有限数学示例

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

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 $1$ 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 $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} - 2 & - 1 \\ 1 & 1 \end{array} |$

解题步骤 1.1.4

Multiply element $a_{11}$ by its cofactor.

$0 | \begin{array}{c} - 2 & - 1 \\ 1 & 1 \end{array} |$

解题步骤 1.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 2 & - 1 \\ - 1 & 1 \end{array} |$

解题步骤 1.1.6

Multiply element $a_{12}$ by its cofactor.

$- 1 | \begin{array}{c} 2 & - 1 \\ - 1 & 1 \end{array} |$

解题步骤 1.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.1.8

Multiply element $a_{13}$ by its cofactor.

$- 1 | \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.1.9

Add the terms together.

$0 | \begin{array}{c} - 2 & - 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & - 1 \\ - 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.2

将 $0$ 乘以 $| \begin{array}{c} - 2 & - 1 \\ 1 & 1 \end{array} |$ 。

$0 - 1 | \begin{array}{c} 2 & - 1 \\ - 1 & 1 \end{array} | - 1 | \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.3

计算 $| \begin{array}{c} 2 & - 1 \\ - 1 & 1 \end{array} |$ 。

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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。

$0 - 1 (2 \cdot 1 - - - 1) - 1 | \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$

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 $2$ 乘以 $1$ 。

$0 - 1 (2 - - - 1) - 1 | \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.3.2.1.2

乘以 $- - - 1$ 。

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解题步骤 1.3.2.1.2.1

将 $- 1$ 乘以 $- 1$ 。

$0 - 1 (2 - 1 \cdot 1) - 1 | \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.3.2.1.2.2

将 $- 1$ 乘以 $1$ 。

$0 - 1 (2 - 1) - 1 | \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.3.2.2

从 $2$ 中减去 $1$ 。

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

解题步骤 1.4

计算 $| \begin{array}{c} 2 & - 2 \\ - 1 & 1 \end{array} |$ 。

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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。

$0 - 1 \cdot 1 - 1 (2 \cdot 1 - - - 2)$

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 $2$ 乘以 $1$ 。

$0 - 1 \cdot 1 - 1 (2 - - - 2)$

解题步骤 1.4.2.1.2

乘以 $- - - 2$ 。

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解题步骤 1.4.2.1.2.1

将 $- 1$ 乘以 $- 2$ 。

$0 - 1 \cdot 1 - 1 (2 - 1 \cdot 2)$

解题步骤 1.4.2.1.2.2

将 $- 1$ 乘以 $2$ 。

$0 - 1 \cdot 1 - 1 (2 - 2)$

解题步骤 1.4.2.2

从 $2$ 中减去 $2$ 。

$0 - 1 \cdot 1 - 1 \cdot 0$

解题步骤 1.5

化简行列式。

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解题步骤 1.5.1

将 $- 1$ 乘以 $1$ 。

$0 - 1 - 1 \cdot 0$

解题步骤 1.5.2

从 $0$ 中减去 $1$ 。

$- 1 - 1 \cdot 0$

解题步骤 1.5.3

从 $- 1$ 中减去 $0$ 。

$- 1$

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.

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

解题步骤 4

求行简化阶梯形矩阵。

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解题步骤 4.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

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

解题步骤 4.2

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

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解题步骤 4.2.1

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

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

解题步骤 4.2.2

化简 $R_{1}$ 。

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

解题步骤 4.3

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

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解题步骤 4.3.1

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

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

解题步骤 4.3.2

化简 $R_{3}$ 。

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

解题步骤 4.4

Multiply each element of $R_{3}$ by $2$ to make the entry at $3, 3$ a $1$ .

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解题步骤 4.4.1

Multiply each element of $R_{3}$ by $2$ to make the entry at $3, 3$ a $1$ .

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

解题步骤 4.4.2

化简 $R_{3}$ 。

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

解题步骤 4.5

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

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解题步骤 4.5.1

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

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

解题步骤 4.5.2

化简 $R_{2}$ 。

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

解题步骤 4.6

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

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解题步骤 4.6.1

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

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

解题步骤 4.6.2

化简 $R_{1}$ 。

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

解题步骤 4.7

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

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解题步骤 4.7.1

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

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

解题步骤 4.7.2

化简 $R_{1}$ 。

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

解题步骤 5

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

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

有限数学 示例

有限数学示例