PX153 - K4 - row-reduced echelon form (gaussian elimination)

\begin{aligned} x_{1} + 2 x_{1} & = 3 & L_{1} \\ x_{1} + x_{2} & = 7 & L_{2} \end{aligned}

\begin{aligned} L_{1} \to L_{1} & : x_{1} + 2 x_{1} = 3 \\ L_{2} \to L_{2} - L_{1} & : - x_{2} = - 4 \end{aligned}

\begin{aligned} L_{1} \to L_{1} + 2 L_{2} & : x_{1} = 11 \\ L_{2} \to - L_{2} & : - x_{1} - x_{2} = - 7 \end{aligned}

in the general case, for a system of $m$ linear equations, and $n$ unknowns of the form:

\begin{aligned} a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} & = b_{1} & L_{1} \\ a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} & = b_{2} & L_{2} \\ ⋮ \\ a_{m 1} x_{1} + a_{m 2} x_{2} + \dots + a_{m n} x_{n} & = b_{m} & L_{m} \end{aligned}

$a_{i j}$ and $b_{i}$ are all known coefficients
let, $A - (a_{i j})_{m \times n}$ , $\vec{x} = (x_{j})_{n}$ , $\vec{b} = (b_{i})_{m}$

A \vec{x} = \vec{b}

usually, $m = n$ , but sometimes, $m \neq n$ , so solution may be unique, incomplete, or non-existent
eg:

\begin{aligned} 2 x_{1} + x_{2} + x_{3} & = 1 \\ 3 x_{1} + (- x_{2}) - 2 x_{3} & = 1 \end{aligned}

- $L_{1} \to \frac{L_{1}}{2}$ ; $L_{2} \to L_{2} :$

[\begin{matrix} 2 & 1 & 1 & | & 1 \\ 3 & - 1 & - 2 & | & 1 \end{matrix}]

- $L_{1} \to L_{1}$ ; $L_{2} \to L_{2} - 3 L_{1} :$

[\begin{matrix} 1 & 1 / 2 & 1 / 2 & | & 1 / 2 \\ 0 & - 5 / 2 & - 7 / 2 & | & - 1 / 2 \end{matrix}]

- $L_{1} \to L_{1}$ ; $L_{2} \to \frac{- 2}{5} L_{2} :$

[\begin{matrix} 1 & 1 / 2 & 1 / 2 & | & 1 / 2 \\ 0 & 1 & 7 / 5 & | & 1 / 5 \end{matrix}]

- $L_{1} \to L_{1} - \frac{1}{2} L_{2}$ ; $L_{2} \to L_{2} :$

[\begin{matrix} 1 & 0 & \frac{1}{2} (1 - \frac{7}{5}) & | & \frac{1}{2} (1 - \frac{1}{5}) \\ 0 & 1 & 7 / 5 & | & 1 / 5 \end{matrix}]

\begin{aligned} x_{1} + \frac{1}{5} x_{3} & = \frac{2}{5} \\ x_{2} + \frac{7}{5} x_{3} & = \frac{1}{5} \end{aligned}

- setting $x_{3} = λ$ , solution set:

[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} \frac{2}{5} - \frac{1}{5} λ \\ \frac{1}{5} - \frac{7}{5} λ \\ λ \end{matrix}]

\begin{aligned} x_{1} - 2 x_{2} + x_{3} & = 1 \\ 2 x_{1} + x_{2} + x_{3} & = 1 \\ 5 x_{2} - x_{3} & = - 1 \end{aligned}

[\begin{matrix} 1 & - 2 & 1 & | & 1 \\ 2 & 1 & 1 & | & 1 \\ 0 & 5 & - 1 & | & - 1 \end{matrix}]

- $L_{2} \to L_{2} - 2 L_{1} :$

[\begin{matrix} 1 & - 2 & 1 & | & 1 \\ 0 & 5 & - 1 & | & - 1 \\ 0 & 5 & - 1 & | & - 1 \end{matrix}]

- $L_{3} \to L_{3} - L_{2} :$

[\begin{matrix} 1 & - 2 & 1 & | & 1 \\ 0 & 5 & - 1 & | & - 1 \\ 0 & 0 & 0 & | & 0 \end{matrix}]

- $L_{2} \to \frac{1}{5} L_{2} :$

[\begin{matrix} 1 & - 2 & 1 & | & 1 \\ 0 & 1 & - \frac{1}{5} & | & - \frac{1}{5} \end{matrix}]

- $L_{1} \to L_{1} + 2 L_{2} :$

[\begin{matrix} 1 & 0 & \frac{3}{5} & | & \frac{3}{5} \\ 0 & 1 & - \frac{1}{5} & | & - \frac{1}{5} \end{matrix}]

[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} \frac{3}{5} (1 - λ) \\ \frac{1}{5} (λ - 1) \\ λ \end{matrix}]

\begin{aligned} α x_{1} + β x_{2} & = a & L_{1} \\ α x_{1} + β x_{2} & = b & L_{2} \end{aligned}

- only consistent if $a = b$

\begin{aligned} α x_{1} + β x_{2} + γ x_{3} & = a & P_{1} \\ α x_{1} + β x_{2} + γ x_{3} & = b & P_{2} \end{aligned}

- if there are three equations with one linearly independent from the others, they can be parallel planes