线性代数示例

x + 2 y - z = 4

$x + 2 y - z = 4$ ,

2 x + y + z = - 2

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

x + 2 y + z = 2

$x + 2 y + z = 2$

解题步骤 1

Write the system as a matrix.

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 2 & 1 & 1 & - 2 1 & 2 & 1 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

$R_{2} = R_{2} - 2 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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解题步骤 2.1.1

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

$R_{2} = R_{2} - 2 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 2 - 2 \cdot 1 & 1 - 2 \cdot 2 & 1 - 2 \cdot - 1 & - 2 - 2 \cdot 4 1 & 2 & 1 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.1.2

化简

R_{2}

$R_{2}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & - 3 & 3 & - 10 1 & 2 & 1 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & - 3 & 3 & - 10 1 & 2 & 1 & 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.2

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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解题步骤 2.2.1

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & - 3 & 3 & - 10 1 - 1 & 2 - 2 & 1 + 1 & 2 - 4 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.2.2

化简

R_{3}

$R_{3}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & - 3 & 3 & - 10 0 & 0 & 2 & - 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & - 3 & 3 & - 10 0 & 0 & 2 & - 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{3}

$- \frac{1}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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解题步骤 2.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{3}

$- \frac{1}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 3 & - \frac{1}{3} \cdot - 10 0 & 0 & 2 & - 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.3.2

化简

R_{2}

$R_{2}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & 1 & - 1 & \frac{10}{3} 0 & 0 & 2 & - 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & 1 & - 1 & \frac{10}{3} 0 & 0 & 2 & - 2 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.4

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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解题步骤 2.4.1

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & 1 & - 1 & \frac{10}{3} \frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{- 2}{2} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

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

解题步骤 2.4.2

化简

R_{3}

$R_{3}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & 1 & - 1 & \frac{10}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & 1 & - 1 & \frac{10}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.5

Perform the row operation

R_{2} = R_{2} + R_{3}

$R_{2} = R_{2} + R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

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解题步骤 2.5.1

Perform the row operation

R_{2} = R_{2} + R_{3}

$R_{2} = R_{2} + R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & \frac{10}{3} - 1 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.5.2

化简

R_{2}

$R_{2}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 & 4 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.6

Perform the row operation

R_{1} = R_{1} + R_{3}

$R_{1} = R_{1} + R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

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解题步骤 2.6.1

Perform the row operation

R_{1} = R_{1} + R_{3}

$R_{1} = R_{1} + R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 + 0 & 2 + 0 & - 1 + 1 \cdot 1 & 4 - 1 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.6.2

化简

R_{1}

$R_{1}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 0 & 3 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 0 & 3 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 2.7

Perform the row operation

R_{1} = R_{1} - 2 R_{2}

$R_{1} = R_{1} - 2 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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解题步骤 2.7.1

Perform the row operation

R_{1} = R_{1} - 2 R_{2}

$R_{1} = R_{1} - 2 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & 3 - 2 (\frac{7}{3}) 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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

解题步骤 2.7.2

化简

R_{1}

$R_{1}$ 。

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & - \frac{5}{3} 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 & - \frac{5}{3} \\ 0 & 1 & 0 & \frac{7}{3} \\ 0 & 0 & 1 & - 1 \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & - \frac{5}{3} 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 & - \frac{5}{3} \\ 0 & 1 & 0 & \frac{7}{3} \\ 0 & 0 & 1 & - 1 \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & - \frac{5}{3} 0 & 1 & 0 & \frac{7}{3} 0 & 0 & 1 & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 & - \frac{5}{3} \\ 0 & 1 & 0 & \frac{7}{3} \\ 0 & 0 & 1 & - 1 \end{array}]$

解题步骤 3

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

x = - \frac{5}{3}

$x = - \frac{5}{3}$

y = \frac{7}{3}

$y = \frac{7}{3}$

z = - 1

$z = - 1$

解题步骤 4

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

(- \frac{5}{3}, \frac{7}{3}, - 1)

$(- \frac{5}{3}, \frac{7}{3}, - 1)$

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线性代数 示例

线性代数示例