Đại số tuyến tính Ví dụ

$[\begin{array}{c} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}]$

Bước 1

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

$[\begin{array}{c} 1 & 2 & 3 & 0 \\ 4 & 5 & 6 & 0 \\ 7 & 8 & 9 & 0 \end{array}]$

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

Nhấp để xem thêm các bước...

Bước 2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

Nhấp để xem thêm các bước...

Bước 2.1.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 2 & 3 & 0 \\ 4 - 4 \cdot 1 & 5 - 4 \cdot 2 & 6 - 4 \cdot 3 & 0 - 4 \cdot 0 \\ 7 & 8 & 9 & 0 \end{array}]$

Bước 2.1.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 0 \\ 0 & - 3 & - 6 & 0 \\ 7 & 8 & 9 & 0 \end{array}]$

Bước 2.2

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

Nhấp để xem thêm các bước...

Bước 2.2.1

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

$[\begin{array}{c} 1 & 2 & 3 & 0 \\ 0 & - 3 & - 6 & 0 \\ 7 - 7 \cdot 1 & 8 - 7 \cdot 2 & 9 - 7 \cdot 3 & 0 - 7 \cdot 0 \end{array}]$

Bước 2.2.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 & 0 \\ 0 & - 3 & - 6 & 0 \\ 0 & - 6 & - 12 & 0 \end{array}]$

Bước 2.3

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

$[\begin{array}{c} 1 & 2 & 3 & 0 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 6 & - \frac{1}{3} \cdot 0 \\ 0 & - 6 & - 12 & 0 \end{array}]$

Bước 2.3.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & - 6 & - 12 & 0 \end{array}]$

Bước 2.4

Perform the row operation $R_{3} = R_{3} + 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

Nhấp để xem thêm các bước...

Bước 2.4.1

Perform the row operation $R_{3} = R_{3} + 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 2 & 3 & 0 \\ 0 & 1 & 2 & 0 \\ 0 + 6 \cdot 0 & - 6 + 6 \cdot 1 & - 12 + 6 \cdot 2 & 0 + 6 \cdot 0 \end{array}]$

Bước 2.4.2

Rút gọn $R_{3}$ .

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

Bước 2.5

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

Nhấp để xem thêm các bước...

Bước 2.5.1

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

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

Bước 2.5.2

Rút gọn $R_{1}$ .

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

Bước 3

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

$x - z = 0$

$y + 2 z = 0$

$0 = 0$

Bước 4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} z \\ - 2 z \\ z \end{array}]$

Bước 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} 1 \\ - 2 \\ 1 \end{array}]$

Bước 6

Write as a solution set.

${z [\begin{array}{c} 1 \\ - 2 \\ 1 \end{array}] | z \in R}$