Đại số tuyến tính Ví dụ

$x + y + z = 6$ , $2 x - y + 3 z = 8$ , $x + 2 y = 4$

Bước 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 2 & - 1 & 3 & 8 \\ 1 & 2 & 0 & 4 \end{array}]$

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

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Bước 2.1

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

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Bước 2.1.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 2 - 2 \cdot 1 & - 1 - 2 \cdot 1 & 3 - 2 \cdot 1 & 8 - 2 \cdot 6 \\ 1 & 2 & 0 & 4 \end{array}]$

Bước 2.1.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & - 3 & 1 & - 4 \\ 1 & 2 & 0 & 4 \end{array}]$

Bước 2.2

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

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Bước 2.2.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & - 3 & 1 & - 4 \\ 1 - 1 & 2 - 1 & 0 - 1 & 4 - 6 \end{array}]$

Bước 2.2.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & - 3 & 1 & - 4 \\ 0 & 1 & - 1 & - 2 \end{array}]$

Bước 2.3

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

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Bước 2.3.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 1 & - \frac{1}{3} \cdot - 4 \\ 0 & 1 & - 1 & - 2 \end{array}]$

Bước 2.3.2

Rút gọn $R_{2}$ .

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

Bước 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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Bước 2.4.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & 1 & - \frac{1}{3} & \frac{4}{3} \\ 0 - 0 & 1 - 1 & - 1 + \frac{1}{3} & - 2 - \frac{4}{3} \end{array}]$

Bước 2.4.2

Rút gọn $R_{3}$ .

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

Bước 2.5

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

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Bước 2.5.1

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

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

Bước 2.5.2

Rút gọn $R_{3}$ .

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

Bước 2.6

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

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Bước 2.6.1

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

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 + \frac{1}{3} \cdot 0 & 1 + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & \frac{4}{3} + \frac{1}{3} \cdot 5 \\ 0 & 0 & 1 & 5 \end{array}]$

Bước 2.6.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 6 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 5 \end{array}]$

Bước 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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Bước 2.7.1

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

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

Bước 2.7.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 5 \end{array}]$

Bước 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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Bước 2.8.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 1 - 3 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 5 \end{array}]$

Bước 2.8.2

Rút gọn $R_{1}$ .

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

Bước 3

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

$x = - 2$

$y = 3$

$z = 5$

Bước 4

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

$(- 2, 3, 5)$