Two pictures of a linear system

Two equations, two unknowns. Watch one answer appear twice: where two lines cross, and as a mix of two arrows.

Rows: each equation is a line

Here is a small puzzle. Find two numbers $x$ and $y$ with

Take the first equation alone. Many pairs satisfy it: $(2,0)$, $(3,2)$, $(0,-4)$ and every point on the straight line through them. So one equation in two unknowns is a line in the plane. The second equation is another line, through $(5,0)$ and $(0,5)$.

A pair that satisfies both equations lies on both lines. That is the crossing point, $(3,2)$, in the left panel at the top of the page. You can find it without the picture: add the two equations, and $y$ cancels, leaving $3x=9$. So $x=3$, and then $y=5-3=2$.

Each line has a direction it cannot slide off. Read the first equation as a dot product, from Length and the dot product: $(2,-1)\cdot(x,y)=4$. The line is every point whose dot product with $(2,-1)$ is 4, and the arrow $(2,-1)$ stands at a right angle to it. Changing the 4 slides the line along that arrow, so the line stays parallel to itself. That is what you see when you drag $\mathbf{b}$ after the clip.

Columns: the same question as arrows

Now read the same two equations down the columns instead of across the rows. The numbers next to $x$ are $2$ and $1$; the numbers next to $y$ are $-1$ and $1$. So the two equations together say

This is a different question with the same answer: how much of the arrow $(2,1)$ and how much of the arrow $(-1,1)$ add up to $(4,5)$? The right panel answers it. Three of the first and two of the second, tip to tail, end exactly on $(4,5)$.

You have met this before. In A matrix moves the plane, the matrix with columns $(2,1)$ and $(-1,1)$ sent the point $(3,2)$ to $(4,5)$. Solving the equations runs that map backwards: we know where the point landed, and we ask where it started.

Drag $\mathbf{b}$ after the clip. The arrows stretch or shrink to reach it, the lines on the left slide, and the two readouts always show the same pair.

The matrix form

Both pictures come from one table of numbers. Write the unknowns as $x_1=x$ and $x_2=y$, and the right-hand sides as $b_1=4$ and $b_2=5$. The entry in row $p$ and column $q$ is $a_{pq}$, and both indices count from 1. Then the system is

The two pictures are the two ways to read $\mathbf{A}\mathbf{x}$. Row by row, each row of $\mathbf{A}$ dotted with $\mathbf{x}$ must give one number of $\mathbf{b}$: that is a line. Column by column, $\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+x_2\mathbf{a}_2$ must equal $\mathbf{b}$: that is a mix of arrows.

One answer, none, or infinitely many

Two lines do not always cross at one point. Try $x+y=2$ and $2x+2y=b_2$, with $b_2$ still to choose. The columns are now $(1,2)$ and $(1,2)$, the same arrow twice.

Drag $\mathbf{b}$ and watch both panels. Off the dashed line no mix of the columns reaches $\mathbf{b}$, because every mix of $(1,2)$ and $(1,2)$ lies on that line. On the left the same fact looks different: the lines are parallel, with no crossing. At $b_2=4$ the second equation is the first one doubled, so both lines are the same line, and every point on it is a solution.

So a system can have one solution, none, or infinitely many. It can never have exactly two. Suppose $\mathbf{x}$ and $\mathbf{x}'$ both solve $\mathbf{A}\mathbf{x}=\mathbf{b}$. Then every point $\mathbf{x}+c\,(\mathbf{x}'-\mathbf{x})$ on the line through them solves it too, because

Two solutions bring a whole line of them.

Having as many equations as unknowns does not settle it either. Both examples above have two of each; one has a single answer and the other has none or infinitely many. What decides is whether the columns point in different directions.

With three unknowns, each equation is a flat plane in space, and three planes usually meet in one point. They can also meet along a whole line, or have no point in common at all.

Worked example

1. One answer. $2x-y=4$ and $x+y=5$. Add them: $3x=9$, so $x=3$ and $y=2$. Column check: $3\,(2,1)+2\,(-1,1)=(4,5)$.
2. Matrix form. The same system is $\mathbf{A}\mathbf{x}=\mathbf{b}$ with $\mathbf{A}=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$ and $\mathbf{b}=(4,5)$.
3. No answer. $x+y=1$ and $x+y=3$. Predict before reading on: where do these lines meet? Nowhere, they are parallel. In columns, $x\,(1,1)+y\,(1,1)$ always has two equal numbers, so it can never be $(1,3)$.
4. Infinitely many. $x+y=2$ and $2x+2y=4$ are the same line. The points $(2,0)$, $(1,1)$ and $(0,2)$ all solve both, and so does every other point on that line.
5. A mixing problem. How many litres of a 10 % solution and of a 30 % solution make 10 litres at 22 %? Call them $x$ and $y$. The volumes give $x+y=10$, and the dissolved amounts give $0.1x+0.3y=2.2$. Take 0.1 times the first equation from the second: $0.2y=1.2$, so $y=6$ and $x=4$. Check: $0.4+1.8=2.2$.

Where you’ll meet this

Mixing problems like the last one are linear systems: one equation per thing you measure, one unknown per ingredient. So are the equations of an electric circuit, where the currents flowing into each junction must add up to zero.

Balancing a chemical equation is one too. To burn propane, find whole numbers in $a\,\mathrm{C_3H_8}+b\,\mathrm{O_2}\to c\,\mathrm{CO_2}+d\,\mathrm{H_2O}$. Each element gives one equation: carbon $3a=c$, hydrogen $8a=2d$, oxygen $2b=2c+d$. With $a=1$, these give $c=3$, $d=4$ and $b=5$: $\mathrm{C_3H_8}+5\,\mathrm{O_2}\to3\,\mathrm{CO_2}+4\,\mathrm{H_2O}$.

Reference card