Aljabar Contoh

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}]

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

Langkah 1

Kernel dari transformasi adalah vektor yang membuat transformasinya sama dengan vektor nol (prabayangan dari transformasi).

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

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

Langkah 2

Buat sistem persamaan dari persamaan vektor.

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

$a + b + 2 c = 0$

Langkah 3

Write the system as a matrix.

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

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

Langkah 4

Tentukan bentuk eselon baris yang dikurangi.

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Langkah 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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Langkah 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{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}]

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

Langkah 4.1.2

Sederhanakan

R_{2}

$R_{2}$ .

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

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

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

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

Langkah 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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Langkah 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 & - 2 & - 1 & 0 \\ 0 & 5 & 5 & 0 \\ 1 - 1 & 1 + 2 & 2 + 1 & 0 - 0 \end{array}]

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

Langkah 4.2.2

Sederhanakan

R_{3}

$R_{3}$ .

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

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

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

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

Langkah 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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Langkah 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 & - 2 & - 1 & 0 \\ \frac{0}{5} & \frac{5}{5} & \frac{5}{5} & \frac{0}{5} \\ 0 & 3 & 3 & 0 \end{array}]

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

Langkah 4.3.2

Sederhanakan

R_{2}

$R_{2}$ .

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

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

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

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

Langkah 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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Langkah 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{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}]

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

Langkah 4.4.2

Sederhanakan

R_{3}

$R_{3}$ .

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

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

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

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

Langkah 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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Langkah 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{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}]

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

Langkah 4.5.2

Sederhanakan

R_{1}

$R_{1}$ .

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

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

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

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

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

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

Langkah 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

$0 = 0$

Langkah 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} - c \\ - c \\ c \end{array}]

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

Langkah 7

Write the solution as a linear combination of vectors.

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

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

Langkah 8

Write as a solution set.

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

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

Langkah 9

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

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

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

Langkah 10

Kernel dari

S

$S$ adalah subruang

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

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

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

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

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