Đại số tuyến tính Ví dụ

$[\begin{array}{c} 1 & 6 & 5 & - 7 \\ 0 & 1 & 2 & - 1 \\ 0 & 0 & 1 & 2 \end{array}]$

Bước 1

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

$[\begin{array}{c} 1 & 6 & 5 & - 7 & 0 \\ 0 & 1 & 2 & - 1 & 0 \\ 0 & 0 & 1 & 2 & 0 \end{array}]$

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

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Bước 2.1

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

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Bước 2.1.1

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

$[\begin{array}{c} 1 & 6 & 5 & - 7 & 0 \\ 0 - 2 \cdot 0 & 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & - 1 - 2 \cdot 2 & 0 - 2 \cdot 0 \\ 0 & 0 & 1 & 2 & 0 \end{array}]$

Bước 2.1.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & 6 & 5 & - 7 & 0 \\ 0 & 1 & 0 & - 5 & 0 \\ 0 & 0 & 1 & 2 & 0 \end{array}]$

Bước 2.2

Perform the row operation $R_{1} = R_{1} - 5 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Bước 2.2.1

Perform the row operation $R_{1} = R_{1} - 5 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 5 \cdot 0 & 6 - 5 \cdot 0 & 5 - 5 \cdot 1 & - 7 - 5 \cdot 2 & 0 - 5 \cdot 0 \\ 0 & 1 & 0 & - 5 & 0 \\ 0 & 0 & 1 & 2 & 0 \end{array}]$

Bước 2.2.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & 6 & 0 & - 17 & 0 \\ 0 & 1 & 0 & - 5 & 0 \\ 0 & 0 & 1 & 2 & 0 \end{array}]$

Bước 2.3

Perform the row operation $R_{1} = R_{1} - 6 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Bước 2.3.1

Perform the row operation $R_{1} = R_{1} - 6 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 6 \cdot 0 & 6 - 6 \cdot 1 & 0 - 6 \cdot 0 & - 17 - 6 \cdot - 5 & 0 - 6 \cdot 0 \\ 0 & 1 & 0 & - 5 & 0 \\ 0 & 0 & 1 & 2 & 0 \end{array}]$

Bước 2.3.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 13 & 0 \\ 0 & 1 & 0 & - 5 & 0 \\ 0 & 0 & 1 & 2 & 0 \end{array}]$

Bước 3

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

$x_{1} + 13 x_{4} = 0$

$x_{2} - 5 x_{4} = 0$

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

Bước 4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} - 13 x_{4} \\ 5 x_{4} \\ - 2 x_{4} \\ x_{4} \end{array}]$

Bước 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{4} [\begin{array}{c} - 13 \\ 5 \\ - 2 \\ 1 \end{array}]$

Bước 6

Write as a solution set.

${x_{4} [\begin{array}{c} - 13 \\ 5 \\ - 2 \\ 1 \end{array}] | x_{4} \in R}$