Giải tích sơ cấp Ví dụ

$- x + y + z = - 4$ , $- x + 3 y - 3 z = - 18$ , $7 x - 3 y - 15 z = 0$

Bước 1

Write the system as a matrix.

$[\begin{array}{c} - 1 & 1 & 1 & - 4 \\ - 1 & 3 & - 3 & - 18 \\ 7 & - 3 & - 15 & 0 \end{array}]$

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

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Bước 2.1

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

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Bước 2.1.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 1 & - 1 \cdot 1 & - - 4 \\ - 1 & 3 & - 3 & - 18 \\ 7 & - 3 & - 15 & 0 \end{array}]$

Bước 2.1.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ - 1 & 3 & - 3 & - 18 \\ 7 & - 3 & - 15 & 0 \end{array}]$

Bước 2.2

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

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Bước 2.2.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ - 1 + 1 \cdot 1 & 3 - 1 & - 3 - 1 & - 18 + 1 \cdot 4 \\ 7 & - 3 & - 15 & 0 \end{array}]$

Bước 2.2.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ 0 & 2 & - 4 & - 14 \\ 7 & - 3 & - 15 & 0 \end{array}]$

Bước 2.3

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Bước 2.3.1

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ 0 & 2 & - 4 & - 14 \\ 7 - 7 \cdot 1 & - 3 - 7 \cdot - 1 & - 15 - 7 \cdot - 1 & 0 - 7 \cdot 4 \end{array}]$

Bước 2.3.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ 0 & 2 & - 4 & - 14 \\ 0 & 4 & - 8 & - 28 \end{array}]$

Bước 2.4

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

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Bước 2.4.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 4}{2} & \frac{- 14}{2} \\ 0 & 4 & - 8 & - 28 \end{array}]$

Bước 2.4.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ 0 & 1 & - 2 & - 7 \\ 0 & 4 & - 8 & - 28 \end{array}]$

Bước 2.5

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Bước 2.5.1

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ 0 & 1 & - 2 & - 7 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & - 8 - 4 \cdot - 2 & - 28 - 4 \cdot - 7 \end{array}]$

Bước 2.5.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 4 \\ 0 & 1 & - 2 & - 7 \\ 0 & 0 & 0 & 0 \end{array}]$

Bước 2.6

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

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Bước 2.6.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & - 1 - 2 & 4 - 7 \\ 0 & 1 & - 2 & - 7 \\ 0 & 0 & 0 & 0 \end{array}]$

Bước 2.6.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 3 & - 3 \\ 0 & 1 & - 2 & - 7 \\ 0 & 0 & 0 & 0 \end{array}]$

Bước 3

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

$x - 3 z = - 3$

$y - 2 z = - 7$

$0 = 0$

Bước 4

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

$(- 3 + 3 z, - 7 + 2 z, z)$