Toán hữu hạn Ví dụ

$w + x + 3 y + z = 2$ , $7 w + 7 x + 21 y + 7 z = 14$ , $4 w + 4 x + 12 y + 4 z = 8$ , $5 w + 5 x + 15 y + 5 z = 10$

Bước 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 3 & 1 & 2 \\ 7 & 7 & 21 & 7 & 14 \\ 4 & 4 & 12 & 4 & 8 \\ 5 & 5 & 15 & 5 & 10 \end{array}]$

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

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Bước 2.1

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

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Bước 2.1.1

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

$[\begin{array}{c} 1 & 1 & 3 & 1 & 2 \\ 7 - 7 \cdot 1 & 7 - 7 \cdot 1 & 21 - 7 \cdot 3 & 7 - 7 \cdot 1 & 14 - 7 \cdot 2 \\ 4 & 4 & 12 & 4 & 8 \\ 5 & 5 & 15 & 5 & 10 \end{array}]$

Bước 2.1.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 3 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 4 & 4 & 12 & 4 & 8 \\ 5 & 5 & 15 & 5 & 10 \end{array}]$

Bước 2.2

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Bước 2.2.1

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 3 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 4 - 4 \cdot 1 & 4 - 4 \cdot 1 & 12 - 4 \cdot 3 & 4 - 4 \cdot 1 & 8 - 4 \cdot 2 \\ 5 & 5 & 15 & 5 & 10 \end{array}]$

Bước 2.2.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 3 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 5 & 5 & 15 & 5 & 10 \end{array}]$

Bước 2.3

Perform the row operation $R_{4} = R_{4} - 5 R_{1}$ to make the entry at $4, 1$ a $0$ .

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Bước 2.3.1

Perform the row operation $R_{4} = R_{4} - 5 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 3 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 5 - 5 \cdot 1 & 5 - 5 \cdot 1 & 15 - 5 \cdot 3 & 5 - 5 \cdot 1 & 10 - 5 \cdot 2 \end{array}]$

Bước 2.3.2

Rút gọn $R_{4}$ .

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

Bước 3

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

$w + x + 3 y + z = 2$

$0 = 0$

Bước 4

The solution is the set of ordered pairs that make the system true.

$(2 - x - 3 y - z, x, y, z)$