Aljabar Linear Contoh

⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 & 3 & 0 0 & 0 & 1 & - 3 & 0 0 & 0 & 0 & 0 & - 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 4 & 0 & 3 & 0 \\ 0 & 0 & 1 & - 3 & 0 \\ 0 & 0 & 0 & 0 & - 1 \end{array}]$

Langkah 1

Write as an augmented matrix for

A x = 0

$A x = 0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 & 3 & 0 & 0 0 & 0 & 1 & - 3 & 0 & 0 0 & 0 & 0 & 0 & - 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 4 & 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & - 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 1 & 0 \end{array}]$

Langkah 2

Tentukan bentuk eselon baris yang dikurangi.

Ketuk untuk lebih banyak langkah...

Langkah 2.1

Multiply each element of

R_{3}

$R_{3}$ by

- 1

$- 1$ to make the entry at

3, 5

$3, 5$ a

1

$1$ .

Ketuk untuk lebih banyak langkah...

Langkah 2.1.1

Multiply each element of

R_{3}

$R_{3}$ by

- 1

$- 1$ to make the entry at

3, 5

$3, 5$ a

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 & 3 & 0 & 0 0 & 0 & 1 & - 3 & 0 & 0 - 0 & - 0 & - 0 & - 0 & - - 1 & - 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 4 & 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & - 3 & 0 & 0 \\ - 0 & - 0 & - 0 & - 0 & - - 1 & - 0 \end{array}]$

Langkah 2.1.2

Sederhanakan

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 & 3 & 0 & 0 0 & 0 & 1 & - 3 & 0 & 0 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 4 & 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & - 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 & 3 & 0 & 0 0 & 0 & 1 & - 3 & 0 & 0 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 4 & 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & - 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 & 3 & 0 & 0 0 & 0 & 1 & - 3 & 0 & 0 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 4 & 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & - 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

Langkah 3

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

x_{1} + 4 x_{2} + 3 x_{4} = 0

$x_{1} + 4 x_{2} + 3 x_{4} = 0$

x_{3} - 3 x_{4} = 0

$x_{3} - 3 x_{4} = 0$

x_{5} = 0

$x_{5} = 0$

Langkah 4

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

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} x_{1} x_{2} x_{3} x_{4} x_{5} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - 4 x_{2} - 3 x_{4} x_{2} 3 x_{4} x_{4} 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = [\begin{array}{c} - 4 x_{2} - 3 x_{4} \\ x_{2} \\ 3 x_{4} \\ x_{4} \\ 0 \end{array}]$

Langkah 5

Write the solution as a linear combination of vectors.

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} x_{1} x_{2} x_{3} x_{4} x_{5} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = x_{2} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - 4 1000 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ + x_{4} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - 3 0310 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = x_{2} [\begin{array}{c} - 4 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} - 3 \\ 0 \\ 3 \\ 1 \\ 0 \end{array}]$

Langkah 6

Write as a solution set.

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x_{2} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - 4 1000 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ + x_{4} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - 3 0310 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ x_{2}, x_{4} \in R ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

${x_{2} [\begin{array}{c} - 4 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} - 3 \\ 0 \\ 3 \\ 1 \\ 0 \end{array}] | x_{2}, x_{4} \in R}$