Regla de Cramer: 8 sistemas de ecuaciones lineales resueltos paso a paso
Published on November 6, 2026
Resolución de ocho sistemas de ecuaciones lineales de dos, tres y cuatro incógnitas mediante la regla de Cramer, calculando el determinante principal y los determinantes de cada variable.
Question
\text{Problemas 3.4 – Regla de Cramer} 1. Sistema \[ \begin{cases} 7x_1 - 8x_2 = 3 \\ 9x_1 + 9x_2 = -8 \end{cases} \] \text{Paso 1. Determinante principal} \[ D = \begin{vmatrix} 7 & -8 \\ 9 & 9 \end{vmatrix} = (7)(9) - (-8)(9) = 63 + 72 = 135 \] \text{Paso 2. Determinante de } x_1 \[ D_1 = \begin{vmatrix} 3 & -8 \\ -8 & 9 \end{vmatrix} = (3)(9) - (-8)(-8) = 27 - 64 = -37 \] \text{Paso 3. Determinante de } x_2 \[ D_2 = \begin{vmatrix} 7 & 3 \\ 9 & -8 \end{vmatrix} = (7)(-8) - (3)(9) = -56 - 27 = -83 \] \text{Paso 4. Resultados} \[ x_1 = \frac{D_1}{D} = \frac{-37}{135}, \quad x_2 = \frac{D_2}{D} = \frac{-83}{135} \] 2. Sistema \[ \begin{cases} 3x_1 - x_2 = 0 \\ 4x_1 + 2x_2 = 5 \end{cases} \] \text{Paso 1. Determinante principal} \[ D = \begin{vmatrix} 3 & -1 \\ 4 & 2 \end{vmatrix} = (3)(2) - (-1)(4) = 6 + 4 = 10 \] \text{Paso 2. Determinante de } x_1 \[ D_1 = \begin{vmatrix} 0 & -1 \\ 5 & 2 \end{vmatrix} = (0)(2) - (-1)(5) = 5 \] \text{Paso 3. Determinante de } x_2 \[ D_2 = \begin{vmatrix} 3 & 0 \\ 4 & 5 \end{vmatrix} = (3)(5) - (0)(4) = 15 \] \text{Paso 4. Resultados} \[ x_1 = \frac{5}{10} = \frac{1}{2}, \quad x_2 = \frac{15}{10} = \frac{3}{2} \] 3. Sistema \[ \begin{cases} 2x_1 + x_2 + x_3 = 6 \\ 3x_1 - 2x_2 - 3x_3 = 5 \\ 8x_1 + 2x_2 + 5x_3 = 11 \end{cases} \] \text{Paso 1. Determinante principal} \[ D = \begin{vmatrix} 2 & 1 & 1 \\ 3 & -2 & -3 \\ 8 & 2 & 5 \end{vmatrix} = 2(-10 + 6) - 1(15 + 24) + 1(6 + 16) = -8 - 39 + 22 = -25 \] \text{Paso 2. Determinante de } x_1 \[ D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 5 & -2 & -3 \\ 11 & 2 & 5 \end{vmatrix} = 6(-10 + 6) - 1(25 + 33) + 1(10 + 22) = -24 - 58 + 32 = -50 \] \text{Paso 3. Determinante de } x_2 \[ D_2 = \begin{vmatrix} 2 & 6 & 1 \\ 3 & 5 & -3 \\ 8 & 11 & 5 \end{vmatrix} = 2(25 + 33) - 6(15 + 24) + 1(33 - 40) = 116 - 234 - 7 = -125 \] \text{Paso 4. Determinante de } x_3 \[ D_3 = \begin{vmatrix} 2 & 1 & 6 \\ 3 & -2 & 5 \\ 8 & 2 & 11 \end{vmatrix} = 2(-22 - 10) - 1(33 - 40) + 6(6 + 16) = -64 + 7 + 132 = 75 \] \text{Paso 5. Resultados} \[ x_1 = \frac{-50}{-25} = 2, \quad x_2 = \frac{-125}{-25} = 5, \quad x_3 = \frac{75}{-25} = -3 \] 4. Sistema \[ \begin{cases} -5x_1 + 8x_2 + 10x_3 = -8 \\ x_1 - 7x_2 = -2 \\ 10x_1 + 10x_2 + 6x_3 = 9 \end{cases} \] \text{Paso 1. Determinante principal} \[ D = \begin{vmatrix} -5 & 8 & 10 \\ 1 & -7 & 0 \\ 10 & 10 & 6 \end{vmatrix} = -5(-42) - 8(6) + 10(80) = 210 - 48 + 800 = 962 \] \text{Paso 2. Determinante de } x_1 \[ D_1 = \begin{vmatrix} -8 & 8 & 10 \\ -2 & -7 & 0 \\ 9 & 10 & 6 \end{vmatrix} = -8(-42) - 8(-12) + 10(43) = 336 + 96 + 430 = 862 \] \text{Paso 3. Determinante de } x_2 \[ D_2 = \begin{vmatrix} -5 & -8 & 10 \\ 1 & -2 & 0 \\ 10 & 9 & 6 \end{vmatrix} = -5(-12) - (-8)(6) + 10(29) = 60 + 48 + 290 = 398 \] \text{Paso 4. Determinante de } x_3 \[ D_3 = \begin{vmatrix} -5 & 8 & -8 \\ 1 & -7 & -2 \\ 10 & 10 & 9 \end{vmatrix} = -5(-43) - 8(29) - 8(80) = 215 - 232 - 640 = -657 \] \text{Paso 5. Resultados} \[ x_1 = \frac{862}{962} = \frac{431}{481}, \quad x_2 = \frac{398}{962} = \frac{199}{481}, \quad x_3 = \frac{-657}{962} \] 5. Sistema \[ \begin{cases} 2x_1 + 2x_2 + x_3 = 7 \\ x_1 + 2x_2 + x_3 = 0 \\ -x_1 + x_2 + 3x_3 = 1 \end{cases} \] \text{Paso 1. Determinante principal} \[ D = \begin{vmatrix} 2 & 2 & 1 \\ 1 & 2 & 1 \\ -1 & 1 & 3 \end{vmatrix} = 2(6 - 1) - 2(3 + 1) + 1(1 + 2) = 10 - 8 + 3 = 5 \] \text{Paso 2. Determinante de } x_1 \[ D_1 = \begin{vmatrix} 7 & 2 & 1 \\ 0 & 2 & 1 \\ 1 & 1 & 3 \end{vmatrix} = 7(6 - 1) - 2(0 - 1) + 1(0 - 2) = 35 + 2 - 2 = 35 \] \text{Paso 3. Determinante de } x_2 \[ D_2 = \begin{vmatrix} 2 & 7 & 1 \\ 1 & 0 & 1 \\ -1 & 1 & 3 \end{vmatrix} = 2(0 - 1) - 7(3 + 1) + 1(1) = -2 - 28 + 1 = -29 \] \text{Paso 4. Determinante de } x_3 \[ D_3 = \begin{vmatrix} 2 & 2 & 7 \\ 1 & 2 & 0 \\ -1 & 1 & 1 \end{vmatrix} = 2(2 - 0) - 2(1 - 0) + 7(1 + 2) = 4 - 2 + 21 = 23 \] \text{Paso 5. Resultados} \[ x_1 = \frac{35}{5} = 7, \quad x_2 = \frac{-29}{5}, \quad x_3 = \frac{23}{5} \] 6. Sistema \[ \begin{cases} 2x_1 + 5x_2 - x_3 = -1 \\ 4x_1 + x_2 + 3x_3 = 3 \\ -2x_1 + 2x_2 = 0 \end{cases} \] \text{Paso 1. Determinante principal} \[ D = \begin{vmatrix} 2 & 5 & -1 \\ 4 & 1 & 3 \\ -2 & 2 & 0 \end{vmatrix} = 2(-6) - 5(6) + (-1)(10) = -12 - 30 - 10 = -52 \] \text{Paso 2. Determinante de } x_1 \[ D_1 = \begin{vmatrix} -1 & 5 & -1 \\ 3 & 1 & 3 \\ 0 & 2 & 0 \end{vmatrix} = (-1)(-6) - 5(0) + (-1)(6) = 6 - 6 = 0 \] \text{Paso 3. Determinante de } x_2 \[ D_2 = \begin{vmatrix} 2 & -1 & -1 \\ 4 & 3 & 3 \\ -2 & 0 & 0 \end{vmatrix} = 2(0) - (-1)(6) + (-1)(6) = 6 - 6 = 0 \] \text{Paso 4. Determinante de } x_3 \[ D_3 = \begin{vmatrix} 2 & 5 & -1 \\ 4 & 1 & 3 \\ -2 & 2 & 0 \end{vmatrix} = 2(-6) - 5(6) + (-1)(10) = -12 - 30 - 10 = -52 \] \text{Paso 5. Resultados} \[ x_1 = \frac{0}{-52} = 0, \quad x_2 = \frac{0}{-52} = 0, \quad x_3 = \frac{-52}{-52} = 1 \] 7. Sistema \[ \begin{cases} 6x_1 - 10x_2 + 4x_3 = -2 \\ 10x_1 + 7x_2 + 5x_3 = 3 \\ 3x_1 + 9x_2 + 5x_3 = -7 \end{cases} \] \text{Paso 1. Determinante principal} \[ D = \begin{vmatrix} 6 & -10 & 4 \\ 10 & 7 & 5 \\ 3 & 9 & 5 \end{vmatrix} = 6(35 - 45) + 10(50 - 15) + 4(90 - 21) = -60 + 350 + 276 = 566 \] \text{Paso 2. Determinante de } x_1 \[ D_1 = \begin{vmatrix} -2 & -10 & 4 \\ 3 & 7 & 5 \\ -7 & 9 & 5 \end{vmatrix} = -2(35 - 45) + 10(15 + 35) + 4(27 + 49) = 20 + 500 + 304 = 824 \] \text{Paso 3. Determinante de } x_2 \[ D_2 = \begin{vmatrix} 6 & -2 & 4 \\ 10 & 3 & 5 \\ 3 & -7 & 5 \end{vmatrix} = 6(15 + 35) + 2(50 - 15) + 4(-70 - 9) = 300 + 70 - 316 = 54 \] \text{Paso 4. Determinante de } x_3 \[ D_3 = \begin{vmatrix} 6 & -10 & -2 \\ 10 & 7 & 3 \\ 3 & 9 & -7 \end{vmatrix} = 6(-49 - 27) + 10(-70 - 9) - 2(90 - 21) = -456 - 790 - 138 = -1384 \] \text{Paso 5. Resultados} \[ x_1 = \frac{824}{566} = \frac{412}{283}, \quad x_2 = \frac{54}{566} = \frac{27}{283}, \quad x_3 = \frac{-1384}{566} = \frac{-692}{283} \] 8. Sistema \[ \begin{cases} x_1 + x_2 + x_3 + x_4 = 6 \\ 2x_1 - x_3 - x_4 = 4 \\ 3x_3 + 6x_4 = 3 \\ x_1 - x_4 = 5 \end{cases} \] \text{Paso 1. Determinante principal} \[ D = \begin{vmatrix} 1 & 1 & 1 & 1 \\ 2 & 0 & -1 & -1 \\ 0 & 0 & 3 & 6 \\ 1 & 0 & 0 & -1 \end{vmatrix} = 9 \] \text{Paso 2. Determinante de } x_1 \[ D_1 = \begin{vmatrix} 6 & 1 & 1 & 1 \\ 4 & 0 & -1 & -1 \\ 3 & 0 & 3 & 6 \\ 5 & 0 & 0 & -1 \end{vmatrix} = 30 \] \text{Paso 3. Determinante de } x_2 \[ D_2 = \begin{vmatrix} 1 & 6 & 1 & 1 \\ 2 & 4 & -1 & -1 \\ 0 & 3 & 3 & 6 \\ 1 & 5 & 0 & -1 \end{vmatrix} = 0 \] \text{Paso 4. Determinante de } x_3 \[ D_3 = \begin{vmatrix} 1 & 1 & 6 & 1 \\ 2 & 0 & 4 & -1 \\ 0 & 0 & 3 & 6 \\ 1 & 0 & 5 & -1 \end{vmatrix} = 39 \] \text{Paso 5. Determinante de } x_4 \[ D_4 = \begin{vmatrix} 1 & 1 & 1 & 6 \\ 2 & 0 & -1 & 4 \\ 0 & 0 & 3 & 3 \\ 1 & 0 & 0 & 5 \end{vmatrix} = -15 \] \text{Paso 6. Resultados} \[ x_1 = \frac{30}{9} = \frac{10}{3}, \quad x_2 = \frac{0}{9} = 0, \quad x_3 = \frac{39}{9} = \frac{13}{3}, \quad x_4 = \frac{-15}{9} = -\frac{5}{3}