代数示例

[\begin{matrix} 7 & 2 & 2 2 & 2 & 7 \end{matrix}]

$[\begin{array}{c} 7 & 2 & 2 \\ 2 & 2 & 7 \end{array}]$

解题步骤 1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{7}

$\frac{1}{7}$ 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}{7}

$\frac{1}{7}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{matrix} \frac{7}{7} & \frac{2}{7} & \frac{2}{7} 2 & 2 & 7 \end{matrix}]

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

解题步骤 1.2

化简

R_{1}

$R_{1}$ 。

[\begin{matrix} 1 & \frac{2}{7} & \frac{2}{7} 2 & 2 & 7 \end{matrix}]

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

[\begin{matrix} 1 & \frac{2}{7} & \frac{2}{7} 2 & 2 & 7 \end{matrix}]

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

解题步骤 2

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

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 & \frac{2}{7} & \frac{2}{7} 2 - 2 \cdot 1 & 2 - 2 (\frac{2}{7}) & 7 - 2 (\frac{2}{7}) \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 2.2

化简

R_{2}

$R_{2}$ 。

[\begin{matrix} 1 & \frac{2}{7} & \frac{2}{7} 0 & \frac{10}{7} & \frac{45}{7} \end{matrix}]

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ 0 & \frac{10}{7} & \frac{45}{7} \end{array}]$

[\begin{matrix} 1 & \frac{2}{7} & \frac{2}{7} 0 & \frac{10}{7} & \frac{45}{7} \end{matrix}]

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ 0 & \frac{10}{7} & \frac{45}{7} \end{array}]$

解题步骤 3

Multiply each element of

R_{2}

$R_{2}$ by

\frac{7}{10}

$\frac{7}{10}$ 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{7}{10}

$\frac{7}{10}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{matrix} 1 & \frac{2}{7} & \frac{2}{7} \frac{7}{10} \cdot 0 & \frac{7}{10} \cdot \frac{10}{7} & \frac{7}{10} \cdot \frac{45}{7} \end{matrix}]

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ \frac{7}{10} \cdot 0 & \frac{7}{10} \cdot \frac{10}{7} & \frac{7}{10} \cdot \frac{45}{7} \end{array}]$

解题步骤 3.2

化简

R_{2}

$R_{2}$ 。

[\begin{matrix} 1 & \frac{2}{7} & \frac{2}{7} 0 & 1 & \frac{9}{2} \end{matrix}]

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ 0 & 1 & \frac{9}{2} \end{array}]$

[\begin{matrix} 1 & \frac{2}{7} & \frac{2}{7} 0 & 1 & \frac{9}{2} \end{matrix}]

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ 0 & 1 & \frac{9}{2} \end{array}]$

解题步骤 4

Perform the row operation

R_{1} = R_{1} - \frac{2}{7} R_{2}

$R_{1} = R_{1} - \frac{2}{7} R_{2}$ 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{2}{7} R_{2}

$R_{1} = R_{1} - \frac{2}{7} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{matrix} 1 - \frac{2}{7} \cdot 0 & \frac{2}{7} - \frac{2}{7} \cdot 1 & \frac{2}{7} - \frac{2}{7} \cdot \frac{9}{2} 0 & 1 & \frac{9}{2} \end{matrix}]

$[\begin{array}{c} 1 - \frac{2}{7} \cdot 0 & \frac{2}{7} - \frac{2}{7} \cdot 1 & \frac{2}{7} - \frac{2}{7} \cdot \frac{9}{2} \\ 0 & 1 & \frac{9}{2} \end{array}]$

解题步骤 4.2

化简

R_{1}

$R_{1}$ 。

[\begin{matrix} 1 & 0 & - 1 0 & 1 & \frac{9}{2} \end{matrix}]

$[\begin{array}{c} 1 & 0 & - 1 \\ 0 & 1 & \frac{9}{2} \end{array}]$

[\begin{matrix} 1 & 0 & - 1 0 & 1 & \frac{9}{2} \end{matrix}]

$[\begin{array}{c} 1 & 0 & - 1 \\ 0 & 1 & \frac{9}{2} \end{array}]$

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