c) = (t,\ 60 - 2t,\ t) with 0 \le t \le 30. The rows 1 \;\; 1 \;\; 1 \mid 60 and 4 \;\; 3 \;\; 2 \mid 180 (protein times 20) reduce by R_2 - 4R_1 \to R_2, -R_2 \to R_2 and R_1 - R_2 \to R_1 to 1 \;\; 0 \;\; -1 \mid 0 and 0 \;\; 1 \;\; 2 \mid 60. So a = c always, and b \ge 0 needs t \le 30.
  19. 10 kilograms of alfalfa, 40 of bran and 10 of corn. The cost equation times 20 is 8a + 5b + 4c = 320. With exercise 18’s rows, R_3 - 8R_1 \to R_3 gives 0 \;\; -3 \;\; -4 \mid -160, and after R_2 - 4R_1 \to R_2, R_3 - 3R_2 \to R_3 gives 0 \;\; 0 \;\; 2 \mid 20. So c = 10, b = 40, a = 10. Check: 4 + 10 + 2 = 16 dollars.
  20. The zero row is 0 = 0, true for every triple, so it removes no triple from the solution set. The row 0 = 3 is false for every triple, so no triple satisfies the whole system, whatever the other rows say.
  21. Each pivot sits in its own row, so two rows hold at most two pivots and one of the three variable columns has none. If the system is consistent, that variable is free and there are infinitely many solutions; otherwise there are none. Example with none: x + y + z = 1, x + y + z = 2.
  22. The difference is (3, 1, 1) - (1, 2, 0) = (2, -1, 1), and (1, 2, 0) + 2(2, -1, 1) = (5, 0, 2). Each left side is linear, so at (1, 2, 0) + t(2, -1, 1) it equals b + t(b - b) = b. Every point (1 + 2t,\ 2 - t,\ t) is a solution.
  23. Yes. Put t = 2 - u in (2 - 2t,\ 2 - t,\ t): 2 - 2(2 - u) = -2 + 2u, 2 - (2 - u) = u, and t = 2 - u. Each u gives exactly one t and each t one u, so the two descriptions list the same triples, with y as the parameter instead of z.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0