Giải tích sơ cấp Ví dụ

$2 x + y = - 4$ , $- 2 y + 4 z = 0$ , $3 x - 2 z = - 11$

Bước 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 1 & 0 & - 4 \\ 0 & - 2 & 4 & 0 \\ 3 & 0 & - 2 & - 11 \end{array}]$

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 $\frac{1}{2}$ 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 $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{1}{2} & \frac{0}{2} & \frac{- 4}{2} \\ 0 & - 2 & 4 & 0 \\ 3 & 0 & - 2 & - 11 \end{array}]$

Bước 2.1.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - 2 \\ 0 & - 2 & 4 & 0 \\ 3 & 0 & - 2 & - 11 \end{array}]$

Bước 2.2

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

Nhấp để xem thêm các bước...

Bước 2.2.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - 2 \\ 0 & - 2 & 4 & 0 \\ 3 - 3 \cdot 1 & 0 - 3 (\frac{1}{2}) & - 2 - 3 \cdot 0 & - 11 - 3 \cdot - 2 \end{array}]$

Bước 2.2.2

Rút gọn $R_{3}$ .

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

Bước 2.3

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

Nhấp để xem thêm các bước...

Bước 2.3.1

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

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

Bước 2.3.2

Rút gọn $R_{2}$ .

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

Bước 2.4

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

Nhấp để xem thêm các bước...

Bước 2.4.1

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

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

Bước 2.4.2

Rút gọn $R_{3}$ .

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

Bước 2.5

Multiply each element of $R_{3}$ by $- \frac{1}{5}$ to make the entry at $3, 3$ a $1$ .

Nhấp để xem thêm các bước...

Bước 2.5.1

Multiply each element of $R_{3}$ by $- \frac{1}{5}$ to make the entry at $3, 3$ a $1$ .

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

Bước 2.5.2

Rút gọn $R_{3}$ .

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

Bước 2.6

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

Nhấp để xem thêm các bước...

Bước 2.6.1

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

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

Bước 2.6.2

Rút gọn $R_{2}$ .

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

Bước 2.7

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

Nhấp để xem thêm các bước...

Bước 2.7.1

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

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

Bước 2.7.2

Rút gọn $R_{1}$ .

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

Bước 3

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

$x = - 3$

$y = 2$

$z = 1$

Bước 4

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

$(- 3, 2, 1)$