线性代数示例

$[\begin{array}{c} 19 & 11 & 21 \\ 9 & 0 & 10 \\ 10 & 12 & 11 \end{array}]$

解题步骤 1

Find the determinant.

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解题步骤 1.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $2$ by its cofactor and add.

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解题步骤 1.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 1.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

解题步骤 1.1.3

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

$| \begin{array}{c} 11 & 21 \\ 12 & 11 \end{array} |$

解题步骤 1.1.4

Multiply element $a_{21}$ by its cofactor.

$- 9 | \begin{array}{c} 11 & 21 \\ 12 & 11 \end{array} |$

解题步骤 1.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

$| \begin{array}{c} 19 & 21 \\ 10 & 11 \end{array} |$

解题步骤 1.1.6

Multiply element $a_{22}$ by its cofactor.

$0 | \begin{array}{c} 19 & 21 \\ 10 & 11 \end{array} |$

解题步骤 1.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

$| \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$

解题步骤 1.1.8

Multiply element $a_{23}$ by its cofactor.

$- 10 | \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$

解题步骤 1.1.9

Add the terms together.

$- 9 | \begin{array}{c} 11 & 21 \\ 12 & 11 \end{array} | + 0 | \begin{array}{c} 19 & 21 \\ 10 & 11 \end{array} | - 10 | \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$

解题步骤 1.2

将 $0$ 乘以 $| \begin{array}{c} 19 & 21 \\ 10 & 11 \end{array} |$ 。

$- 9 | \begin{array}{c} 11 & 21 \\ 12 & 11 \end{array} | + 0 - 10 | \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$

解题步骤 1.3

计算 $| \begin{array}{c} 11 & 21 \\ 12 & 11 \end{array} |$ 。

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解题步骤 1.3.1

可以使用公式 $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求 $2 \times 2$ 矩阵的行列式。

$- 9 (11 \cdot 11 - 12 \cdot 21) + 0 - 10 | \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$

解题步骤 1.3.2

化简行列式。

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解题步骤 1.3.2.1

化简每一项。

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解题步骤 1.3.2.1.1

将 $11$ 乘以 $11$ 。

$- 9 (121 - 12 \cdot 21) + 0 - 10 | \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$

解题步骤 1.3.2.1.2

将 $- 12$ 乘以 $21$ 。

$- 9 (121 - 252) + 0 - 10 | \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$

解题步骤 1.3.2.2

从 $121$ 中减去 $252$ 。

$- 9 \cdot - 131 + 0 - 10 | \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$

解题步骤 1.4

计算 $| \begin{array}{c} 19 & 11 \\ 10 & 12 \end{array} |$ 。

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解题步骤 1.4.1

可以使用公式 $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求 $2 \times 2$ 矩阵的行列式。

$- 9 \cdot - 131 + 0 - 10 (19 \cdot 12 - 10 \cdot 11)$

解题步骤 1.4.2

化简行列式。

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解题步骤 1.4.2.1

化简每一项。

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解题步骤 1.4.2.1.1

将 $19$ 乘以 $12$ 。

$- 9 \cdot - 131 + 0 - 10 (228 - 10 \cdot 11)$

解题步骤 1.4.2.1.2

将 $- 10$ 乘以 $11$ 。

$- 9 \cdot - 131 + 0 - 10 (228 - 110)$

解题步骤 1.4.2.2

从 $228$ 中减去 $110$ 。

$- 9 \cdot - 131 + 0 - 10 \cdot 118$

解题步骤 1.5

化简行列式。

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解题步骤 1.5.1

化简每一项。

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解题步骤 1.5.1.1

将 $- 9$ 乘以 $- 131$ 。

$1179 + 0 - 10 \cdot 118$

解题步骤 1.5.1.2

将 $- 10$ 乘以 $118$ 。

$1179 + 0 - 1180$

解题步骤 1.5.2

将 $1179$ 和 $0$ 相加。

$1179 - 1180$

解题步骤 1.5.3

从 $1179$ 中减去 $1180$ 。

$- 1$

解题步骤 2

Since the determinant is non-zero, the inverse exists.

解题步骤 3

Set up a $3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

$[\begin{array}{c} 19 & 11 & 21 & 1 & 0 & 0 \\ 9 & 0 & 10 & 0 & 1 & 0 \\ 10 & 12 & 11 & 0 & 0 & 1 \end{array}]$

解题步骤 4

求行简化阶梯形矩阵。

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解题步骤 4.1

Multiply each element of $R_{1}$ by $\frac{1}{19}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{19}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{19}{19} & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & \frac{0}{19} & \frac{0}{19} \\ 9 & 0 & 10 & 0 & 1 & 0 \\ 10 & 12 & 11 & 0 & 0 & 1 \end{array}]$

解题步骤 4.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 9 & 0 & 10 & 0 & 1 & 0 \\ 10 & 12 & 11 & 0 & 0 & 1 \end{array}]$

解题步骤 4.2

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 4.2.1

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 9 - 9 \cdot 1 & 0 - 9 (\frac{11}{19}) & 10 - 9 (\frac{21}{19}) & 0 - 9 (\frac{1}{19}) & 1 - 9 \cdot 0 & 0 - 9 \cdot 0 \\ 10 & 12 & 11 & 0 & 0 & 1 \end{array}]$

解题步骤 4.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & - \frac{99}{19} & \frac{1}{19} & - \frac{9}{19} & 1 & 0 \\ 10 & 12 & 11 & 0 & 0 & 1 \end{array}]$

解题步骤 4.3

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 4.3.1

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & - \frac{99}{19} & \frac{1}{19} & - \frac{9}{19} & 1 & 0 \\ 10 - 10 \cdot 1 & 12 - 10 (\frac{11}{19}) & 11 - 10 (\frac{21}{19}) & 0 - 10 (\frac{1}{19}) & 0 - 10 \cdot 0 & 1 - 10 \cdot 0 \end{array}]$

解题步骤 4.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & - \frac{99}{19} & \frac{1}{19} & - \frac{9}{19} & 1 & 0 \\ 0 & \frac{118}{19} & - \frac{1}{19} & - \frac{10}{19} & 0 & 1 \end{array}]$

解题步骤 4.4

Multiply each element of $R_{2}$ by $- \frac{19}{99}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 4.4.1

Multiply each element of $R_{2}$ by $- \frac{19}{99}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ - \frac{19}{99} \cdot 0 & - \frac{19}{99} (- \frac{99}{19}) & - \frac{19}{99} \cdot \frac{1}{19} & - \frac{19}{99} (- \frac{9}{19}) & - \frac{19}{99} \cdot 1 & - \frac{19}{99} \cdot 0 \\ 0 & \frac{118}{19} & - \frac{1}{19} & - \frac{10}{19} & 0 & 1 \end{array}]$

解题步骤 4.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & 1 & - \frac{1}{99} & \frac{1}{11} & - \frac{19}{99} & 0 \\ 0 & \frac{118}{19} & - \frac{1}{19} & - \frac{10}{19} & 0 & 1 \end{array}]$

解题步骤 4.5

Perform the row operation $R_{3} = R_{3} - \frac{118}{19} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 4.5.1

Perform the row operation $R_{3} = R_{3} - \frac{118}{19} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & 1 & - \frac{1}{99} & \frac{1}{11} & - \frac{19}{99} & 0 \\ 0 - \frac{118}{19} \cdot 0 & \frac{118}{19} - \frac{118}{19} \cdot 1 & - \frac{1}{19} - \frac{118}{19} (- \frac{1}{99}) & - \frac{10}{19} - \frac{118}{19} \cdot \frac{1}{11} & 0 - \frac{118}{19} (- \frac{19}{99}) & 1 - \frac{118}{19} \cdot 0 \end{array}]$

解题步骤 4.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & 1 & - \frac{1}{99} & \frac{1}{11} & - \frac{19}{99} & 0 \\ 0 & 0 & \frac{1}{99} & - \frac{12}{11} & \frac{118}{99} & 1 \end{array}]$

解题步骤 4.6

Multiply each element of $R_{3}$ by $99$ to make the entry at $3, 3$ a $1$ .

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解题步骤 4.6.1

Multiply each element of $R_{3}$ by $99$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & 1 & - \frac{1}{99} & \frac{1}{11} & - \frac{19}{99} & 0 \\ 99 \cdot 0 & 99 \cdot 0 & 99 (\frac{1}{99}) & 99 (- \frac{12}{11}) & 99 (\frac{118}{99}) & 99 \cdot 1 \end{array}]$

解题步骤 4.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & 1 & - \frac{1}{99} & \frac{1}{11} & - \frac{19}{99} & 0 \\ 0 & 0 & 1 & - 108 & 118 & 99 \end{array}]$

解题步骤 4.7

Perform the row operation $R_{2} = R_{2} + \frac{1}{99} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 4.7.1

Perform the row operation $R_{2} = R_{2} + \frac{1}{99} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 + \frac{1}{99} \cdot 0 & 1 + \frac{1}{99} \cdot 0 & - \frac{1}{99} + \frac{1}{99} \cdot 1 & \frac{1}{11} + \frac{1}{99} \cdot - 108 & - \frac{19}{99} + \frac{1}{99} \cdot 118 & 0 + \frac{1}{99} \cdot 99 \\ 0 & 0 & 1 & - 108 & 118 & 99 \end{array}]$

解题步骤 4.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{11}{19} & \frac{21}{19} & \frac{1}{19} & 0 & 0 \\ 0 & 1 & 0 & - 1 & 1 & 1 \\ 0 & 0 & 1 & - 108 & 118 & 99 \end{array}]$

解题步骤 4.8

Perform the row operation $R_{1} = R_{1} - \frac{21}{19} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 4.8.1

Perform the row operation $R_{1} = R_{1} - \frac{21}{19} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{21}{19} \cdot 0 & \frac{11}{19} - \frac{21}{19} \cdot 0 & \frac{21}{19} - \frac{21}{19} \cdot 1 & \frac{1}{19} - \frac{21}{19} \cdot - 108 & 0 - \frac{21}{19} \cdot 118 & 0 - \frac{21}{19} \cdot 99 \\ 0 & 1 & 0 & - 1 & 1 & 1 \\ 0 & 0 & 1 & - 108 & 118 & 99 \end{array}]$

解题步骤 4.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{11}{19} & 0 & \frac{2269}{19} & - \frac{2478}{19} & - \frac{2079}{19} \\ 0 & 1 & 0 & - 1 & 1 & 1 \\ 0 & 0 & 1 & - 108 & 118 & 99 \end{array}]$

解题步骤 4.9

Perform the row operation $R_{1} = R_{1} - \frac{11}{19} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 4.9.1

Perform the row operation $R_{1} = R_{1} - \frac{11}{19} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{11}{19} \cdot 0 & \frac{11}{19} - \frac{11}{19} \cdot 1 & 0 - \frac{11}{19} \cdot 0 & \frac{2269}{19} - \frac{11}{19} \cdot - 1 & - \frac{2478}{19} - \frac{11}{19} \cdot 1 & - \frac{2079}{19} - \frac{11}{19} \cdot 1 \\ 0 & 1 & 0 & - 1 & 1 & 1 \\ 0 & 0 & 1 & - 108 & 118 & 99 \end{array}]$

解题步骤 4.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 120 & - 131 & - 110 \\ 0 & 1 & 0 & - 1 & 1 & 1 \\ 0 & 0 & 1 & - 108 & 118 & 99 \end{array}]$

解题步骤 5

The right half of the reduced row echelon form is the inverse.

$[\begin{array}{c} 120 & - 131 & - 110 \\ - 1 & 1 & 1 \\ - 108 & 118 & 99 \end{array}]$

线性代数 示例

线性代数示例