Đại số tuyến tính Ví dụ

$[\begin{array}{c} 3 & 7 & 1 & 0 \\ 5 & 2 & 3 & 1 \\ 8 & 5 & 2 & 1 \end{array}]$

Bước 1

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

$[\begin{array}{c} 3 & 7 & 1 & 0 & 0 \\ 5 & 2 & 3 & 1 & 0 \\ 8 & 5 & 2 & 1 & 0 \end{array}]$

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

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Bước 2.1

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

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Bước 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{7}{3} & \frac{1}{3} & \frac{0}{3} & \frac{0}{3} \\ 5 & 2 & 3 & 1 & 0 \\ 8 & 5 & 2 & 1 & 0 \end{array}]$

Bước 2.1.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 5 & 2 & 3 & 1 & 0 \\ 8 & 5 & 2 & 1 & 0 \end{array}]$

Bước 2.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Bước 2.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 5 - 5 \cdot 1 & 2 - 5 (\frac{7}{3}) & 3 - 5 (\frac{1}{3}) & 1 - 5 \cdot 0 & 0 - 5 \cdot 0 \\ 8 & 5 & 2 & 1 & 0 \end{array}]$

Bước 2.2.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & - \frac{29}{3} & \frac{4}{3} & 1 & 0 \\ 8 & 5 & 2 & 1 & 0 \end{array}]$

Bước 2.3

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Bước 2.3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & - \frac{29}{3} & \frac{4}{3} & 1 & 0 \\ 8 - 8 \cdot 1 & 5 - 8 (\frac{7}{3}) & 2 - 8 (\frac{1}{3}) & 1 - 8 \cdot 0 & 0 - 8 \cdot 0 \end{array}]$

Bước 2.3.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & - \frac{29}{3} & \frac{4}{3} & 1 & 0 \\ 0 & - \frac{41}{3} & - \frac{2}{3} & 1 & 0 \end{array}]$

Bước 2.4

Multiply each element of $R_{2}$ by $- \frac{3}{29}$ to make the entry at $2, 2$ a $1$ .

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Bước 2.4.1

Multiply each element of $R_{2}$ by $- \frac{3}{29}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ - \frac{3}{29} \cdot 0 & - \frac{3}{29} (- \frac{29}{3}) & - \frac{3}{29} \cdot \frac{4}{3} & - \frac{3}{29} \cdot 1 & - \frac{3}{29} \cdot 0 \\ 0 & - \frac{41}{3} & - \frac{2}{3} & 1 & 0 \end{array}]$

Bước 2.4.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{4}{29} & - \frac{3}{29} & 0 \\ 0 & - \frac{41}{3} & - \frac{2}{3} & 1 & 0 \end{array}]$

Bước 2.5

Perform the row operation $R_{3} = R_{3} + \frac{41}{3} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Bước 2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{41}{3} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{4}{29} & - \frac{3}{29} & 0 \\ 0 + \frac{41}{3} \cdot 0 & - \frac{41}{3} + \frac{41}{3} \cdot 1 & - \frac{2}{3} + \frac{41}{3} (- \frac{4}{29}) & 1 + \frac{41}{3} (- \frac{3}{29}) & 0 + \frac{41}{3} \cdot 0 \end{array}]$

Bước 2.5.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{4}{29} & - \frac{3}{29} & 0 \\ 0 & 0 & - \frac{74}{29} & - \frac{12}{29} & 0 \end{array}]$

Bước 2.6

Multiply each element of $R_{3}$ by $- \frac{29}{74}$ to make the entry at $3, 3$ a $1$ .

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Bước 2.6.1

Multiply each element of $R_{3}$ by $- \frac{29}{74}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{4}{29} & - \frac{3}{29} & 0 \\ - \frac{29}{74} \cdot 0 & - \frac{29}{74} \cdot 0 & - \frac{29}{74} (- \frac{74}{29}) & - \frac{29}{74} (- \frac{12}{29}) & - \frac{29}{74} \cdot 0 \end{array}]$

Bước 2.6.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - \frac{4}{29} & - \frac{3}{29} & 0 \\ 0 & 0 & 1 & \frac{6}{37} & 0 \end{array}]$

Bước 2.7

Perform the row operation $R_{2} = R_{2} + \frac{4}{29} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Bước 2.7.1

Perform the row operation $R_{2} = R_{2} + \frac{4}{29} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 + \frac{4}{29} \cdot 0 & 1 + \frac{4}{29} \cdot 0 & - \frac{4}{29} + \frac{4}{29} \cdot 1 & - \frac{3}{29} + \frac{4}{29} \cdot \frac{6}{37} & 0 + \frac{4}{29} \cdot 0 \\ 0 & 0 & 1 & \frac{6}{37} & 0 \end{array}]$

Bước 2.7.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 & - \frac{3}{37} & 0 \\ 0 & 0 & 1 & \frac{6}{37} & 0 \end{array}]$

Bước 2.8

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

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Bước 2.8.1

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

$[\begin{array}{c} 1 - \frac{1}{3} \cdot 0 & \frac{7}{3} - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & 0 - \frac{1}{3} \cdot \frac{6}{37} & 0 - \frac{1}{3} \cdot 0 \\ 0 & 1 & 0 & - \frac{3}{37} & 0 \\ 0 & 0 & 1 & \frac{6}{37} & 0 \end{array}]$

Bước 2.8.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & 0 & - \frac{2}{37} & 0 \\ 0 & 1 & 0 & - \frac{3}{37} & 0 \\ 0 & 0 & 1 & \frac{6}{37} & 0 \end{array}]$

Bước 2.9

Perform the row operation $R_{1} = R_{1} - \frac{7}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Bước 2.9.1

Perform the row operation $R_{1} = R_{1} - \frac{7}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{7}{3} \cdot 0 & \frac{7}{3} - \frac{7}{3} \cdot 1 & 0 - \frac{7}{3} \cdot 0 & - \frac{2}{37} - \frac{7}{3} (- \frac{3}{37}) & 0 - \frac{7}{3} \cdot 0 \\ 0 & 1 & 0 & - \frac{3}{37} & 0 \\ 0 & 0 & 1 & \frac{6}{37} & 0 \end{array}]$

Bước 2.9.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{5}{37} & 0 \\ 0 & 1 & 0 & - \frac{3}{37} & 0 \\ 0 & 0 & 1 & \frac{6}{37} & 0 \end{array}]$

Bước 3

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

$x_{1} + \frac{5}{37} x_{4} = 0$

$x_{2} - \frac{3}{37} x_{4} = 0$

$x_{3} + \frac{6}{37} x_{4} = 0$

Bước 4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} - \frac{5 x_{4}}{37} \\ \frac{3 x_{4}}{37} \\ - \frac{6 x_{4}}{37} \\ x_{4} \end{array}]$

Bước 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{4} [\begin{array}{c} - \frac{5}{37} \\ \frac{3}{37} \\ - \frac{6}{37} \\ 1 \end{array}]$

Bước 6

Write as a solution set.

${x_{4} [\begin{array}{c} - \frac{5}{37} \\ \frac{3}{37} \\ - \frac{6}{37} \\ 1 \end{array}] | x_{4} \in R}$