初级微积分示例

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

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.

$| \begin{array}{c} - 1 & - 2 \\ - 2 & - 3 \end{array} |$

解题步骤 1.1.4

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

$3 | \begin{array}{c} - 1 & - 2 \\ - 2 & - 3 \end{array} |$

解题步骤 1.1.5

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

$| \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} |$

解题步骤 1.1.6

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

$2 | \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} |$

解题步骤 1.1.7

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

$| \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.1.8

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

$1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.1.9

Add the terms together.

$3 | \begin{array}{c} - 1 & - 2 \\ - 2 & - 3 \end{array} | + 2 | \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} | + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.2

计算 $| \begin{array}{c} - 1 & - 2 \\ - 2 & - 3 \end{array} |$ 。

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解题步骤 1.2.1

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。

$3 (- - 3 - (- 2 \cdot - 2)) + 2 | \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} | + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.2.2

化简行列式。

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解题步骤 1.2.2.1

化简每一项。

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解题步骤 1.2.2.1.1

将 $- 1$ 乘以 $- 3$ 。

$3 (3 - (- 2 \cdot - 2)) + 2 | \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} | + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.2.2.1.2

乘以 $- (- 2 \cdot - 2)$ 。

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解题步骤 1.2.2.1.2.1

将 $- 2$ 乘以 $- 2$ 。

$3 (3 - 1 \cdot 4) + 2 | \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} | + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.2.2.1.2.2

将 $- 1$ 乘以 $4$ 。

$3 (3 - 4) + 2 | \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} | + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.2.2.2

从 $3$ 中减去 $4$ 。

$3 \cdot - 1 + 2 | \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} | + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.3

计算 $| \begin{array}{c} 3 & - 2 \\ 3 & - 3 \end{array} |$ 。

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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。

$3 \cdot - 1 + 2 (3 \cdot - 3 - 3 \cdot - 2) + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

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 $3$ 乘以 $- 3$ 。

$3 \cdot - 1 + 2 (- 9 - 3 \cdot - 2) + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.3.2.1.2

将 $- 3$ 乘以 $- 2$ 。

$3 \cdot - 1 + 2 (- 9 + 6) + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.3.2.2

将 $- 9$ 和 $6$ 相加。

$3 \cdot - 1 + 2 \cdot - 3 + 1 | \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$

解题步骤 1.4

计算 $| \begin{array}{c} 3 & - 1 \\ 3 & - 2 \end{array} |$ 。

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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。

$3 \cdot - 1 + 2 \cdot - 3 + 1 (3 \cdot - 2 - 3 \cdot - 1)$

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 $3$ 乘以 $- 2$ 。

$3 \cdot - 1 + 2 \cdot - 3 + 1 (- 6 - 3 \cdot - 1)$

解题步骤 1.4.2.1.2

将 $- 3$ 乘以 $- 1$ 。

$3 \cdot - 1 + 2 \cdot - 3 + 1 (- 6 + 3)$

解题步骤 1.4.2.2

将 $- 6$ 和 $3$ 相加。

$3 \cdot - 1 + 2 \cdot - 3 + 1 \cdot - 3$

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 $3$ 乘以 $- 1$ 。

$- 3 + 2 \cdot - 3 + 1 \cdot - 3$

解题步骤 1.5.1.2

将 $2$ 乘以 $- 3$ 。

$- 3 - 6 + 1 \cdot - 3$

解题步骤 1.5.1.3

将 $- 3$ 乘以 $1$ 。

$- 3 - 6 - 3$

解题步骤 1.5.2

从 $- 3$ 中减去 $6$ 。

$- 9 - 3$

解题步骤 1.5.3

从 $- 9$ 中减去 $3$ 。

$- 12$

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.

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

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 $\frac{1}{3}$ 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 $\frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

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

解题步骤 4.1.2

化简 $R_{1}$ 。

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

解题步骤 4.2

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

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解题步骤 4.2.1

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

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

解题步骤 4.2.2

化简 $R_{2}$ 。

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

解题步骤 4.3

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

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解题步骤 4.3.1

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

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

解题步骤 4.3.2

化简 $R_{3}$ 。

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

解题步骤 4.4

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ 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 $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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

解题步骤 4.4.2

化简 $R_{3}$ 。

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

解题步骤 4.5

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

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解题步骤 4.5.1

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

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

解题步骤 4.5.2

化简 $R_{2}$ 。

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

解题步骤 4.6

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

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解题步骤 4.6.1

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

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

解题步骤 4.6.2

化简 $R_{1}$ 。

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

解题步骤 4.7

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

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解题步骤 4.7.1

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

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

解题步骤 4.7.2

化简 $R_{1}$ 。

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

解题步骤 5

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

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

初级微积分 示例

初级微积分示例