선형 대수 예제

S ([\begin{array}{c} a \\ b \\ c \end{array}]) = [\begin{array}{c} a + 3 b - 6 c \\ 2 a + b + c \\ a + 5 b + c \end{array}]

$S ([\begin{array}{c} a \\ b \\ c \end{array}]) = [\begin{array}{c} a + 3 b - 6 c \\ 2 a + b + c \\ a + 5 b + c \end{array}]$

단계 1

변환의 핵(커널)은 변환 결과 영벡터가 되는 벡터를 말합니다(변환의 원상).

[\begin{array}{c} a + 3 b - 6 c \\ 2 a + b + c \\ a + 5 b + c \end{array}] = 0

$[\begin{array}{c} a + 3 b - 6 c \\ 2 a + b + c \\ a + 5 b + c \end{array}] = 0$

단계 2

벡터 방정식으로부터 연립 방정식을 세웁니다.

a + 3 b - 6 c = 0

$a + 3 b - 6 c = 0$

2 a + b + c = 0

$2 a + b + c = 0$

a + 5 b + c = 0

$a + 5 b + c = 0$

단계 3

Write the system as a matrix.

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 2 & 1 & 1 & 0 \\ 1 & 5 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 2 & 1 & 1 & 0 \\ 1 & 5 & 1 & 0 \end{array}]$

단계 4

기약 행 사다리꼴을 구합니다.

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단계 4.1

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

$R_{2} = R_{2} - 2 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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단계 4.1.1

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

$R_{2} = R_{2} - 2 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot 3 & 1 - 2 \cdot - 6 & 0 - 2 \cdot 0 \\ 1 & 5 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot 3 & 1 - 2 \cdot - 6 & 0 - 2 \cdot 0 \\ 1 & 5 & 1 & 0 \end{array}]$

단계 4.1.2

R_{2}

$R_{2}$ 을 간단히 합니다.

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 1 & 5 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 1 & 5 & 1 & 0 \end{array}]$

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 1 & 5 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 1 & 5 & 1 & 0 \end{array}]$

단계 4.2

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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단계 4.2.1

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 1 - 1 & 5 - 3 & 1 + 6 & 0 - 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 1 - 1 & 5 - 3 & 1 + 6 & 0 - 0 \end{array}]$

단계 4.2.2

R_{3}

$R_{3}$ 을 간단히 합니다.

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 0 & 2 & 7 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 0 & 2 & 7 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

단계 4.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{5}

$- \frac{1}{5}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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단계 4.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{5}

$- \frac{1}{5}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot 13 & - \frac{1}{5} \cdot 0 \\ 0 & 2 & 7 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot 13 & - \frac{1}{5} \cdot 0 \\ 0 & 2 & 7 & 0 \end{array}]$

단계 4.3.2

R_{2}

$R_{2}$ 을 간단히 합니다.

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 2 & 7 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 2 & 7 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

단계 4.4

Perform the row operation

R_{3} = R_{3} - 2 R_{2}

$R_{3} = R_{3} - 2 R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

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단계 4.4.1

Perform the row operation

R_{3} = R_{3} - 2 R_{2}

$R_{3} = R_{3} - 2 R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 - 2 \cdot 0 & 2 - 2 \cdot 1 & 7 - 2 (- \frac{13}{5}) & 0 - 2 \cdot 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 - 2 \cdot 0 & 2 - 2 \cdot 1 & 7 - 2 (- \frac{13}{5}) & 0 - 2 \cdot 0 \end{array}]$

단계 4.4.2

R_{3}

$R_{3}$ 을 간단히 합니다.

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & \frac{61}{5} & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & \frac{61}{5} & 0 \end{array}]$

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & \frac{61}{5} & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & \frac{61}{5} & 0 \end{array}]$

단계 4.5

Multiply each element of

R_{3}

$R_{3}$ by

\frac{5}{61}

$\frac{5}{61}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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단계 4.5.1

Multiply each element of

R_{3}

$R_{3}$ by

\frac{5}{61}

$\frac{5}{61}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ \frac{5}{61} \cdot 0 & \frac{5}{61} \cdot 0 & \frac{5}{61} \cdot \frac{61}{5} & \frac{5}{61} \cdot 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ \frac{5}{61} \cdot 0 & \frac{5}{61} \cdot 0 & \frac{5}{61} \cdot \frac{61}{5} & \frac{5}{61} \cdot 0 \end{array}]$

단계 4.5.2

R_{3}

$R_{3}$ 을 간단히 합니다.

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

단계 4.6

Perform the row operation

R_{2} = R_{2} + \frac{13}{5} R_{3}

$R_{2} = R_{2} + \frac{13}{5} R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

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단계 4.6.1

Perform the row operation

R_{2} = R_{2} + \frac{13}{5} R_{3}

$R_{2} = R_{2} + \frac{13}{5} R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 + \frac{13}{5} \cdot 0 & 1 + \frac{13}{5} \cdot 0 & - \frac{13}{5} + \frac{13}{5} \cdot 1 & 0 + \frac{13}{5} \cdot 0 \\ 0 & 0 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 + \frac{13}{5} \cdot 0 & 1 + \frac{13}{5} \cdot 0 & - \frac{13}{5} + \frac{13}{5} \cdot 1 & 0 + \frac{13}{5} \cdot 0 \\ 0 & 0 & 1 & 0 \end{array}]$

단계 4.6.2

R_{2}

$R_{2}$ 을 간단히 합니다.

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

단계 4.7

Perform the row operation

R_{1} = R_{1} + 6 R_{3}

$R_{1} = R_{1} + 6 R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

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단계 4.7.1

Perform the row operation

R_{1} = R_{1} + 6 R_{3}

$R_{1} = R_{1} + 6 R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

[\begin{array}{c} 1 + 6 \cdot 0 & 3 + 6 \cdot 0 & - 6 + 6 \cdot 1 & 0 + 6 \cdot 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]

$[\begin{array}{c} 1 + 6 \cdot 0 & 3 + 6 \cdot 0 & - 6 + 6 \cdot 1 & 0 + 6 \cdot 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

단계 4.7.2

R_{1}

$R_{1}$ 을 간단히 합니다.

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

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

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

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

단계 4.8

Perform the row operation

R_{1} = R_{1} - 3 R_{2}

$R_{1} = R_{1} - 3 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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단계 4.8.1

Perform the row operation

R_{1} = R_{1} - 3 R_{2}

$R_{1} = R_{1} - 3 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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

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

단계 4.8.2

R_{1}

$R_{1}$ 을 간단히 합니다.

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

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

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

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

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

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

단계 5

Use the result matrix to declare the final solution to the system of equations.

a = 0

$a = 0$

b = 0

$b = 0$

c = 0

$c = 0$

단계 6

Write a solution vector by solving in terms of the free variables in each row.

[\begin{array}{c} a \\ b \\ c \end{array}] = [\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]

$[\begin{array}{c} a \\ b \\ c \end{array}] = [\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]$

단계 7

Write as a solution set.

{[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]}

${[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]}$

단계 8

S

$S$ 의 핵(커널)은 부분공간

{[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]}

${[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]}$ 입니다.

K (S) = {[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]}

$K (S) = {[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]}$