Matematika Berhingga Contoh

$2 x + y - 3 z = 12$ , $5 x - 4 y + 7 = 27$ , $10 x + 3 y - z = 40$

Langkah 1

Pindahkan semua suku yang tidak mengandung variabel ke sisi kanan dari persamaan.

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Langkah 1.1

Kurangkan $7$ dari kedua sisi persamaan tersebut.

$2 x + y - 3 z = 12$

$5 x - 4 y = 27 - 7$

$10 x + 3 y - z = 40$

Langkah 1.2

Kurangi $7$ dengan $27$ .

$2 x + y - 3 z = 12$

$5 x - 4 y = 20$

$10 x + 3 y - z = 40$

$2 x + y - 3 z = 12$

$5 x - 4 y = 20$

$10 x + 3 y - z = 40$

Langkah 2

Write the system as a matrix.

$[\begin{array}{c} 2 & 1 & - 3 & 12 \\ 5 & - 4 & 0 & 20 \\ 10 & 3 & - 1 & 40 \end{array}]$

Langkah 3

Tentukan bentuk eselon baris yang dikurangi.

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Langkah 3.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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Langkah 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{1}{2} & \frac{- 3}{2} & \frac{12}{2} \\ 5 & - 4 & 0 & 20 \\ 10 & 3 & - 1 & 40 \end{array}]$

Langkah 3.1.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 5 & - 4 & 0 & 20 \\ 10 & 3 & - 1 & 40 \end{array}]$

Langkah 3.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Langkah 3.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 5 - 5 \cdot 1 & - 4 - 5 (\frac{1}{2}) & 0 - 5 (- \frac{3}{2}) & 20 - 5 \cdot 6 \\ 10 & 3 & - 1 & 40 \end{array}]$

Langkah 3.2.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & - \frac{13}{2} & \frac{15}{2} & - 10 \\ 10 & 3 & - 1 & 40 \end{array}]$

Langkah 3.3

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Langkah 3.3.1

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & - \frac{13}{2} & \frac{15}{2} & - 10 \\ 10 - 10 \cdot 1 & 3 - 10 (\frac{1}{2}) & - 1 - 10 (- \frac{3}{2}) & 40 - 10 \cdot 6 \end{array}]$

Langkah 3.3.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & - \frac{13}{2} & \frac{15}{2} & - 10 \\ 0 & - 2 & 14 & - 20 \end{array}]$

Langkah 3.4

Multiply each element of $R_{2}$ by $- \frac{2}{13}$ to make the entry at $2, 2$ a $1$ .

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Langkah 3.4.1

Multiply each element of $R_{2}$ by $- \frac{2}{13}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ - \frac{2}{13} \cdot 0 & - \frac{2}{13} (- \frac{13}{2}) & - \frac{2}{13} \cdot \frac{15}{2} & - \frac{2}{13} \cdot - 10 \\ 0 & - 2 & 14 & - 20 \end{array}]$

Langkah 3.4.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & 1 & - \frac{15}{13} & \frac{20}{13} \\ 0 & - 2 & 14 & - 20 \end{array}]$

Langkah 3.5

Perform the row operation $R_{3} = R_{3} + 2 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Langkah 3.5.1

Perform the row operation $R_{3} = R_{3} + 2 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & 1 & - \frac{15}{13} & \frac{20}{13} \\ 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 14 + 2 (- \frac{15}{13}) & - 20 + 2 (\frac{20}{13}) \end{array}]$

Langkah 3.5.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & 1 & - \frac{15}{13} & \frac{20}{13} \\ 0 & 0 & \frac{152}{13} & - \frac{220}{13} \end{array}]$

Langkah 3.6

Multiply each element of $R_{3}$ by $\frac{13}{152}$ to make the entry at $3, 3$ a $1$ .

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Langkah 3.6.1

Multiply each element of $R_{3}$ by $\frac{13}{152}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & 1 & - \frac{15}{13} & \frac{20}{13} \\ \frac{13}{152} \cdot 0 & \frac{13}{152} \cdot 0 & \frac{13}{152} \cdot \frac{152}{13} & \frac{13}{152} (- \frac{220}{13}) \end{array}]$

Langkah 3.6.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & 1 & - \frac{15}{13} & \frac{20}{13} \\ 0 & 0 & 1 & - \frac{55}{38} \end{array}]$

Langkah 3.7

Perform the row operation $R_{2} = R_{2} + \frac{15}{13} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Langkah 3.7.1

Perform the row operation $R_{2} = R_{2} + \frac{15}{13} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 + \frac{15}{13} \cdot 0 & 1 + \frac{15}{13} \cdot 0 & - \frac{15}{13} + \frac{15}{13} \cdot 1 & \frac{20}{13} + \frac{15}{13} (- \frac{55}{38}) \\ 0 & 0 & 1 & - \frac{55}{38} \end{array}]$

Langkah 3.7.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{3}{2} & 6 \\ 0 & 1 & 0 & - \frac{5}{38} \\ 0 & 0 & 1 & - \frac{55}{38} \end{array}]$

Langkah 3.8

Perform the row operation $R_{1} = R_{1} + \frac{3}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Langkah 3.8.1

Perform the row operation $R_{1} = R_{1} + \frac{3}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{3}{2} \cdot 0 & \frac{1}{2} + \frac{3}{2} \cdot 0 & - \frac{3}{2} + \frac{3}{2} \cdot 1 & 6 + \frac{3}{2} (- \frac{55}{38}) \\ 0 & 1 & 0 & - \frac{5}{38} \\ 0 & 0 & 1 & - \frac{55}{38} \end{array}]$

Langkah 3.8.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & \frac{291}{76} \\ 0 & 1 & 0 & - \frac{5}{38} \\ 0 & 0 & 1 & - \frac{55}{38} \end{array}]$

Langkah 3.9

Perform the row operation $R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Langkah 3.9.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 0 & \frac{291}{76} - \frac{1}{2} (- \frac{5}{38}) \\ 0 & 1 & 0 & - \frac{5}{38} \\ 0 & 0 & 1 & - \frac{55}{38} \end{array}]$

Langkah 3.9.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{74}{19} \\ 0 & 1 & 0 & - \frac{5}{38} \\ 0 & 0 & 1 & - \frac{55}{38} \end{array}]$

Langkah 4

Use the result matrix to declare the final solution to the system of equations.

$x = \frac{74}{19}$

$y = - \frac{5}{38}$

$z = - \frac{55}{38}$

Langkah 5

The solution is the set of ordered pairs that make the system true.

$(\frac{74}{19}, - \frac{5}{38}, - \frac{55}{38})$