Giải tích sơ cấp Ví dụ

$x - z = 2$ , $x + 2 y = 1$ , $y - 3 z = 3$

Bước 1

Write the system as a matrix.

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

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

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Bước 2.1

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

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Bước 2.1.1

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

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

Bước 2.1.2

Rút gọn $R_{2}$ .

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

Bước 2.2

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

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Bước 2.2.1

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

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

Bước 2.2.2

Rút gọn $R_{2}$ .

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

Bước 2.3

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

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Bước 2.3.1

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

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

Bước 2.3.2

Rút gọn $R_{3}$ .

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

Bước 2.4

Multiply each element of $R_{3}$ by $- \frac{2}{7}$ to make the entry at $3, 3$ a $1$ .

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Bước 2.4.1

Multiply each element of $R_{3}$ by $- \frac{2}{7}$ to make the entry at $3, 3$ a $1$ .

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

Bước 2.4.2

Rút gọn $R_{3}$ .

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

Bước 2.5

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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Bước 2.5.1

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

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

Bước 2.5.2

Rút gọn $R_{2}$ .

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

Bước 2.6

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

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Bước 2.6.1

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

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

Bước 2.6.2

Rút gọn $R_{1}$ .

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

Bước 3

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

$x = 1$

$y = 0$

$z = - 1$

Bước 4

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

$(1, 0, - 1)$