Solving linear systems

Rewrite a system without changing its solutions. Use rows to uncover a single solution, a family of solutions, or a contradiction.

Keep the equations; shorten the writing

Suppose two unknown real numbers must satisfy $x+2y=5$ and $3x+5y=13$ together. Each equation is linear: its unknowns are multiplied by fixed numbers and added. There is no $x^2$, product $xy$, or unknown in a denominator. A solution of the system is one pair $(x,y)$ that makes both equations true.

For these chosen equations, $(1,2)$ works: $1+2(2)=5$ and $3(1)+5(2)=13$. Guessing becomes awkward with more equations. We will organize the coefficients, the numbers multiplying the unknowns, so that a repeatable calculation can find the solutions.

$$\begin{cases}x+2y=5\\3x+5y=13\end{cases}\qquad A=\begin{bmatrix}1&2\\3&5\end{bmatrix}\qquad [A\mid b]=\left[\begin{array}{cc|c}1&2&5\\3&5&13\end{array}\right]$$

A stores the coefficients. The added column b stores the right-hand sides. The bar separates coefficients from constants; it is not division.

The augmented matrix keeps one equation per row and one unknown per coefficient column. Fix the order first: here column 1 means x and column 2 means y. In entry notation $a_{ij}$, i is the row and j the column, counting from 1; thus $a_{21}=3$. The coefficient matrix is $2\times2$, while the augmented matrix is $2\times3$.

Prepare an equation before copying its row

An equation such as $y=5-x$ first becomes $x+y=5$. In the order x, y, its augmented row is $[1\;1\mid5]$. Moving x left changes its sign.
With three unknowns in the order x, y, z, the equation $2x+z=3$ gives $[2\;0\;1\mid3]$. The absent y needs a zero placeholder.
Decode the row $[0\;-2\;1\mid4]$ in that same three-variable order: it says $-2y+z=4$. Rows never choose their own column order.

Writing a matrix has not solved or approximated anything. It records the same equations with the variable labels supplied by the agreed column order. Next, look for an arrangement that makes those equations easy to solve.