Toán hữu hạn Ví dụ

$- 21 x + 6 y + 12 z = - 6$ , $- 7 x + 2 y + 4 z = - 2$

Bước 1

Write the system as a matrix.

$[\begin{array}{c} - 21 & 6 & 12 & - 6 \\ - 7 & 2 & 4 & - 2 \end{array}]$

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

Nhấp để xem thêm các bước...

Bước 2.1

Multiply each element of $R_{1}$ by $- \frac{1}{21}$ to make the entry at $1, 1$ a $1$ .

Nhấp để xem thêm các bước...

Bước 2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{21}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{21} \cdot - 21 & - \frac{1}{21} \cdot 6 & - \frac{1}{21} \cdot 12 & - \frac{1}{21} \cdot - 6 \\ - 7 & 2 & 4 & - 2 \end{array}]$

Bước 2.1.2

Rút gọn $R_{1}$ .

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

Bước 2.2

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

Nhấp để xem thêm các bước...

Bước 2.2.1

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

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

Bước 2.2.2

Rút gọn $R_{2}$ .

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

Bước 3

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

$x - \frac{2}{7} y - \frac{4}{7} z = \frac{2}{7}$

$0 = 0$

Bước 4

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

$(\frac{2}{7} + \frac{2 y}{7} + \frac{4 z}{7}, y, z)$