Algebra Beispiele

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

Schritt 1

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

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

Schritt 2

Erzeuge aus der Vektorgleichung ein Gleichungssystem.

$a - b - c = 0$

$a - b + c = 0$

Schritt 3

Write the system as a matrix.

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

Schritt 4

Ermittele die normierte Zeilenstufenform.

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Schritt 4.1

Perform the row operation $R_{2} = R_{2} - R_{1}$ to make the entry at $2, 1$ a $0$ .

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Schritt 4.1.1

Perform the row operation $R_{2} = R_{2} - R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Schritt 4.1.2

Vereinfache $R_{2}$ .

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

Schritt 4.2

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

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Schritt 4.2.1

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

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

Schritt 4.2.2

Vereinfache $R_{3}$ .

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

Schritt 4.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 3$ .

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

Schritt 4.4

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 3$ a $1$ .

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Schritt 4.4.1

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 3$ a $1$ .

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

Schritt 4.4.2

Vereinfache $R_{2}$ .

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

Schritt 4.5

Perform the row operation $R_{1} = R_{1} + R_{2}$ to make the entry at $1, 3$ a $0$ .

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Schritt 4.5.1

Perform the row operation $R_{1} = R_{1} + R_{2}$ to make the entry at $1, 3$ a $0$ .

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

Schritt 4.5.2

Vereinfache $R_{1}$ .

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

Schritt 5

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

$a - b = 0$

$c = 0$

$0 = 0$

Schritt 6

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

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

Schritt 7

Write the solution as a linear combination of vectors.

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

Schritt 8

Write as a solution set.

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

Schritt 9

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

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

Schritt 10

Der Nullraum von $S$ ist der Teilraum ${[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]}$ .

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

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