Ejercicios resueltos de sistemas 2×2

$\begin{pmatrix} 3 & 2 & \mid & 11 \\ 1 & 4 & \mid & 9 \end{pmatrix}$

$\begin{pmatrix} 1 & 4 & \mid & 9 \\ 3 & 2 & \mid & 11 \end{pmatrix}$

$\begin{pmatrix} 1 & 4 & \mid & 9 \\ 0 & -10 & \mid & -16 \end{pmatrix}$

$-10y = -16 \Rightarrow y = \frac{8}{5}$

$x + 4 \cdot \frac{8}{5} = 9 \Rightarrow x = 9 - \frac{32}{5} = \frac{13}{5}$

Solución: $x = \frac{13}{5}$, $y = \frac{8}{5}$

$\begin{pmatrix} 2 & -3 & \mid & -7 \\ 4 & 1 & \mid & 6 \end{pmatrix}$

$\begin{pmatrix} 1 & -\frac{3}{2} & \mid & -\frac{7}{2} \\ 4 & 1 & \mid & 6 \end{pmatrix}$

$\begin{pmatrix} 1 & -\frac{3}{2} & \mid & -\frac{7}{2} \\ 0 & 7 & \mid & 20 \end{pmatrix}$

$7y = 20 \Rightarrow y = \frac{20}{7}$

$x - \frac{3}{2} \cdot \frac{20}{7} = -\frac{7}{2} \Rightarrow x = -\frac{7}{2} + \frac{30}{7} = \frac{11}{14}$

$2 \cdot \frac{11}{14} - 3 \cdot \frac{20}{7} = \frac{11}{7} - \frac{60}{7} = -7$ ✓

$4 \cdot \frac{11}{14} + \frac{20}{7} = \frac{22}{7} + \frac{20}{7} = 6$ ✓

Solución: $x = \frac{11}{14}$, $y = \frac{20}{7}$

$\begin{pmatrix} 5 & -2 & \mid & 8 \\ 3 & 4 & \mid & 14 \end{pmatrix}$

$\begin{pmatrix} 1 & -\frac{2}{5} & \mid & \frac{8}{5} \\ 0 & \frac{26}{5} & \mid & \frac{46}{5} \end{pmatrix}$

$\frac{26}{5}y = \frac{46}{5} \Rightarrow y = \frac{46}{26} = \frac{23}{13}$

$x = \frac{8}{5} + \frac{2}{5} \cdot \frac{23}{13} = \frac{8}{5} + \frac{46}{65} = \frac{150}{65} = \frac{30}{13}$

Las rectas $5x - 2y = 8$ y $3x + 4y = 14$ se intersectan en el punto $\left(\frac{30}{13}, \frac{23}{13}\right)$

• Pendiente recta 1: $m_1 = \frac{5}{2}$
• Pendiente recta 2: $m_2 = -\frac{3}{4}$

Como $m_1 \neq m_2$, las rectas son secantes (solución única)

Solución: $x = \frac{30}{13}$, $y = \frac{23}{13}$

El sistema tiene solución única si y solo si $\text{det}\begin{pmatrix} a & b \\ d & e \end{pmatrix} = ae - bd \neq 0$

Matriz ampliada: $\begin{pmatrix} a & b & \mid & c \\ d & e & \mid & f \end{pmatrix}$

Si $a \neq 0$: $f_{21}(-d/a)$ da $\begin{pmatrix} a & b & \mid & c \\ 0 & e-\frac{bd}{a} & \mid & f-\frac{cd}{a} \end{pmatrix}$

$y = \frac{f - \frac{cd}{a}}{e - \frac{bd}{a}} = \frac{af - cd}{ae - bd}$

$x = \frac{c - by}{a} = \frac{ce - bf}{ae - bd}$

Con $a=7, b=-3, c=1, d=2, e=5, f=16$:

$\text{det} = 7 \cdot 5 - (-3) \cdot 2 = 35 + 6 = 41 \neq 0$ ✓

$x = \frac{1 \cdot 5 - (-3) \cdot 16}{41} = \frac{5 + 48}{41} = \frac{53}{41}$

$y = \frac{7 \cdot 16 - 1 \cdot 2}{41} = \frac{112 - 2}{41} = \frac{110}{41}$

Solución: $x = \frac{53}{41}$, $y = \frac{110}{41}$