线性代数示例

$5 x + y + z = 12$ , $- x + y - 4 z = - 20$ , $x + 4 y + 4 z = 10$

解题步骤 1

从方程组中求 $A X = B$ 。

$[\begin{array}{c} 5 & 1 & 1 \\ - 1 & 1 & - 4 \\ 1 & 4 & 4 \end{array}] \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 12 \\ - 20 \\ 10 \end{array}]$

解题步骤 2

求系数矩阵的逆矩阵。

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解题步骤 2.1

Find the determinant.

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解题步骤 2.1.1

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 2.1.1.1

Consider the corresponding sign chart.

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

解题步骤 2.1.1.2

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

解题步骤 2.1.1.3

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

$| \begin{array}{c} 1 & - 4 \\ 4 & 4 \end{array} |$

解题步骤 2.1.1.4

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

$5 | \begin{array}{c} 1 & - 4 \\ 4 & 4 \end{array} |$

解题步骤 2.1.1.5

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

$| \begin{array}{c} - 1 & - 4 \\ 1 & 4 \end{array} |$

解题步骤 2.1.1.6

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

$- 1 | \begin{array}{c} - 1 & - 4 \\ 1 & 4 \end{array} |$

解题步骤 2.1.1.7

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

$| \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.1.8

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

$1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.1.9

Add the terms together.

$5 | \begin{array}{c} 1 & - 4 \\ 4 & 4 \end{array} | - 1 | \begin{array}{c} - 1 & - 4 \\ 1 & 4 \end{array} | + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.2

计算 $| \begin{array}{c} 1 & - 4 \\ 4 & 4 \end{array} |$ 。

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解题步骤 2.1.2.1

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。

$5 (1 \cdot 4 - 4 \cdot - 4) - 1 | \begin{array}{c} - 1 & - 4 \\ 1 & 4 \end{array} | + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.2.2

化简行列式。

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解题步骤 2.1.2.2.1

化简每一项。

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解题步骤 2.1.2.2.1.1

将 $4$ 乘以 $1$ 。

$5 (4 - 4 \cdot - 4) - 1 | \begin{array}{c} - 1 & - 4 \\ 1 & 4 \end{array} | + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.2.2.1.2

将 $- 4$ 乘以 $- 4$ 。

$5 (4 + 16) - 1 | \begin{array}{c} - 1 & - 4 \\ 1 & 4 \end{array} | + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.2.2.2

将 $4$ 和 $16$ 相加。

$5 \cdot 20 - 1 | \begin{array}{c} - 1 & - 4 \\ 1 & 4 \end{array} | + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.3

计算 $| \begin{array}{c} - 1 & - 4 \\ 1 & 4 \end{array} |$ 。

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解题步骤 2.1.3.1

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。

$5 \cdot 20 - 1 (- 1 \cdot 4 - 1 \cdot - 4) + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.3.2

化简行列式。

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解题步骤 2.1.3.2.1

化简每一项。

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解题步骤 2.1.3.2.1.1

将 $- 1$ 乘以 $4$ 。

$5 \cdot 20 - 1 (- 4 - 1 \cdot - 4) + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.3.2.1.2

将 $- 1$ 乘以 $- 4$ 。

$5 \cdot 20 - 1 (- 4 + 4) + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.3.2.2

将 $- 4$ 和 $4$ 相加。

$5 \cdot 20 - 1 \cdot 0 + 1 | \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$

解题步骤 2.1.4

计算 $| \begin{array}{c} - 1 & 1 \\ 1 & 4 \end{array} |$ 。

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解题步骤 2.1.4.1

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。

$5 \cdot 20 - 1 \cdot 0 + 1 (- 1 \cdot 4 - 1 \cdot 1)$

解题步骤 2.1.4.2

化简行列式。

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解题步骤 2.1.4.2.1

化简每一项。

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解题步骤 2.1.4.2.1.1

将 $- 1$ 乘以 $4$ 。

$5 \cdot 20 - 1 \cdot 0 + 1 (- 4 - 1 \cdot 1)$

解题步骤 2.1.4.2.1.2

将 $- 1$ 乘以 $1$ 。

$5 \cdot 20 - 1 \cdot 0 + 1 (- 4 - 1)$

解题步骤 2.1.4.2.2

从 $- 4$ 中减去 $1$ 。

$5 \cdot 20 - 1 \cdot 0 + 1 \cdot - 5$

解题步骤 2.1.5

化简行列式。

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解题步骤 2.1.5.1

化简每一项。

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解题步骤 2.1.5.1.1

将 $5$ 乘以 $20$ 。

$100 - 1 \cdot 0 + 1 \cdot - 5$

解题步骤 2.1.5.1.2

将 $- 1$ 乘以 $0$ 。

$100 + 0 + 1 \cdot - 5$

解题步骤 2.1.5.1.3

将 $- 5$ 乘以 $1$ 。

$100 + 0 - 5$

解题步骤 2.1.5.2

将 $100$ 和 $0$ 相加。

$100 - 5$

解题步骤 2.1.5.3

从 $100$ 中减去 $5$ 。

$95$

解题步骤 2.2

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

解题步骤 2.3

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

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

解题步骤 2.4

求行简化阶梯形矩阵。

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解题步骤 2.4.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 2.4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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

解题步骤 2.4.1.2

化简 $R_{1}$ 。

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

解题步骤 2.4.2

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

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解题步骤 2.4.2.1

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

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ - 1 + 1 \cdot 1 & 1 + \frac{1}{5} & - 4 + \frac{1}{5} & 0 + \frac{1}{5} & 1 + 0 & 0 + 0 \\ 1 & 4 & 4 & 0 & 0 & 1 \end{array}]$

解题步骤 2.4.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & \frac{6}{5} & - \frac{19}{5} & \frac{1}{5} & 1 & 0 \\ 1 & 4 & 4 & 0 & 0 & 1 \end{array}]$

解题步骤 2.4.3

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 2.4.3.1

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

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & \frac{6}{5} & - \frac{19}{5} & \frac{1}{5} & 1 & 0 \\ 1 - 1 & 4 - \frac{1}{5} & 4 - \frac{1}{5} & 0 - \frac{1}{5} & 0 - 0 & 1 - 0 \end{array}]$

解题步骤 2.4.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & \frac{6}{5} & - \frac{19}{5} & \frac{1}{5} & 1 & 0 \\ 0 & \frac{19}{5} & \frac{19}{5} & - \frac{1}{5} & 0 & 1 \end{array}]$

解题步骤 2.4.4

Multiply each element of $R_{2}$ by $\frac{5}{6}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.4.4.1

Multiply each element of $R_{2}$ by $\frac{5}{6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ \frac{5}{6} \cdot 0 & \frac{5}{6} \cdot \frac{6}{5} & \frac{5}{6} (- \frac{19}{5}) & \frac{5}{6} \cdot \frac{1}{5} & \frac{5}{6} \cdot 1 & \frac{5}{6} \cdot 0 \\ 0 & \frac{19}{5} & \frac{19}{5} & - \frac{1}{5} & 0 & 1 \end{array}]$

解题步骤 2.4.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{19}{6} & \frac{1}{6} & \frac{5}{6} & 0 \\ 0 & \frac{19}{5} & \frac{19}{5} & - \frac{1}{5} & 0 & 1 \end{array}]$

解题步骤 2.4.5

Perform the row operation $R_{3} = R_{3} - \frac{19}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.4.5.1

Perform the row operation $R_{3} = R_{3} - \frac{19}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{19}{6} & \frac{1}{6} & \frac{5}{6} & 0 \\ 0 - \frac{19}{5} \cdot 0 & \frac{19}{5} - \frac{19}{5} \cdot 1 & \frac{19}{5} - \frac{19}{5} (- \frac{19}{6}) & - \frac{1}{5} - \frac{19}{5} \cdot \frac{1}{6} & 0 - \frac{19}{5} \cdot \frac{5}{6} & 1 - \frac{19}{5} \cdot 0 \end{array}]$

解题步骤 2.4.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{19}{6} & \frac{1}{6} & \frac{5}{6} & 0 \\ 0 & 0 & \frac{95}{6} & - \frac{5}{6} & - \frac{19}{6} & 1 \end{array}]$

解题步骤 2.4.6

Multiply each element of $R_{3}$ by $\frac{6}{95}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 2.4.6.1

Multiply each element of $R_{3}$ by $\frac{6}{95}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{19}{6} & \frac{1}{6} & \frac{5}{6} & 0 \\ \frac{6}{95} \cdot 0 & \frac{6}{95} \cdot 0 & \frac{6}{95} \cdot \frac{95}{6} & \frac{6}{95} (- \frac{5}{6}) & \frac{6}{95} (- \frac{19}{6}) & \frac{6}{95} \cdot 1 \end{array}]$

解题步骤 2.4.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{19}{6} & \frac{1}{6} & \frac{5}{6} & 0 \\ 0 & 0 & 1 & - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}]$

解题步骤 2.4.7

Perform the row operation $R_{2} = R_{2} + \frac{19}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.4.7.1

Perform the row operation $R_{2} = R_{2} + \frac{19}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 + \frac{19}{6} \cdot 0 & 1 + \frac{19}{6} \cdot 0 & - \frac{19}{6} + \frac{19}{6} \cdot 1 & \frac{1}{6} + \frac{19}{6} (- \frac{1}{19}) & \frac{5}{6} + \frac{19}{6} (- \frac{1}{5}) & 0 + \frac{19}{6} \cdot \frac{6}{95} \\ 0 & 0 & 1 & - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}]$

解题步骤 2.4.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{5} & \frac{1}{5} \\ 0 & 0 & 1 & - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}]$

解题步骤 2.4.8

Perform the row operation $R_{1} = R_{1} - \frac{1}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.4.8.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{5} \cdot 0 & \frac{1}{5} - \frac{1}{5} \cdot 0 & \frac{1}{5} - \frac{1}{5} \cdot 1 & \frac{1}{5} - \frac{1}{5} (- \frac{1}{19}) & 0 - \frac{1}{5} (- \frac{1}{5}) & 0 - \frac{1}{5} \cdot \frac{6}{95} \\ 0 & 1 & 0 & 0 & \frac{1}{5} & \frac{1}{5} \\ 0 & 0 & 1 & - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}]$

解题步骤 2.4.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{5} & 0 & \frac{4}{19} & \frac{1}{25} & - \frac{6}{475} \\ 0 & 1 & 0 & 0 & \frac{1}{5} & \frac{1}{5} \\ 0 & 0 & 1 & - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}]$

解题步骤 2.4.9

Perform the row operation $R_{1} = R_{1} - \frac{1}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.4.9.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{5} \cdot 0 & \frac{1}{5} - \frac{1}{5} \cdot 1 & 0 - \frac{1}{5} \cdot 0 & \frac{4}{19} - \frac{1}{5} \cdot 0 & \frac{1}{25} - \frac{1}{5} \cdot \frac{1}{5} & - \frac{6}{475} - \frac{1}{5} \cdot \frac{1}{5} \\ 0 & 1 & 0 & 0 & \frac{1}{5} & \frac{1}{5} \\ 0 & 0 & 1 & - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}]$

解题步骤 2.4.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & \frac{4}{19} & 0 & - \frac{1}{19} \\ 0 & 1 & 0 & 0 & \frac{1}{5} & \frac{1}{5} \\ 0 & 0 & 1 & - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}]$

解题步骤 2.5

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

$[\begin{array}{c} \frac{4}{19} & 0 & - \frac{1}{19} \\ 0 & \frac{1}{5} & \frac{1}{5} \\ - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}]$

解题步骤 3

对矩阵方程的两边同时左乘逆矩阵。

$([\begin{array}{c} \frac{4}{19} & 0 & - \frac{1}{19} \\ 0 & \frac{1}{5} & \frac{1}{5} \\ - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}] \cdot [\begin{array}{c} 5 & 1 & 1 \\ - 1 & 1 & - 4 \\ 1 & 4 & 4 \end{array}]) \cdot [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{4}{19} & 0 & - \frac{1}{19} \\ 0 & \frac{1}{5} & \frac{1}{5} \\ - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}] \cdot [\begin{array}{c} 12 \\ - 20 \\ 10 \end{array}]$

解题步骤 4

任何矩阵与其逆矩阵的乘积始终等于 $1$ 。 $A \cdot A^{- 1} = 1$ 。

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{4}{19} & 0 & - \frac{1}{19} \\ 0 & \frac{1}{5} & \frac{1}{5} \\ - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}] \cdot [\begin{array}{c} 12 \\ - 20 \\ 10 \end{array}]$

解题步骤 5

乘以 $[\begin{array}{c} \frac{4}{19} & 0 & - \frac{1}{19} \\ 0 & \frac{1}{5} & \frac{1}{5} \\ - \frac{1}{19} & - \frac{1}{5} & \frac{6}{95} \end{array}] [\begin{array}{c} 12 \\ - 20 \\ 10 \end{array}]$ 。

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解题步骤 5.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $3 \times 3$ and the second matrix is $3 \times 1$ .

解题步骤 5.2

将第一个矩阵中的每一行乘以第二个矩阵中的每一列。

$[\begin{array}{c} \frac{4}{19} \cdot 12 + 0 \cdot - 20 - \frac{1}{19} \cdot 10 \\ 0 \cdot 12 + \frac{1}{5} \cdot - 20 + \frac{1}{5} \cdot 10 \\ - \frac{1}{19} \cdot 12 - \frac{1}{5} \cdot - 20 + \frac{6}{95} \cdot 10 \end{array}]$

解题步骤 5.3

通过展开所有表达式化简矩阵的每一个元素。

$[\begin{array}{c} 2 \\ - 2 \\ 4 \end{array}]$

解题步骤 6

化简左右两边。

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 2 \\ - 2 \\ 4 \end{array}]$

解题步骤 7

求解。

$x = 2$

$y = - 2$

$z = 4$

线性代数 示例

线性代数示例