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Löser für lineare Gleichungssysteme

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

Aktualisiert Geprüfte Beispiele: 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.
Ausprobieren
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
So wird gerechnet 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

Über Löser für lineare Gleichungssysteme

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.

Durchgerechnete Beispiele

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

Prüfquelle: 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

Prüfquelle: 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

Prüfquelle: 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

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

Fragen

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.

Wie genau arbeitet „Löser für lineare Gleichungssysteme“?

Die Genauigkeit hängt von Ihren Eingaben und den Annahmen der Methode ab. Die Dezimalrechnung nutzt 50 signifikante Stellen, doch Schätzungen, numerische Verfahren und Quelldaten können ungenauer sein. Die angezeigte Rundung beseitigt diese Grenzen nicht. Anhand unabhängiger Quellen geprüfte Rechenbeispiele: 5. Beispielsweise wird „Wikipedia 3×3 example“ anhand von Wikipedia “Gaussian elimination” example; Python fractions: det = −1, ‖A‖∞ = 6, ‖A⁻¹‖∞ = 10 geprüft.

Woher stammt die Methode?

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

Über diesen Rechner

[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}

Quellen

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

Anhand von Quellen geprüft

Dieser Rechner enthält 5 Rechenbeispiele mit Ergebnissen aus unabhängigen Quellen. Sie laufen in der Testsuite und können auch hier ausgeführt werden.

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