Writing a System as Ax = b
Every linear system folds up into one compact matrix equation. The coefficients become a matrix
A, the unknowns a vector \vec{x}, and the
right-hand sides a vector \vec{b}:
\begin{aligned} 2x + y &= 5 \\ x - y &= 1 \end{aligned} \quad\Longleftrightarrow\quad \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}.
Check it: the matrix
times the vector reproduces the two equations exactly. Solving the system now reads
as one clean question — find the vector \vec{x} that
A sends to \vec{b}.
The column picture
Because A\vec{x} is a
combination of
A's columns, solving
A\vec{x}=\vec{b} means: how much of each column do I add to reach
\vec{b}? The two faint arrows are the columns; the star is
\vec{b}. Dial x and
y until the bold combination lands on the star — that's the solution.
Two ways to see the same answer
The row picture (lines crossing) and the column picture
(columns combining to reach \vec{b}) are two views of the identical
solution. The row picture is friendlier in 2D; the column picture is the one that scales — and it
makes the big truths obvious. If the columns are
independent
(so A is
invertible),
there's exactly one way to mix them — one unique solution.