Đại số tuyến tính Ví dụ

$[\begin{array}{c} 4 & 3 & - 3 & - 7 \\ - 8 & - 3 & 15 & - 1 \\ 16 & 3 & - 39 & 17 \end{array}]$

Bước 1

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

$[\begin{array}{c} 4 & 3 & - 3 & - 7 & 0 \\ - 8 & - 3 & 15 & - 1 & 0 \\ 16 & 3 & - 39 & 17 & 0 \end{array}]$

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 $\frac{1}{4}$ 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 $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{3}{4} & \frac{- 3}{4} & \frac{- 7}{4} & \frac{0}{4} \\ - 8 & - 3 & 15 & - 1 & 0 \\ 16 & 3 & - 39 & 17 & 0 \end{array}]$

Bước 2.1.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & - \frac{3}{4} & - \frac{7}{4} & 0 \\ - 8 & - 3 & 15 & - 1 & 0 \\ 16 & 3 & - 39 & 17 & 0 \end{array}]$

Bước 2.2

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

Nhấp để xem thêm các bước...

Bước 2.2.1

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & - \frac{3}{4} & - \frac{7}{4} & 0 \\ - 8 + 8 \cdot 1 & - 3 + 8 (\frac{3}{4}) & 15 + 8 (- \frac{3}{4}) & - 1 + 8 (- \frac{7}{4}) & 0 + 8 \cdot 0 \\ 16 & 3 & - 39 & 17 & 0 \end{array}]$

Bước 2.2.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & - \frac{3}{4} & - \frac{7}{4} & 0 \\ 0 & 3 & 9 & - 15 & 0 \\ 16 & 3 & - 39 & 17 & 0 \end{array}]$

Bước 2.3

Perform the row operation $R_{3} = R_{3} - 16 R_{1}$ to make the entry at $3, 1$ a $0$ .

Nhấp để xem thêm các bước...

Bước 2.3.1

Perform the row operation $R_{3} = R_{3} - 16 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & - \frac{3}{4} & - \frac{7}{4} & 0 \\ 0 & 3 & 9 & - 15 & 0 \\ 16 - 16 \cdot 1 & 3 - 16 (\frac{3}{4}) & - 39 - 16 (- \frac{3}{4}) & 17 - 16 (- \frac{7}{4}) & 0 - 16 \cdot 0 \end{array}]$

Bước 2.3.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & - \frac{3}{4} & - \frac{7}{4} & 0 \\ 0 & 3 & 9 & - 15 & 0 \\ 0 & - 9 & - 27 & 45 & 0 \end{array}]$

Bước 2.4

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 2.4.1

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

$[\begin{array}{c} 1 & \frac{3}{4} & - \frac{3}{4} & - \frac{7}{4} & 0 \\ \frac{0}{3} & \frac{3}{3} & \frac{9}{3} & \frac{- 15}{3} & \frac{0}{3} \\ 0 & - 9 & - 27 & 45 & 0 \end{array}]$

Bước 2.4.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & - \frac{3}{4} & - \frac{7}{4} & 0 \\ 0 & 1 & 3 & - 5 & 0 \\ 0 & - 9 & - 27 & 45 & 0 \end{array}]$

Bước 2.5

Perform the row operation $R_{3} = R_{3} + 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

Nhấp để xem thêm các bước...

Bước 2.5.1

Perform the row operation $R_{3} = R_{3} + 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & - \frac{3}{4} & - \frac{7}{4} & 0 \\ 0 & 1 & 3 & - 5 & 0 \\ 0 + 9 \cdot 0 & - 9 + 9 \cdot 1 & - 27 + 9 \cdot 3 & 45 + 9 \cdot - 5 & 0 + 9 \cdot 0 \end{array}]$

Bước 2.5.2

Rút gọn $R_{3}$ .

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

Bước 2.6

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

Nhấp để xem thêm các bước...

Bước 2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Bước 2.6.2

Rút gọn $R_{1}$ .

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

Bước 3

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

$x_{1} - 3 x_{3} + 2 x_{4} = 0$

$x_{2} + 3 x_{3} - 5 x_{4} = 0$

$0 = 0$

Bước 4

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

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

Bước 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{3} [\begin{array}{c} 3 \\ - 3 \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} - 2 \\ 5 \\ 0 \\ 1 \end{array}]$

Bước 6

Write as a solution set.

${x_{3} [\begin{array}{c} 3 \\ - 3 \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} - 2 \\ 5 \\ 0 \\ 1 \end{array}] | x_{3}, x_{4} \in R}$