Aljabar Linear Contoh

$[\begin{array}{c} 1 & 2 & - 3 & - 1 \\ - 2 & - 4 & 6 & 3 \end{array}]$

Langkah 1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 1 & 2 & - 3 & - 1 & 0 \\ - 2 & - 4 & 6 & 3 & 0 \end{array}]$

Langkah 2

Tentukan bentuk eselon baris yang dikurangi.

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Langkah 2.1

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

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Langkah 2.1.1

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

$[\begin{array}{c} 1 & 2 & - 3 & - 1 & 0 \\ - 2 + 2 \cdot 1 & - 4 + 2 \cdot 2 & 6 + 2 \cdot - 3 & 3 + 2 \cdot - 1 & 0 + 2 \cdot 0 \end{array}]$

Langkah 2.1.2

Sederhanakan $R_{2}$ .

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

Langkah 2.2

Perform the row operation $R_{1} = R_{1} + R_{2}$ to make the entry at $1, 4$ a $0$ .

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Langkah 2.2.1

Perform the row operation $R_{1} = R_{1} + R_{2}$ to make the entry at $1, 4$ a $0$ .

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

Langkah 2.2.2

Sederhanakan $R_{1}$ .

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

Langkah 3

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

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

$x_{4} = 0$

Langkah 4

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

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

Langkah 5

Write the solution as a linear combination of vectors.

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

Langkah 6

Write as a solution set.

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