선형 대수 예제

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

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

단계 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{array}{c} d & - b \\ - c & a \end{array}]

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

a d - b c

$a d - b c$ is the determinant.

단계 2

2 \times 2

$2 \times 2$ 행렬의 행렬식은

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

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 공식을 이용해 계산합니다.

a d - c b

$a d - c b$

단계 3

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

단계 4

Substitute the known values into the formula for the inverse.

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

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

단계 5

행렬의 각 원소에

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

$\frac{1}{a d - c b}$ 을 곱합니다.

[\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}]

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

단계 6

행렬의 각 원소를 간단히 합니다.

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단계 6.1

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

$\frac{1}{a d - c b}$ 와

d

$d$ 을 묶습니다.

[\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}]

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

단계 6.2

곱셈의 교환법칙을 사용하여 다시 씁니다.

[\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}]

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

단계 6.3

b

$b$ 와

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

$\frac{1}{a d - c b}$ 을 묶습니다.

[\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}]

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

단계 6.4

곱셈의 교환법칙을 사용하여 다시 씁니다.

[\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}]

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

단계 6.5

c

$c$ 와

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

$\frac{1}{a d - c b}$ 을 묶습니다.

[\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}]

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

단계 6.6

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

$\frac{1}{a d - c b}$ 와

a

$a$ 을 묶습니다.

[\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}]

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

[\begin{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}]

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