In Algebra 1 you solved two equations in two unknowns by eliminating one of them. Linear algebra starts from that same skill, then asks a bigger question: what does a system look like, and what kinds of answers can it have? There are two useful pictures, and you will use both for the rest of the course.
A linear equation in the unknowns $x_1, \dots, x_n$ has the form $a_1x_1 + a_2x_2 + \cdots + a_nx_n = b$, where the $a_i$ and $b$ are numbers. No products of unknowns, no powers, no square roots. A system is a list of such equations, and a solution is a list of numbers $(s_1, \dots, s_n)$ that makes every equation true at once. The set of all solutions is the solution set.
The row picture: lines and planes
Take the system $2x - y = 1$ and $x + y = 5$. Each equation, on its own, is a line in the $xy$-plane: all the points that satisfy it. A solution must lie on both lines, so it is where they cross. Drag the probe point and watch which equations it satisfies.
With three unknowns, each equation is a plane in space. Three planes can meet in a single point, share a whole line, or have no point in common (for example, two of them parallel).
The column picture: combining vectors
Now read the same system down the columns. The coefficients of $x$ form the vector $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ and those of $y$ form $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$. The system says $$x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 5 \end{bmatrix}.$$ So the question becomes: how much of each column vector do we need to reach the right-hand side? This uses exactly the scaling and adding of vectors from precalculus.
Worked example: Both pictures for one system
- Add the two equations: the $y$ terms cancel, leaving $3x = 6$, so $x = 2$.
- Substitute into $x + y = 5$: $y = 3$.
- Row picture check: $(2, 3)$ is on both lines, since $2(2) - 3 = 1$ and $2 + 3 = 5$.
- Column picture check: $2\begin{bmatrix} 2 \\ 1 \end{bmatrix} + 3\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ $= \begin{bmatrix} 4 \\ 2 \end{bmatrix} + \begin{bmatrix} -3 \\ 3 \end{bmatrix}$ $= \begin{bmatrix} 1 \\ 5 \end{bmatrix}$.
Answer: $(x, y) = (2, 3)$
Worked example: A system with no solution
- Double the first equation: $2x + 4y = 6$. The second equation has the same left side but says it equals $7$.
- No pair $(x, y)$ can make $2x + 4y$ equal both $6$ and $7$. The lines have the same slope and different intercepts: they are parallel.
- Column picture: the columns $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 4 \end{bmatrix}$ point along the same line through the origin, and $\begin{bmatrix} 3 \\ 7 \end{bmatrix}$ is not on that line, so no combination reaches it.
Answer: No solution.
Why it works
| Row picture (2 unknowns) | Column picture | Number of solutions |
|---|---|---|
| Lines cross at one point | Exactly one combination of the columns gives $\mathbf{b}$ | Exactly one |
| Lines are parallel | $\mathbf{b}$ cannot be reached | None |
| Lines are the same line | $\mathbf{b}$ can be reached in many ways | Infinitely many |
Key idea