Augmented matrices and row operations

Why bother with a matrix

• Elimination involves writing $x$, $y$ and $z$ over and over. The letters carry no information — only the coefficients change.
• An augmented matrix strips out the letters and keeps the numbers in position:

$$\begin{cases} x + 2y - z = 3 \\ 2x - y + 3z = 6 \\ x + y + z = 4 \end{cases} \qquad \longrightarrow \qquad \left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & -1 & 3 & 6 \\ 1 & 1 & 1 & 4 \end{array}\right]$$

• Each row is one equation. Each column is one variable, and the column after the bar holds the constants.
• The vertical bar matters. It marks where the equals signs were; without it the constants look like a fourth variable.
• This is bookkeeping, not new mathematics. Every row operation is exactly a step of elimination written more compactly.

Setting the matrix up correctly

• Every equation must be in the same order of variables before you start: $x$ terms, then $y$, then $z$, then the constant.
• A missing variable is a ZERO, not a gap. $2x + 3z = 7$ becomes the row $[\,2 \ \ 0 \ \ 3 \ | \ 7\,]$.
• Keep the signs. A coefficient of $-1$ is written $-1$, not 1.

The three row operations

• Only three operations are allowed, and each corresponds to something you already do in elimination:

• Every one of these leaves the solution unchanged, because each is reversible and each corresponds to a valid algebraic step.
• Multiplying by ZERO is not allowed. It destroys an equation and can turn an inconsistent system into a consistent-looking one.
• Label every operation. "$R_2 - 2R_1$" written beside the new row is the working, and without it the method is invisible.

Row echelon form

• The goal is a staircase of zeros in the lower left:

$$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -5 & 5 & 0 \\ 0 & 0 & 5 & 5 \end{array}\right]$$

• Once in this form, the last row is a single equation in one unknown, and the system unwinds by back-substitution from the bottom up.

• The procedure:

  • Use row 1 to make zeros below the first entry of column 1.
  • Use row 2 to make a zero below the leading entry of column 2.
  • Read off the bottom row, then back-substitute upwards.

• Worked through — from the matrix above:

  • Bottom row: $5z = 5$, so $z = 1$
  • Middle row: $-5y + 5(1) = 0$, so $y = 1$
  • Top row: $x + 2(1) - 1 = 3$, so $x = 2$
  • Solution $(2, 1, 1)$ — the same answer as by ordinary elimination ✓

Reading the nature of the solution off the matrix

• The bottom row tells you everything, before any back-substitution:

• This is the fastest way to classify a system, and it is why the matrix form is worth learning even though it changes no mathematics.
• A row of zeros including the constant means one equation carried no new information — it was a combination of the others.

What NOT to do with the matrix

• Do not swap columns. Rows are equations and may be reordered freely; columns are variables and reordering them silently renames your unknowns.
• Do not treat the augmented matrix as an object of matrix algebra. At this level it is a layout for elimination — there is no need for determinants or inverses, and this standard does not require them.