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Công cụ giải hệ phương trình tuyến tính

Solve 2 to 5 simultaneous linear equations exactly by Gauss–Jordan elimination, with every row operation shown, and detect no-solution or infinite cases.

Cập nhật Ví dụ đã kiểm tra: 5

One equation per line: the coefficients, then the constant. 2 1 -1 8 means 2x + y − z = 8. A | or = before the constant is fine; fractions like 1/2 work.
Thử
Solution
x = 2, y = 3, z = −1
Solution: x = 2, y = 3, z = −1
Type of system
Unique solution
x₁ (x)
2
x₂ (y)
3
x₃ (z)
−1
Determinant of the coefficients
−1
Condition number (∞-norm)
60

The 3 equations meet at exactly one point: x = 2, y = 3, z = −1. The determinant is −1, which is non-zero, so the solution is unique.

Solution values

x2y3z−1
Cách tính S

  1. Augmented matrix

    [A∣b]=[21−18−3−12−11−212−3]\left[A \mid \mathbf b\right] = \left[\begin{array}{ccc|c}2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3\end{array}\right]

    Unknowns: x, y, z.

  2. Column 1: make the pivot 1 and clear the rest of the column

    R1←12 R1, R2←R2+3R1, R3←R3+2R1  ⟶  [112−1240121210215]R_{1} \leftarrow \frac{1}{2}\,R_{1},\ R_{2} \leftarrow R_{2} + 3R_{1},\ R_{3} \leftarrow R_{3} + 2R_{1} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 2 & 1 & 5\end{array}\right]

  3. Column 2: make the pivot 1 and clear the rest of the column

    R2←2 R2, R1←R1−12R2, R3←R3−2R2  ⟶  [10−13011200−11]R_{2} \leftarrow 2\,R_{2},\ R_{1} \leftarrow R_{1} - \frac{1}{2}R_{2},\ R_{3} \leftarrow R_{3} - 2R_{2} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -1 & 1\end{array}\right]

  4. Column 3: make the pivot 1 and clear the rest of the column

    R3←−R3, R1←R1+R3, R2←R2−R3  ⟶  [10020103001−1]R_{3} \leftarrow -R_{3},\ R_{1} \leftarrow R_{1} + R_{3},\ R_{2} \leftarrow R_{2} - R_{3} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1\end{array}\right]

  5. Read off the solution

    x=2,y=3,z=−1x = 2,\quad y = 3,\quad z = -1

Giới thiệu Công cụ giải hệ phương trình tuyến tính

A system of linear equations is written as an augmented matrix [A | b], one row per equation. Gauss–Jordan elimination scales each pivot row so the pivot is 1 and subtracts multiples of it from every other row until the coefficients form the identity matrix, leaving the solution in the last column. The arithmetic uses exact fractions, so 2x + 3y = 7, 4x − y = 1 gives x = 5/7 and y = 13/7 rather than rounded decimals.

Simultaneous equations come up in circuit analysis, mixture and pricing problems, and algebra courses. The default is the 3 × 3 example from Wikipedia's Gaussian elimination article, 2x + y − z = 8, −3x − y + 2z = −11, −2x + y + 2z = −3, with the unique solution x = 2, y = 3, z = −1.

By the Rouché–Capelli theorem the system has one solution when the rank of A equals the number of unknowns, infinitely many when A and [A | b] share a smaller rank, and none when their ranks differ.

Ví dụ có lời giải

Wikipedia 3×3 example

Equations as rows of numbers
2 1 -1 8 -3 -1 2 -11 -2 1 2 -3
Solution
x = 2, y = 3, z = −1
x₁ (x)
2
x₂ (y)
3
x₃ (z)
-1
Determinant of the coefficients
-1
Condition number (∞-norm)
60

Nguồn đối chiếu: Wikipedia “Gaussian elimination” example; Python fractions: det = −1, ‖A‖∞ = 6, ‖A⁻¹‖∞ = 10

Fractional answer: 2x + 3y = 7, 4x − y = 1

Equations as rows of numbers
2 3 7 4 -1 1
Solution
x = 5/7, y = 13/7
x₁ (x)
0.7142857143
Determinant of the coefficients
-14

Nguồn đối chiếu: Python fractions: x = 10/14, y = 4x − 1 = 13/7

Inconsistent: x + y = 1, 2x + 2y = 3

Equations as rows of numbers
1 1 1 2 2 3
Type of system
No solution — the equations are inconsistent
Solution
No solution

Nguồn đối chiếu: 2 × (first) gives 2x + 2y = 2 ≠ 3

Dependent equations (edge case)

Equations as rows of numbers
1 1 1 6 2 2 2 12 1 -1 0 0
Type of system
Infinitely many solutions
Solution
x = 3 − t/2, y = 3 − t/2, z = t
Determinant of the coefficients
0

Nguồn đối chiếu: Row 2 = 2 × row 1; x = y and 2y + z = 6 (hand elimination)

Câu hỏi

How do you solve a system of three equations with three unknowns?

Eliminate one variable at a time. Gauss–Jordan elimination writes the equations as a matrix, turns the first coefficient into 1, subtracts multiples of that row to clear x from the other equations, then repeats for y and z. For the default system three rounds of row operations leave x = 2, y = 3, z = −1; substituting into 2x + y − z = 8 gives 4 + 3 + 1 = 8.

How can you tell if a system has no solution or infinitely many?

Row-reduce it and look at rows whose coefficients all become 0. A row reading 0 = 1, or 0 equal to any non-zero number, means no solution: x + y = 1 and 2x + 2y = 3 are parallel lines. A row reading 0 = 0 means one equation repeats another, leaving a free variable and infinitely many solutions. With as many equations as unknowns, a determinant of 0 signals one of these two cases.

What is Cramer's rule?

Cramer's rule gives each unknown as a ratio of determinants: x = det(Aₓ)/det(A), where Aₓ is A with its x column replaced by the constants. For 2x + 3y = 7, 4x − y = 1, det(A) = 2 × (−1) − 3 × 4 = −14 and det(Aₓ) = 7 × (−1) − 3 × 1 = −10, so x = 10/14 = 5/7. It needs det(A) ≠ 0 and much more arithmetic than elimination beyond 3 × 3.

What is the difference between Gaussian and Gauss–Jordan elimination?

Gaussian elimination stops at row echelon form, a triangular matrix, and finds the unknowns by back-substitution from the last equation upward. Gauss–Jordan continues until the coefficients form the identity matrix, so each row reads off one unknown directly. Gauss–Jordan takes about n³ arithmetic operations for n equations against about 2n³/3, in exchange for skipping the substitution step.

What does the condition number of a system mean?

It bounds how much a small relative change in the coefficients or constants can grow in the solution. The ∞-norm version is κ = ‖A‖∞ × ‖A⁻¹‖∞; for the default system ‖A‖∞ = 6 and ‖A⁻¹‖∞ = 10, so κ = 60. As a rule of thumb about log₁₀ κ significant digits are lost, here about 2, and a value near 10¹² means the equations are nearly dependent.

“Công cụ giải hệ phương trình tuyến tính” chính xác đến mức nào?

Độ chính xác phụ thuộc vào dữ liệu nhập và giả định của phương pháp. Phép tính thập phân dùng 50 chữ số có nghĩa, nhưng ước lượng, phương pháp số và dữ liệu nguồn có thể kém chính xác hơn; làm tròn khi hiển thị không loại bỏ các giới hạn đó. Ví dụ có lời giải đã đối chiếu với nguồn độc lập: 5. Ví dụ, “Wikipedia 3×3 example” được kiểm tra bằng Wikipedia “Gaussian elimination” example; Python fractions: det = −1, ‖A‖∞ = 6, ‖A⁻¹‖∞ = 10.

Phương pháp này lấy từ đâu?

Wikipedia — Gaussian elimination (worked 3×3 example); Wolfram MathWorld — Rouché–Capelli (Kronecker–Capelli) theorem: consistency by rank.

Về công cụ tính này

[A∣b]→row operations[I∣x]κ∞(A)=∥A∥∞ ∥A−1∥∞\begin{gathered} [A \mid \mathbf b] \xrightarrow{\text{row operations}} [I \mid \mathbf x] \\[10pt] \kappa_\infty(A) = \|A\|_\infty \, \|A^{-1}\|_\infty \end{gathered}

Nguồn

  1. Wikipedia — Gaussian elimination (worked 3×3 example)
  2. Wolfram MathWorld — Rouché–Capelli (Kronecker–Capelli) theorem: consistency by rank

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