有限数学示例

[\begin{matrix} 2 & 3 & 3 & - 1 & - 4 & 2 & 5 & 0 & 1 & 4 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{matrix}]

$[\begin{array}{c} 2 & 3 & 3 & - 1 & - 4 & 2 & 5 & 0 & 1 & 4 \\ 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{array}]$

解题步骤 1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

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

1, 1

$1, 1$ a

1

$1$ .

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解题步骤 1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

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

1, 1

$1, 1$ a

1

$1$ .

[\begin{matrix} \frac{2}{2} & \frac{3}{2} & \frac{3}{2} & \frac{- 1}{2} & \frac{- 4}{2} & \frac{2}{2} & \frac{5}{2} & \frac{0}{2} & \frac{1}{2} & \frac{4}{2} 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{matrix}]

$[\begin{array}{c} \frac{2}{2} & \frac{3}{2} & \frac{3}{2} & \frac{- 1}{2} & \frac{- 4}{2} & \frac{2}{2} & \frac{5}{2} & \frac{0}{2} & \frac{1}{2} & \frac{4}{2} \\ 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{array}]$

解题步骤 1.2

化简

R_{1}

$R_{1}$ 。

[\begin{matrix} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{matrix}]

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{array}]$

[\begin{matrix} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{matrix}]

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{array}]$

解题步骤 2

Perform the row operation

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

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

2, 1

$2, 1$ a

0

$0$ .

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解题步骤 2.1

Perform the row operation

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

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

2, 1

$2, 1$ a

0

$0$ .

[\begin{matrix} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 1 - 1 & 5 - \frac{3}{2} & 2 - \frac{3}{2} & 0 + \frac{1}{2} & 0 + 2 & 0 - 1 & 1 - \frac{5}{2} & - 2 - 0 & - 3 - \frac{1}{2} & - 4 - 2 \end{matrix}]

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ 1 - 1 & 5 - \frac{3}{2} & 2 - \frac{3}{2} & 0 + \frac{1}{2} & 0 + 2 & 0 - 1 & 1 - \frac{5}{2} & - 2 - 0 & - 3 - \frac{1}{2} & - 4 - 2 \end{array}]$

解题步骤 2.2

化简

R_{2}

$R_{2}$ 。

[\begin{matrix} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 0 & \frac{7}{2} & \frac{1}{2} & \frac{1}{2} & 2 & - 1 & - \frac{3}{2} & - 2 & - \frac{7}{2} & - 6 \end{matrix}]

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ 0 & \frac{7}{2} & \frac{1}{2} & \frac{1}{2} & 2 & - 1 & - \frac{3}{2} & - 2 & - \frac{7}{2} & - 6 \end{array}]$

[\begin{matrix} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 0 & \frac{7}{2} & \frac{1}{2} & \frac{1}{2} & 2 & - 1 & - \frac{3}{2} & - 2 & - \frac{7}{2} & - 6 \end{matrix}]

解题步骤 3

Multiply each element of

R_{2}

$R_{2}$ by

\frac{2}{7}

$\frac{2}{7}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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解题步骤 3.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{2}{7}

$\frac{2}{7}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \frac{2}{7} \cdot 0 & \frac{2}{7} \cdot \frac{7}{2} & \frac{2}{7} \cdot \frac{1}{2} & \frac{2}{7} \cdot \frac{1}{2} & \frac{2}{7} \cdot 2 & \frac{2}{7} \cdot - 1 & \frac{2}{7} (- \frac{3}{2}) & \frac{2}{7} \cdot - 2 & \frac{2}{7} (- \frac{7}{2}) & \frac{2}{7} \cdot - 6 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ \frac{2}{7} \cdot 0 & \frac{2}{7} \cdot \frac{7}{2} & \frac{2}{7} \cdot \frac{1}{2} & \frac{2}{7} \cdot \frac{1}{2} & \frac{2}{7} \cdot 2 & \frac{2}{7} \cdot - 1 & \frac{2}{7} (- \frac{3}{2}) & \frac{2}{7} \cdot - 2 & \frac{2}{7} (- \frac{7}{2}) & \frac{2}{7} \cdot - 6 \end{array}]$

解题步骤 3.2

化简

R_{2}

$R_{2}$ 。

[\begin{matrix} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{matrix}]

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{array}]$

[\begin{matrix} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{matrix}]

解题步骤 4

Perform the row operation

R_{1} = R_{1} - \frac{3}{2} R_{2}

$R_{1} = R_{1} - \frac{3}{2} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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解题步骤 4.1

Perform the row operation

R_{1} = R_{1} - \frac{3}{2} R_{2}

$R_{1} = R_{1} - \frac{3}{2} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & \frac{3}{2} - \frac{3}{2} \cdot \frac{1}{7} & - \frac{1}{2} - \frac{3}{2} \cdot \frac{1}{7} & - 2 - \frac{3}{2} \cdot \frac{4}{7} & 1 - \frac{3}{2} (- \frac{2}{7}) & \frac{5}{2} - \frac{3}{2} (- \frac{3}{7}) & 0 - \frac{3}{2} (- \frac{4}{7}) & \frac{1}{2} - \frac{3}{2} \cdot - 1 & 2 - \frac{3}{2} (- \frac{12}{7}) 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & \frac{3}{2} - \frac{3}{2} \cdot \frac{1}{7} & - \frac{1}{2} - \frac{3}{2} \cdot \frac{1}{7} & - 2 - \frac{3}{2} \cdot \frac{4}{7} & 1 - \frac{3}{2} (- \frac{2}{7}) & \frac{5}{2} - \frac{3}{2} (- \frac{3}{7}) & 0 - \frac{3}{2} (- \frac{4}{7}) & \frac{1}{2} - \frac{3}{2} \cdot - 1 & 2 - \frac{3}{2} (- \frac{12}{7}) \\ 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{array}]$

解题步骤 4.2

化简

R_{1}

$R_{1}$ 。

[\begin{matrix} 1 & 0 & \frac{9}{7} & - \frac{5}{7} & - \frac{20}{7} & \frac{10}{7} & \frac{22}{7} & \frac{6}{7} & 2 & \frac{32}{7} 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{matrix}]

$[\begin{array}{c} 1 & 0 & \frac{9}{7} & - \frac{5}{7} & - \frac{20}{7} & \frac{10}{7} & \frac{22}{7} & \frac{6}{7} & 2 & \frac{32}{7} \\ 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{array}]$

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有限数学 示例

有限数学示例