线性代数示例

⎡ ⎢ ⎣ \begin{matrix} 2 & 4 & 6 8 & 10 & 12 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

解题步骤 1

Nullity is the dimension of the null space, which is the same as the number of free variables in the system after row reducing. The free variables are the columns without pivot positions.

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

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 2.1.1

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{4}{2} & \frac{6}{2} 8 & 10 & 12 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} \frac{2}{2} & \frac{4}{2} & \frac{6}{2} \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

解题步骤 2.1.2

化简

R_{1}

$R_{1}$ 。

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 8 & 10 & 12 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 8 & 10 & 12 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

解题步骤 2.2

Perform the row operation

R_{2} = R_{2} - 8 R_{1}

$R_{2} = R_{2} - 8 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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解题步骤 2.2.1

Perform the row operation

R_{2} = R_{2} - 8 R_{1}

$R_{2} = R_{2} - 8 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 8 - 8 \cdot 1 & 10 - 8 \cdot 2 & 12 - 8 \cdot 3 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 8 - 8 \cdot 1 & 10 - 8 \cdot 2 & 12 - 8 \cdot 3 \\ 1 & 2 & 0 \end{array}]$

解题步骤 2.2.2

化简

R_{2}

$R_{2}$ 。

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 0 & - 6 & - 12 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ 1 & 2 & 0 \end{array}]$

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 0 & - 6 & - 12 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ 1 & 2 & 0 \end{array}]$

解题步骤 2.3

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 2.3.1

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 & 3 0 & - 6 & - 12 1 - 1 & 2 - 2 & 0 - 3 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ 1 - 1 & 2 - 2 & 0 - 3 \end{array}]$

解题步骤 2.3.2

化简

R_{3}

$R_{3}$ 。

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 0 & - 6 & - 12 0 & 0 & - 3 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ 0 & 0 & - 3 \end{array}]$

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 0 & - 6 & - 12 0 & 0 & - 3 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ 0 & 0 & - 3 \end{array}]$

解题步骤 2.4

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{6}

$- \frac{1}{6}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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解题步骤 2.4.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{6}

$- \frac{1}{6}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot - 12 0 & 0 & - 3 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 2 & 3 \\ - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot - 12 \\ 0 & 0 & - 3 \end{array}]$

解题步骤 2.4.2

化简

R_{2}

$R_{2}$ 。

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

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

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

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

解题步骤 2.5

Multiply each element of

R_{3}

$R_{3}$ by

- \frac{1}{3}

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

3, 3

$3, 3$ a

1

$1$ .

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解题步骤 2.5.1

Multiply each element of

R_{3}

$R_{3}$ by

- \frac{1}{3}

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

3, 3

$3, 3$ a

1

$1$ .

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

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

解题步骤 2.5.2

化简

R_{3}

$R_{3}$ 。

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

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

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

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

解题步骤 2.6

Perform the row operation

R_{2} = R_{2} - 2 R_{3}

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

2, 3

$2, 3$ a

0

$0$ .

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解题步骤 2.6.1

Perform the row operation

R_{2} = R_{2} - 2 R_{3}

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

2, 3

$2, 3$ a

0

$0$ .

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

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

解题步骤 2.6.2

化简

R_{2}

$R_{2}$ 。

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

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

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

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

解题步骤 2.7

Perform the row operation

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

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

1, 3

$1, 3$ a

0

$0$ .

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解题步骤 2.7.1

Perform the row operation

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

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

1, 3

$1, 3$ a

0

$0$ .

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

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

解题步骤 2.7.2

化简

R_{1}

$R_{1}$ 。

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

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

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

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

解题步骤 2.8

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 2.8.1

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 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 2.8.2

化简

R_{1}

$R_{1}$ 。

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 3

The pivot positions are the locations with the leading

1

$1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions:

a_{11}, a_{22},

$a_{11}, a_{22},$ and

a_{33}

$a_{33}$

Pivot Columns:

1, 2,

$1, 2,$ and

3

$3$

解题步骤 4

The nullity is the number of columns without a pivot position in the row reduced matrix.

0

$0$

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线性代数 示例

线性代数示例