Systems of Linear Equations (Three Unknowns)

You already know how to solve two equations in two unknowns by elimination. With three unknowns the idea is the same, just with one more round of eliminating. The new part is understanding what kind of answer you get: exactly one solution, infinitely many, or none at all. Each equation in $x$, $y$ and $z$ is a plane, so this is really a question about how three planes meet.

Key ideas

Three possible outcomes

A system of linear equations has exactly one of these:

Outcome	Name	Geometric meaning (three planes)
unique solution	consistent	the planes meet at a single point
infinitely many solutions	consistent	the planes share a whole line (or are all the same plane)
no solution	inconsistent	there is no point on all three planes

The different ways planes can meet (or fail to) are drawn on intersections of planes.

Augmented matrix notation

Writing $x$, $y$, $z$ and $=$ over and over is tiring. An augmented matrix keeps just the numbers: one row per equation, one column per variable, and a bar before the constants.

$$\begin{aligned} x + y + z &= 6 \\ 2x - y + z &= 3 \\ x + 2y - z &= 2 \end{aligned} \qquad\longleftrightarrow\qquad \left(\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 2 \end{array}\right)$$

It’s only a shorthand: each row is still an equation.

Row operations

These three moves never change the solutions of a system:

• swap two rows;
• multiply a row by a non-zero number;
• add (or subtract) a multiple of one row to another row.

We write them like $R_2 \to R_2 - 2R_1$ (“replace row $2$ with row $2$ minus twice row $1$”).

The goal of row reduction is to make zeros below the leading diagonal (a “triangular” or row echelon form):

$$\left(\begin{array}{ccc|c} * & * & * & * \\ 0 & * & * & * \\ 0 & 0 & * & * \end{array}\right)$$

Then the last row gives $z$, and you back-substitute to find $y$ and $x$.

Reading the last row

Once the matrix is triangular (rows 1 and 2 have leading entries), the last row tells you which outcome you have. In general, any row of the form $(0 \ \ 0 \ \ 0 \mid b)$ with $b \ne 0$, wherever it appears, means there is no solution.

Last row	Equation	Outcome
$(0 \ \ 0 \ \ a \mid b)$ with $a \ne 0$	$az = b$	unique solution
$(0 \ \ 0 \ \ 0 \mid 0)$	$0 = 0$	infinitely many solutions
$(0 \ \ 0 \ \ 0 \mid b)$ with $b \ne 0$	$0 = b$, which is false	no solution

The general solution

When a row becomes $0 = 0$, one equation was really a combination of the others, so you have fewer genuine equations than unknowns. Let one variable be a parameter, usually $z = \lambda$, where $\lambda$ can be any real number, and write the other variables in terms of $\lambda$. This is the general solution. Each value of $\lambda$ gives one particular solution.

With three unknowns and one free parameter, the solutions form a line, and the general solution is its vector equation in disguise:

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} + \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}$$

Using technology

On a GDC you can:

• use the simultaneous equation solver: enter the coefficients and it gives the unique solution, or reports that there are no solutions or infinitely many;
• enter the augmented matrix and use rref (reduced row echelon form). The GDC row reduces all the way, so you can read off the answer, or the general solution, directly.

The guide expects you to solve systems both by algebra and with technology, so practise both. When a question has an unknown constant in it (like $k$), you have to work by hand.

Systems with a parameter

If a coefficient is an unknown $k$, row reduce as usual, keeping $k$ in the working. The last row will look like $(0 \ \ 0 \ \ f(k) \mid g)$:

• if $f(k) \ne 0$: a unique solution;
• if $f(k) = 0$ and the right-hand side is not zero: no solution;
• if $f(k) = 0$ and the right-hand side is also zero: infinitely many solutions.

Worked examples

Example 1: A unique solution

Solve the system

$$\begin{aligned} x + y + z &= 6 \\ 2x - y + z &= 3 \\ x + 2y - z &= 2 \end{aligned}$$

Solution. Write the augmented matrix and clear the first column below the top row:

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 2 \end{array}\right) \quad \begin{array}{l} \\ R_2 \to R_2 - 2R_1 \\ R_3 \to R_3 - R_1 \end{array} \quad \left(\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & -3 & -1 & -9 \\ 0 & 1 & -2 & -4 \end{array}\right)$$

Now clear the second column below row $2$. Using $R_3 \to 3R_3 + R_2$ avoids fractions:

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & -3 & -1 & -9 \\ 0 & 0 & -7 & -21 \end{array}\right)$$

Back-substitute:

$$\begin{aligned} -7z &= -21 &&\Rightarrow\quad z = 3 \\ -3y - 3 &= -9 &&\Rightarrow\quad y = 2 \\ x + 2 + 3 &= 6 &&\Rightarrow\quad x = 1 \end{aligned}$$

The solution is $x = 1$, $y = 2$, $z = 3$: the three planes meet at the point $(1, 2, 3)$.

Check in the original equations: $1 + 2 + 3 = 6$ ✓, $2 - 2 + 3 = 3$ ✓, $1 + 4 - 3 = 2$ ✓.

Example 2: Infinitely many solutions

Solve the system

$$\begin{aligned} x + 2y - z &= 3 \\ 2x + 5y + z &= 9 \\ x + 3y + 2z &= 6 \end{aligned}$$

Solution.

$$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & 5 & 1 & 9 \\ 1 & 3 & 2 & 6 \end{array}\right) \quad \begin{array}{l} \\ R_2 \to R_2 - 2R_1 \\ R_3 \to R_3 - R_1 \end{array} \quad \left(\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & 3 & 3 \\ 0 & 1 & 3 & 3 \end{array}\right) \quad R_3 \to R_3 - R_2 \quad \left(\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & 3 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right)$$

The last row says $0 = 0$, so there are infinitely many solutions. Let $z = \lambda$. Then

$$\begin{aligned} y + 3\lambda &= 3 &&\Rightarrow\quad y = 3 - 3\lambda \\ x + 2(3 - 3\lambda) - \lambda &= 3 &&\Rightarrow\quad x = -3 + 7\lambda \end{aligned}$$

The general solution is $x = -3 + 7\lambda$, $y = 3 - 3\lambda$, $z = \lambda$, for $\lambda \in \mathbb{R}$. In vector form:

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 \\ 3 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 7 \\ -3 \\ 1 \end{pmatrix}$$

Geometrically, the three planes meet in this line. (No two of the equations are multiples of each other, so the planes are all different: they form a sheaf, like pages around a book’s spine.)

Check with $\lambda = 1$, the point $(4, 0, 1)$: $4 + 0 - 1 = 3$ ✓, $8 + 0 + 1 = 9$ ✓, $4 + 0 + 2 = 6$ ✓.

With a GDC: rref of the augmented matrix gives rows $(1 \ \ 0 \ \ {-7} \mid {-3})$ and $(0 \ \ 1 \ \ 3 \mid 3)$ and a row of zeros. That’s $x - 7z = -3$ and $y + 3z = 3$, the same general solution.

Example 3: No solution

Show that this system has no solution, and describe it geometrically.

$$\begin{aligned} x + y + z &= 2 \\ x - y + 2z &= 1 \\ 2x + 3z &= 4 \end{aligned}$$

Solution.

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 1 & -1 & 2 & 1 \\ 2 & 0 & 3 & 4 \end{array}\right) \quad \begin{array}{l} \\ R_2 \to R_2 - R_1 \\ R_3 \to R_3 - 2R_1 \end{array} \quad \left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & -2 & 1 & -1 \\ 0 & -2 & 1 & 0 \end{array}\right) \quad R_3 \to R_3 - R_2 \quad \left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & -2 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{array}\right)$$

The last row says $0 = 1$, which is impossible. So the system is inconsistent: there is no solution.

You can see why in the original equations: adding the first two gives $2x + 3z = 3$, but the third equation says $2x + 3z = 4$.

Geometrically: the normals $(1, 1, 1)$, $(1, -1, 2)$ and $(2, 0, 3)$ are not parallel to each other, so no two planes are parallel. Each pair of planes meets in a line, but there’s no point common to all three: the planes form a triangular prism.

Example 4: A system with parameters

Consider the system

$$\begin{aligned} x + 2y - z &= 1 \\ 2x + 5y + z &= 4 \\ x + 3y + kz &= m \end{aligned}$$

where $k, m \in \mathbb{R}$.

• (a) Find the value of $k$ for which the system does not have a unique solution.
• (b) For this value of $k$, find the value of $m$ for which the system has infinitely many solutions, and find the general solution.
• (c) Describe what happens for the same $k$ and any other value of $m$.

Solution. Row reduce, keeping $k$ and $m$:

$$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 2 & 5 & 1 & 4 \\ 1 & 3 & k & m \end{array}\right) \quad \begin{array}{l} \\ R_2 \to R_2 - 2R_1 \\ R_3 \to R_3 - R_1 \end{array} \quad \left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 1 & k + 1 & m - 1 \end{array}\right)$$ $$R_3 \to R_3 - R_2 \qquad \left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & k - 2 & m - 3 \end{array}\right)$$

(a) The last row is $(k - 2)z = m - 3$. If $k \ne 2$, it gives a single value of $z$, and so a unique solution. The system does not have a unique solution when $k = 2$.

(b) With $k = 2$, the last row is $0 = m - 3$. For infinitely many solutions this must say $0 = 0$, so $m = 3$.

Then let $z = \lambda$:

$$\begin{aligned} y + 3\lambda &= 2 &&\Rightarrow\quad y = 2 - 3\lambda \\ x + 2(2 - 3\lambda) - \lambda &= 1 &&\Rightarrow\quad x = -3 + 7\lambda \end{aligned}$$

The general solution is $x = -3 + 7\lambda$, $y = 2 - 3\lambda$, $z = \lambda$, for $\lambda \in \mathbb{R}$.

(c) If $k = 2$ and $m \ne 3$, the last row says $0 = m - 3$ with $m - 3 \ne 0$, which is false. The system is inconsistent and has no solution.

Check for (b), with $\lambda = 0$, the point $(-3, 2, 0)$: $-3 + 4 - 0 = 1$ ✓, $-6 + 10 + 0 = 4$ ✓, $-3 + 6 + 0 = 3$ ✓.

Common mistakes

Doing a row operation on only part of a row. $R_2 \to R_2 - 2R_1$ applies to every entry in the row, including the number after the bar.

Reading a row of zeros as “no solution”. A row $(0 \ \ 0 \ \ 0 \mid 0)$ means $0 = 0$, which is always true: infinitely many solutions. It’s $(0 \ \ 0 \ \ 0 \mid b)$ with $b \ne 0$ that means no solution.

Giving one particular solution instead of the general solution. When there are infinitely many solutions, the question wants all of them: write $x$, $y$ and $z$ in terms of a parameter $\lambda$, and say $\lambda \in \mathbb{R}$.

Stopping at “no unique solution”. When the last coefficient is zero (for example $k = 2$ in Example 4), you still have to decide between no solution and infinitely many, by looking at the right-hand side.

Dividing by an expression that could be zero. Solving $(k - 2)z = m - 3$ as $z = \dfrac{m - 3}{k - 2}$ is only allowed when $k \ne 2$. Deal with the case $k = 2$ separately.

Arithmetic slips with negatives. Row reduction involves a lot of subtraction. Check your final answer by substituting it into all three original equations, not just one.

Practice

1. (Warm-up) Solve the system by back-substitution:

$$\begin{aligned} x + 2y + z &= 7 \\ y - z &= -1 \\ 2z &= 6 \end{aligned}$$

Solution

$$\begin{aligned} 2z &= 6 &&\Rightarrow\quad z = 3 \\ y - 3 &= -1 &&\Rightarrow\quad y = 2 \\ x + 4 + 3 &= 7 &&\Rightarrow\quad x = 0 \end{aligned}$$

The solution is $x = 0$, $y = 2$, $z = 3$.

2. (Warm-up) Each augmented matrix has been row reduced. Say whether the system has a unique solution, infinitely many solutions, or no solution.

• (a) $\left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 5 \end{array}\right)$
• (b) $\left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right)$
• (c) $\left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 4 & 8 \end{array}\right)$

Solution

(a) The last row says $0 = 5$, which is false: no solution.

(b) The last row says $0 = 0$: infinitely many solutions.

(c) The last row says $4z = 8$, so $z = 2$, and back-substitution gives one value each for $y$ and $x$: a unique solution.

3. (Core) Solve the system by row reduction:

$$\begin{aligned} 2x + y - z &= 3 \\ x - y + 2z &= 5 \\ 3x + 2y + z &= 10 \end{aligned}$$

Solution

Swap $R_1$ and $R_2$ so the top-left entry is $1$:

$$\left(\begin{array}{ccc|c} 1 & -1 & 2 & 5 \\ 2 & 1 & -1 & 3 \\ 3 & 2 & 1 & 10 \end{array}\right) \quad \begin{array}{l} \\ R_2 \to R_2 - 2R_1 \\ R_3 \to R_3 - 3R_1 \end{array} \quad \left(\begin{array}{ccc|c} 1 & -1 & 2 & 5 \\ 0 & 3 & -5 & -7 \\ 0 & 5 & -5 & -5 \end{array}\right)$$$$R_3 \to 3R_3 - 5R_2 \qquad \left(\begin{array}{ccc|c} 1 & -1 & 2 & 5 \\ 0 & 3 & -5 & -7 \\ 0 & 0 & 10 & 20 \end{array}\right)$$

Back-substitute:

$$\begin{aligned} 10z &= 20 &&\Rightarrow\quad z = 2 \\ 3y - 10 &= -7 &&\Rightarrow\quad y = 1 \\ x - 1 + 4 &= 5 &&\Rightarrow\quad x = 2 \end{aligned}$$

The solution is $x = 2$, $y = 1$, $z = 2$.

Check: $4 + 1 - 2 = 3$ ✓, $2 - 1 + 4 = 5$ ✓, $6 + 2 + 2 = 10$ ✓.

4. (Core) Show that this system has infinitely many solutions, and find the general solution.

$$\begin{aligned} x + y + z &= 4 \\ x - y + 3z &= 2 \\ 3x + y + 5z &= 10 \end{aligned}$$

Solution

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 1 & -1 & 3 & 2 \\ 3 & 1 & 5 & 10 \end{array}\right) \quad \begin{array}{l} \\ R_2 \to R_2 - R_1 \\ R_3 \to R_3 - 3R_1 \end{array} \quad \left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & -2 & 2 & -2 \\ 0 & -2 & 2 & -2 \end{array}\right) \quad R_3 \to R_3 - R_2 \quad \left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & -2 & 2 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right)$$

The last row is $0 = 0$, so there are infinitely many solutions. Let $z = \lambda$:

$$\begin{aligned} -2y + 2\lambda &= -2 &&\Rightarrow\quad y = 1 + \lambda \\ x + (1 + \lambda) + \lambda &= 4 &&\Rightarrow\quad x = 3 - 2\lambda \end{aligned}$$

The general solution is $x = 3 - 2\lambda$, $y = 1 + \lambda$, $z = \lambda$, for $\lambda \in \mathbb{R}$.

Check with $\lambda = 1$, the point $(1, 2, 1)$: $1 + 2 + 1 = 4$ ✓, $1 - 2 + 3 = 2$ ✓, $3 + 2 + 5 = 10$ ✓.

5. (Core) Show that this system is inconsistent, and describe the three planes geometrically.

$$\begin{aligned} x + 2y + 3z &= 1 \\ 2x + 4y + 6z &= 5 \\ x - y + z &= 0 \end{aligned}$$

Solution

Apply $R_2 \to R_2 - 2R_1$:

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

The second row says $0 = 3$, which is false, so the system is inconsistent: there’s no solution.

Geometrically, the first two equations have parallel normals, $(2, 4, 6) = 2(1, 2, 3)$, but the second equation is not twice the first ($5 \ne 2 \times 1$). So the first two planes are parallel and distinct. The third plane’s normal $(1, -1, 1)$ isn’t parallel to them, so it cuts both planes, in two parallel lines.

6. (Core) A café sells small coffees, lattes and teas. Three orders cost:

• $2$ coffees, $1$ latte and $3$ teas: $15.50
• $1$ coffee, $2$ lattes and $1$ tea: $13.50
• $3$ coffees, $1$ latte and $2$ teas: $16.00

Use your GDC to find the price of each drink.

Solution

Let $c$, $l$ and $t$ be the prices of a coffee, a latte and a tea, in dollars:

$$\begin{aligned} 2c + l + 3t &= 15.50 \\ c + 2l + t &= 13.50 \\ 3c + l + 2t &= 16.00 \end{aligned}$$

Enter the coefficients in the GDC’s simultaneous equation solver (or find the rref of the augmented matrix). It gives

$$c = 2.5, \qquad l = 4.5, \qquad t = 2$$

A coffee costs $2.50, a latte $4.50 and a tea $2.00.

Check the first order: $2(2.5) + 4.5 + 3(2) = 5 + 4.5 + 6 = 15.5$ ✓.

7. (Core) Consider the system

$$\begin{aligned} x + y + z &= 3 \\ x + 2y + 3z &= 5 \\ 2x + 3y + kz &= 9 \end{aligned}$$

Find the value of $k$ for which the system does not have a unique solution, and show that for this value it has no solution.

Solution

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 1 & 2 & 3 & 5 \\ 2 & 3 & k & 9 \end{array}\right) \quad \begin{array}{l} \\ R_2 \to R_2 - R_1 \\ R_3 \to R_3 - 2R_1 \end{array} \quad \left(\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 1 & k - 2 & 3 \end{array}\right) \quad R_3 \to R_3 - R_2 \quad \left(\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & k - 4 & 1 \end{array}\right)$$

The last row is $(k - 4)z = 1$. If $k \ne 4$, $z = \dfrac{1}{k - 4}$ and there’s a unique solution. So the system does not have a unique solution when $k = 4$.

When $k = 4$, the last row says $0 = 1$, which is false. So for $k = 4$ the system has no solution.

8. (Challenge) Consider the system

$$\begin{aligned} x + y + 2z &= 1 \\ 2x + 3y + 5z &= 4 \\ x + 2y + kz &= m \end{aligned}$$

• (a) Find the values of $k$ and $m$ for which the system has infinitely many solutions.
• (b) For these values, find the general solution.
• (c) For which values of $k$ and $m$ does the system have no solution?

Solution

$$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 2 & 3 & 5 & 4 \\ 1 & 2 & k & m \end{array}\right) \quad \begin{array}{l} \\ R_2 \to R_2 - 2R_1 \\ R_3 \to R_3 - R_1 \end{array} \quad \left(\begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 1 & k - 2 & m - 1 \end{array}\right) \quad R_3 \to R_3 - R_2 \quad \left(\begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & k - 3 & m - 3 \end{array}\right)$$

(a) Infinitely many solutions need the last row to be $0 = 0$: $k - 3 = 0$ and $m - 3 = 0$. So $k = 3$ and $m = 3$.

(b) Let $z = \lambda$:

$$\begin{aligned} y + \lambda &= 2 &&\Rightarrow\quad y = 2 - \lambda \\ x + (2 - \lambda) + 2\lambda &= 1 &&\Rightarrow\quad x = -1 - \lambda \end{aligned}$$

The general solution is $x = -1 - \lambda$, $y = 2 - \lambda$, $z = \lambda$, for $\lambda \in \mathbb{R}$.

Check with $\lambda = 0$, the point $(-1, 2, 0)$: $-1 + 2 + 0 = 1$ ✓, $-2 + 6 + 0 = 4$ ✓, $-1 + 4 + 0 = 3$ ✓.

(c) No solution when the last row reads $0 = m - 3$ with $m - 3 \ne 0$: that is, $k = 3$ and $m \ne 3$.

9. (Challenge) Return to the system in Question 7.

• (a) For $k \ne 4$, find the unique solution in terms of $k$.
• (b) Describe the three planes geometrically when $k = 4$.

Solution

(a) From the reduced matrix in Question 7, $z = \dfrac{1}{k - 4}$. Back-substitute into $y + 2z = 2$:

$$y = 2 - \frac{2}{k - 4}$$

Then from $x + y + z = 3$:

$$x = 3 - \left(2 - \frac{2}{k - 4}\right) - \frac{1}{k - 4} = 1 + \frac{1}{k - 4}$$

So $x = 1 + \dfrac{1}{k - 4}$, $y = 2 - \dfrac{2}{k - 4}$, $z = \dfrac{1}{k - 4}$.

Check with $k = 5$: $(2, 0, 1)$ gives $2 + 0 + 1 = 3$ ✓, $2 + 0 + 3 = 5$ ✓, $4 + 0 + 5 = 9$ ✓.

(b) When $k = 4$ the normals are $(1, 1, 1)$, $(1, 2, 3)$ and $(2, 3, 4)$. No two are parallel (none is a multiple of another), so no two planes are parallel. But the system has no solution, so the three planes have no common point: each pair meets in a line, and the three lines are parallel. The planes form a triangular prism.