Elimination for systems of equations uses three moves: swap two equations, multiply an equation by a nonzero number, and add a multiple of one equation to another. With matrices we do exactly the same moves on rows of numbers, which is faster to write and easy to organize.
Idea clave
Por qué funciona
Worked examples
Ejemplo resuelto: Two equations
- $\left[\begin{array}{cc|c} 2 & 1 & 5 \\ 1 & -1 & 1 \end{array}\right]$. Swap rows to put a $1$ on top: $\left[\begin{array}{cc|c} 1 & -1 & 1 \\ 2 & 1 & 5 \end{array}\right]$.
- $R_2 \to R_2 - 2R_1$: $\left[\begin{array}{cc|c} 1 & -1 & 1 \\ 0 & 3 & 3 \end{array}\right]$.
- $R_2 \to \frac{1}{3}R_2$: row 2 says $y = 1$. Then $R_1 \to R_1 + R_2$: row 1 says $x = 2$.
Respuesta: $(x, y) = (2, 1)$
Ejemplo resuelto: Three equations
- $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 2 \end{array}\right]$. $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$: $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & -3 & -1 & -9 \\ 0 & 1 & -2 & -4 \end{array}\right]$.
- Swap $R_2$ and $R_3$, then $R_3 \to R_3 + 3R_2$: $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -2 & -4 \\ 0 & 0 & -7 & -21 \end{array}\right]$.
- Back-substitute: $z = 3$; $y - 6 = -4$ gives $y = 2$; $x + 2 + 3 = 6$ gives $x = 1$.
Respuesta: $(1, 2, 3)$
Ejemplo resuelto: No solution
- Row 2 says $0x + 0y = 3$, which no numbers satisfy.
Respuesta: The system has no solution (its lines are parallel)
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