例

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

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

ステップ 1

変換のカーネルは、変換を0ベクトルに等しくするベクトルです（変換の原像）

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

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

ステップ 2

ベクトル方程式で連立方程式を作成します。

a - b - c = 0

$a - b - c = 0$

a - b - c = 0

$a - b - c = 0$

a - b + c = 0

$a - b + c = 0$

ステップ 3

Write the system as a matrix.

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

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

ステップ 4

縮小行の階段形を求めます。

タップして手順をさらに表示してください…

ステップ 4.1

Perform the row operation

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

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

2, 1

$2, 1$ a

0

$0$ .

タップして手順をさらに表示してください…

ステップ 4.1.1

Perform the row operation

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

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

2, 1

$2, 1$ a

0

$0$ .

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

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

ステップ 4.1.2

R_{2}

$R_{2}$ を簡約します。

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

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

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

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

ステップ 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

$0$ .

タップして手順をさらに表示してください…

ステップ 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 & - 1 & - 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 - 1 & - 1 + 1 & 1 + 1 & 0 - 0 \end{array}]

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

ステップ 4.2.2

R_{3}

$R_{3}$ を簡約します。

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

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

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

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

ステップ 4.3

Swap

R_{3}

$R_{3}$ with

R_{2}

$R_{2}$ to put a nonzero entry at

2, 3

$2, 3$ .

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

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

ステップ 4.4

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

2, 3

$2, 3$ a

1

$1$ .

タップして手順をさらに表示してください…

ステップ 4.4.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

2, 3

$2, 3$ a

1

$1$ .

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

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

ステップ 4.4.2

R_{2}

$R_{2}$ を簡約します。

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

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

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

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

ステップ 4.5

Perform the row operation

R_{1} = R_{1} + R_{2}

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

1, 3

$1, 3$ a

0

$0$ .

タップして手順をさらに表示してください…

ステップ 4.5.1

Perform the row operation

R_{1} = R_{1} + R_{2}

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

1, 3

$1, 3$ a

0

$0$ .

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

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

ステップ 4.5.2

R_{1}

$R_{1}$ を簡約します。

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

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

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

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

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

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

ステップ 5

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

a - b = 0

$a - b = 0$

c = 0

$c = 0$

0 = 0

$0 = 0$

ステップ 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} b \\ b \\ 0 \end{array}]

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

ステップ 7

Write the solution as a linear combination of vectors.

[\begin{array}{c} a \\ b \\ c \end{array}] = b [\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]

$[\begin{array}{c} a \\ b \\ c \end{array}] = b [\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]$

ステップ 8

Write as a solution set.

{b [\begin{array}{c} 1 \\ 1 \\ 0 \end{array}] | b \in R}

${b [\begin{array}{c} 1 \\ 1 \\ 0 \end{array}] | b \in R}$

ステップ 9

The solution is the set of vectors created from the free variables of the system.

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

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

ステップ 10

S

$S$ の核（カーネル）は部分空間

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

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

K (S) = {[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]}

$K (S) = {[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]}$

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