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What makes an equation linear, why a system has one solution, none or infinitely many, and the augmented matrix every later method operates on.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to say whether an equation is linear and why it matters, turn a described situation into a system of linear equations with its unknowns named and ordered, write the coefficient and augmented matrices of that system with a zero in every place a variable is absent, recognise from a pair of equations which of the three possible solution sets it has, and explain why exactly two solutions is not one of the possibilities.
You have solved a pair of simultaneous equations before, by substitution or by adding a multiple of one to the other. Nothing in this lesson replaces that. What changes is the bookkeeping: the same two steps, written so that they work for twenty equations in thirty unknowns and so that a machine can carry them out.
A linear equation in $x_1, \dots, x_n$ is $a_1x_1 + \dots + a_nx_n = b$, with the $a_i$ and $b$ fixed numbers. A system is a finite list of them, sharing their unknowns. A solution is a list of numbers that satisfies every equation at once, and the solution set is the set of all of them. A system with at least one solution is consistent; one with none is inconsistent. The coefficient matrix holds the $a_i$; the augmented matrix is the coefficient matrix with the column of $b$ attached.
An equation is linear when each unknown appears on its own, to the first power, multiplied only by a constant. $3x - 2y = 7$ is linear; $xy = 1$, $x^2 + y = 0$ and $1/x + y = 4$ are not, and no method in this course applies to them. The restriction looks severe and buys everything: linear equations are the only ones whose solution sets can be described completely, in every dimension, by one algorithm.
Two linear equations in two unknowns are two lines in the plane, and two lines can be arranged in exactly three ways. They cross at one point; they are parallel and never meet; or they coincide and meet everywhere. So the system has exactly one solution, none, or infinitely many.
A linear system never has exactly two solutions, or exactly seventeen. That is not a fact about small systems. If $u$ and $v$ are both solutions of $Ax = b$, then so is $u + t(v - u)$ for every real $t$ — a whole line of solutions through the two you had. Two distinct solutions therefore force infinitely many, in any number of equations and unknowns.
Three unknowns are three planes and the same trichotomy appears with more ways of reaching each case: three planes can miss each other pairwise, meet in a line, meet in a point, or coincide. The count of solutions is still one, none or infinitely many.
Another way: picture
Draw the two lines. A unique solution is a proper crossing, and it is stable: nudge either line and the crossing point moves a little but does not disappear. No solution is two parallel lines, and infinitely many is one line drawn twice — and those two are both unstable. Nudge one of a pair of parallel lines and they cross; nudge one of a coincident pair and they cross. The degenerate cases are exactly the ones a tiny change destroys, which is why numerical work treats a nearly-parallel pair with such suspicion.
Another way: steps
Three operations on a system leave its solution set exactly as it was: swapping two equations, multiplying an equation by a non-zero number, and adding a multiple of one equation to another. Each is reversible, which is the reason — an operation you can undo cannot lose a solution or invent one.
None of those three does anything to the variable names. Swapping two equations moves whole rows; scaling multiplies a whole row; adding a multiple of one to another adds row to row. So the names are dead weight, and dropping them is not an abbreviation but a genuine simplification:
$$\begin{array}{rcl} 2x + y - z &=& 8 \\ -3x - y + 2z &=& -11 \\ -2x + y + 2z &=& -3 \end{array} \qquad \longrightarrow \qquad \left(\begin{array}{rrr|r} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array}\right)$$
The vertical bar is a reminder, not a piece of mathematics: the last column is the right-hand side and the ones before it are coefficients, and the distinction matters when the time comes to read an answer back out.
Write the zeros. An equation with no $z$ in it has a $z$ coefficient of $0$, and a row that silently omits it is a row of the wrong length. Nearly every setting-up error in this subject is a missing zero.
A system is almost never handed over ready-made. It is built, and building it is the step that requires judgement:
In every case the unknowns must be named before anything is written, and the order fixed. The order is arbitrary and the consistency is not.
"No solution" and "the solution is zero" are different answers. $x = 0, y = 0$ is a solution — a list of numbers satisfying every equation. An inconsistent system has no list at all. The confusion is worth killing early, because a homogeneous system always has the zero solution and can never be inconsistent.
A variable that is missing is not absent; it has coefficient zero. Writing $x + z = 4$ as a row of two numbers rather than three makes the matrix ragged, and every operation after it is nonsense.
Linear does not mean "the graph is a straight line". $y = 3x + 2$ is linear in the sense of this course only after it is rearranged to $3x - y = -2$; and $y = 3x^2$ has a perfectly good graph and is not linear at all. The test is algebraic: each unknown alone, first power, constant multiplier.
Two solutions is never the answer. If you find two, you have found a line of them, and the right answer names the whole line.
$x + 2y = 5$ and $3x - y = 1$. Unknowns in the order $x, y$; the augmented matrix is $\left(\begin{array}{rr|r} 1 & 2 & 5 \\ 3 & -1 & 1 \end{array}\right)$.
Fix the order before writing anything.
Subtract $3$ times row 1 from row 2: $\left(\begin{array}{rr|r} 1 & 2 & 5 \\ 0 & -7 & -14 \end{array}\right)$.
One operation, one variable gone.
The second row says $-7y = -14$, so $y = 2$, and the first then gives $x = 1$. One solution: the lines cross.
Read back out, in the fixed order.
$2x + 4y = 6$ with $x + 2y = 3$: the first is twice the second. Row-reducing gives a row of zeros, and the solution set is the whole line $x = 3 - 2y$.
A repeated equation is no equation.
$2x + 4y = 6$ with $x + 2y = 5$: the left-hand sides are still proportional but the right-hand sides are not. Row-reducing gives $0 = -4$.
A row that reads zero equals something non-zero.
The same left-hand sides gave opposite answers, so the constants decide between the two degenerate cases and the coefficients alone never can.
Which is why the augmented column is kept separate.
Compare the left-hand sides: the second is $-2$ times the first.
Proportional, so not a crossing.
Now the constants: $-2 \times 2 = -4$, which is what the second equation says.
Consistent, so not parallel.
One line written twice: infinitely many solutions, namely every $(x, y)$ with $y = 3x - 2$.
Write the augmented matrix of the system $x + 7y = 4$, $6x + y = 2$.
This task has no paper form; do it on a device.
Solve $x + 2y = 8$ and $2x + 5y = 18$ for $x$.
Answer:
Which of these equations is linear in $x$ and $y$?
Match each system to the size of its solution set.
| Exactly one solution | No solution | Infinitely many solutions | |
|---|---|---|---|
| $x + 2y = 7$ and $3x + y = 8$ | |||
| $x + 2y = 7$ and $x + 2y = 8$ | |||
| $x + 2y = 7$ and $2x + 4y = 14$ |
A workshop makes chairs and tables. Select every sentence that states a linear equation in the number of chairs and the number of tables.
This task has no paper form; do it on a device.
One customer bought $2$ apples and $3$ pears for $26$ pence; another bought $1$ apples and $5$ pears for $34$ pence. Fill in the coefficient table of the system this gives.
| Apples | Pears | Pence paid | |
|---|---|---|---|
| First customer | |||
| Second customer |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Write the augmented matrix of the system $x + y = 4$, $7x + 8y = 2$.
This task has no paper form; do it on a device.
You can build a system from a description, write its augmented matrix, and name which of the three cases it falls into. Say in your own words why a linear system cannot have exactly two solutions. Next: the algorithm that settles the question for any system at all.
10. Your turn: how many solutions has $3x - y = 2$ with $-6x + 2y = -4$?, step 3
Name the set, not a count.