Algebra Beispiele

S ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} a b c \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠ = ⎡ ⎢ ⎣ \begin{matrix} a - 2 b - c 3 a - b + 2 c a + b + 2 c \end{matrix} ⎤ ⎥ ⎦

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

Schritt 1

Der Kern einer Transformation ist ein Vektor, der die Transformation gleich dem Nullvektor (dem Urbild der Transformation) macht.

⎡ ⎢ ⎣ \begin{matrix} a - 2 b - c 3 a - b + 2 c a + b + 2 c \end{matrix} ⎤ ⎥ ⎦ = 0

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

Schritt 2

Erzeuge aus der Vektorgleichung ein Gleichungssystem.

a - 2 b - c = 0

$a - 2 b - c = 0$

3 a - b + 2 c = 0

$3 a - b + 2 c = 0$

a + b + 2 c = 0

$a + b + 2 c = 0$

Schritt 3

Write the system as a matrix.

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 3 & - 1 & 2 & 0 1 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 4

Ermittele die normierte Zeilenstufenform.

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Schritt 4.1

Perform the row operation

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

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

2, 1

$2, 1$ a

0

$0$ .

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Schritt 4.1.1

Perform the row operation

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

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

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 3 - 3 \cdot 1 & - 1 - 3 \cdot - 2 & 2 - 3 \cdot - 1 & 0 - 3 \cdot 0 1 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 4.1.2

Vereinfache

R_{2}

$R_{2}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 5 & 5 & 0 1 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 5 & 5 & 0 1 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 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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Schritt 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{matrix} 1 & - 2 & - 1 & 0 0 & 5 & 5 & 0 1 - 1 & 1 + 2 & 2 + 1 & 0 - 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 4.2.2

Vereinfache

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 5 & 5 & 0 0 & 3 & 3 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 5 & 5 & 0 0 & 3 & 3 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 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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Schritt 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{matrix} 1 & - 2 & - 1 & 0 \frac{0}{5} & \frac{5}{5} & \frac{5}{5} & \frac{0}{5} 0 & 3 & 3 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 4.3.2

Vereinfache

R_{2}

$R_{2}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 1 & 1 & 0 0 & 3 & 3 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 1 & 1 & 0 0 & 3 & 3 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 4.4

Perform the row operation

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

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

3, 2

$3, 2$ a

0

$0$ .

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Schritt 4.4.1

Perform the row operation

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

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

3, 2

$3, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 1 & 1 & 0 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & 3 - 3 \cdot 1 & 0 - 3 \cdot 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 4.4.2

Vereinfache

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 1 & 1 & 0 0 & 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 & 0 0 & 1 & 1 & 0 0 & 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 4.5

Perform the row operation

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

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

1, 2

$1, 2$ a

0

$0$ .

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Schritt 4.5.1

Perform the row operation

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

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

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 1 + 2 \cdot 1 & 0 + 2 \cdot 0 0 & 1 & 1 & 0 0 & 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 4.5.2

Vereinfache

R_{1}

$R_{1}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 0 & 1 & 1 & 0 0 & 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 0 & 1 & 1 & 0 0 & 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 1 & 0 0 & 1 & 1 & 0 0 & 0 & 0 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Schritt 5

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

a + c = 0

$a + c = 0$

b + c = 0

$b + c = 0$

0 = 0

$0 = 0$

Schritt 6

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

⎡ ⎢ ⎣ \begin{matrix} a b c \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} - c - c c \end{matrix} ⎤ ⎥ ⎦

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

Schritt 7

Write the solution as a linear combination of vectors.

⎡ ⎢ ⎣ \begin{matrix} a b c \end{matrix} ⎤ ⎥ ⎦ = c ⎡ ⎢ ⎣ \begin{matrix} - 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦

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

Schritt 8

Write as a solution set.

⎧ ⎪ ⎨ ⎪ ⎩ c ⎡ ⎢ ⎣ \begin{matrix} - 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦ ∣ ∣ ∣ ∣ c \in R ⎫ ⎪ ⎬ ⎪ ⎭

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

Schritt 9

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

⎧ ⎪ ⎨ ⎪ ⎩ ⎡ ⎢ ⎣ \begin{matrix} - 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭

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

Schritt 10

Der Nullraum von

S

$S$ ist der Teilraum

⎧ ⎪ ⎨ ⎪ ⎩ ⎡ ⎢ ⎣ \begin{matrix} - 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭

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

K (S) = ⎧ ⎪ ⎨ ⎪ ⎩ ⎡ ⎢ ⎣ \begin{matrix} - 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭

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

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