الرياضيات المتناهية الأمثلة

$x + y + z = - 2$ , $x - y + 2 z = 4$ , $2 x + y + z = - 7$

خطوة 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 & - 2 \\ 1 & - 1 & 2 & 4 \\ 2 & 1 & 1 & - 7 \end{array}]$

خطوة 2

أوجِد الصيغة الدرجية المختزلة صفيًا.

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خطوة 2.1

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

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خطوة 2.1.1

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

$[\begin{array}{c} 1 & 1 & 1 & - 2 \\ 1 - 1 & - 1 - 1 & 2 - 1 & 4 + 2 \\ 2 & 1 & 1 & - 7 \end{array}]$

خطوة 2.1.2

بسّط $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & - 2 \\ 0 & - 2 & 1 & 6 \\ 2 & 1 & 1 & - 7 \end{array}]$

خطوة 2.2

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

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خطوة 2.2.1

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

$[\begin{array}{c} 1 & 1 & 1 & - 2 \\ 0 & - 2 & 1 & 6 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot 1 & 1 - 2 \cdot 1 & - 7 - 2 \cdot - 2 \end{array}]$

خطوة 2.2.2

بسّط $R_{3}$ .

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

خطوة 2.3

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

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خطوة 2.3.1

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

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

خطوة 2.3.2

بسّط $R_{2}$ .

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

خطوة 2.4

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

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خطوة 2.4.1

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

$[\begin{array}{c} 1 & 1 & 1 & - 2 \\ 0 & 1 & - \frac{1}{2} & - 3 \\ 0 + 0 & - 1 + 1 \cdot 1 & - 1 - \frac{1}{2} & - 3 - 3 \end{array}]$

خطوة 2.4.2

بسّط $R_{3}$ .

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

خطوة 2.5

Multiply each element of $R_{3}$ by $- \frac{2}{3}$ to make the entry at $3, 3$ a $1$ .

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خطوة 2.5.1

Multiply each element of $R_{3}$ by $- \frac{2}{3}$ to make the entry at $3, 3$ a $1$ .

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

خطوة 2.5.2

بسّط $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & - 2 \\ 0 & 1 & - \frac{1}{2} & - 3 \\ 0 & 0 & 1 & 4 \end{array}]$

خطوة 2.6

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

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خطوة 2.6.1

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

$[\begin{array}{c} 1 & 1 & 1 & - 2 \\ 0 + \frac{1}{2} \cdot 0 & 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & - 3 + \frac{1}{2} \cdot 4 \\ 0 & 0 & 1 & 4 \end{array}]$

خطوة 2.6.2

بسّط $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & - 2 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 4 \end{array}]$

خطوة 2.7

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

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خطوة 2.7.1

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

$[\begin{array}{c} 1 - 0 & 1 - 0 & 1 - 1 & - 2 - 4 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 4 \end{array}]$

خطوة 2.7.2

بسّط $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 4 \end{array}]$

خطوة 2.8

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

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خطوة 2.8.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & - 6 + 1 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 4 \end{array}]$

خطوة 2.8.2

بسّط $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 5 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 4 \end{array}]$

خطوة 3

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

$x = - 5$

$y = - 1$

$z = 4$

خطوة 4

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

$(- 5, - 1, 4)$