线性代数示例

[\begin{matrix} a & b c & d \end{matrix}]

$[\begin{array}{c} a & b \\ c & d \end{array}]$

解题步骤 1

The inverse of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

\frac{1}{a d - b c} [\begin{matrix} d & - b - c & a \end{matrix}]

$\frac{1}{a d - b c} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where

a d - b c

$a d - b c$ is the determinant.

解题步骤 2

可以使用公式

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

2 \times 2

$2 \times 2$ 矩阵的行列式。

a d - c b

$a d - c b$

解题步骤 3

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

解题步骤 4

Substitute the known values into the formula for the inverse.

\frac{1}{a d - c b} [\begin{matrix} d & - b - c & a \end{matrix}]

$\frac{1}{a d - c b} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

解题步骤 5

将

\frac{1}{a d - c b}

$\frac{1}{a d - c b}$ 乘以矩阵中的每一个元素。

[\begin{matrix} \frac{1}{a d - c b} d & \frac{1}{a d - c b} (- b) \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{matrix}]

$[\begin{array}{c} \frac{1}{a d - c b} d & \frac{1}{a d - c b} (- b) \\ \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{array}]$

解题步骤 6

化简矩阵中的每一个元素。

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解题步骤 6.1

组合

\frac{1}{a d - c b}

$\frac{1}{a d - c b}$ 和

d

$d$ 。

[\begin{matrix} \frac{d}{a d - c b} & \frac{1}{a d - c b} (- b) \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{matrix}]

$[\begin{array}{c} \frac{d}{a d - c b} & \frac{1}{a d - c b} (- b) \\ \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{array}]$

解题步骤 6.2

使用乘法的交换性质重写。

[\begin{matrix} \frac{d}{a d - c b} & - \frac{1}{a d - c b} b \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{matrix}]

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{1}{a d - c b} b \\ \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{array}]$

解题步骤 6.3

组合

b

$b$ 和

\frac{1}{a d - c b}

$\frac{1}{a d - c b}$ 。

[\begin{matrix} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{matrix}]

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{array}]$

解题步骤 6.4

使用乘法的交换性质重写。

[\begin{matrix} \frac{d}{a d - c b} & - \frac{b}{a d - c b} - \frac{1}{a d - c b} c & \frac{1}{a d - c b} a \end{matrix}]

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ - \frac{1}{a d - c b} c & \frac{1}{a d - c b} a \end{array}]$

解题步骤 6.5

组合

c

$c$ 和

\frac{1}{a d - c b}

$\frac{1}{a d - c b}$ 。

[\begin{matrix} \frac{d}{a d - c b} & - \frac{b}{a d - c b} - \frac{c}{a d - c b} & \frac{1}{a d - c b} a \end{matrix}]

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ - \frac{c}{a d - c b} & \frac{1}{a d - c b} a \end{array}]$

解题步骤 6.6

组合

\frac{1}{a d - c b}

$\frac{1}{a d - c b}$ 和

a

$a$ 。

[\begin{matrix} \frac{d}{a d - c b} & - \frac{b}{a d - c b} - \frac{c}{a d - c b} & \frac{a}{a d - c b} \end{matrix}]

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ - \frac{c}{a d - c b} & \frac{a}{a d - c b} \end{array}]$

[\begin{matrix} \frac{d}{a d - c b} & - \frac{b}{a d - c b} - \frac{c}{a d - c b} & \frac{a}{a d - c b} \end{matrix}]

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ - \frac{c}{a d - c b} & \frac{a}{a d - c b} \end{array}]$

线性代数 示例

线性代数示例