# రేఖీయ సమీకరణ వ్యవస్థ సాధనం

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

ఇంటరాక్టివ్ వెర్షన్: https://www.calcopenly.com/te/math/system-of-linear-equations-solver
విషయం: గణిత కాలిక్యులేటర్లు

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.

## ఇన్‌పుట్‌లు

- **Equations as rows of numbers**: 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.

## ఫలితాలు

- Solution — ప్రధాన ఫలితం
- Type of system
- x₁ (x)
- x₂ (y)
- x₃ (z)
- x₄
- x₅
- Determinant of the coefficients
- Condition number (∞-norm)

## సూత్రం

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

## పరిష్కరించిన ఉదాహరణలు

### 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**
- తనిఖీ చేసిన మూలం: 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**
- తనిఖీ చేసిన మూలం: 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**
- తనిఖీ చేసిన మూలం: 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**
- తనిఖీ చేసిన మూలం: Row 2 = 2 × row 1; x = y and 2y + z = 6 (hand elimination)

### 4×4 system

- Equations as rows of numbers: 2 1 -1 3 9 / 1 -3 2 1 17 / 3 2 1 -1 -2 / 1 1 1 1 6
- **Solution: x₁ = 1, x₂ = −2, x₃ = 3, x₄ = 4**
- **x₄: 4**
- తనిఖీ చేసిన మూలం: Constants built from x = (1, −2, 3, 4); Python fractions Gauss–Jordan confirms the unique solution

## ప్రశ్నలు

### 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.

### “రేఖీయ సమీకరణ వ్యవస్థ సాధనం” ఎంత కచ్చితమైనది?

కచ్చితత్వం మీ ఇన్‌పుట్‌లు, పద్ధతిలోని ఊహలపై ఆధారపడి ఉంటుంది. దశాంశ గణన 50 సార్థక అంకెలను ఉపయోగిస్తుంది. అంచనాలు, సంఖ్యా పద్ధతులు, మూల డేటా తక్కువ కచ్చితత్వంతో ఉండవచ్చు; ప్రదర్శనలో విలువలను రౌండ్ చేయడం ఈ పరిమితులను తొలగించదు. స్వతంత్ర మూలాల పరిష్కారాలతో తనిఖీ చేసిన ఉదాహరణలు: 5. ఉదాహరణకు, “Wikipedia 3×3 example”ను Wikipedia “Gaussian elimination” example; Python fractions: det = −1, ‖A‖∞ = 6, ‖A⁻¹‖∞ = 10తో తనిఖీ చేశారు.

### ఈ పద్ధతికి మూలం ఏమిటి?

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

## మూలాలు

- [Wikipedia — Gaussian elimination (worked 3×3 example)](https://en.wikipedia.org/wiki/Gaussian_elimination)
- [Wolfram MathWorld — Rouché–Capelli (Kronecker–Capelli) theorem: consistency by rank](https://mathworld.wolfram.com/Rouche-CapelliTheorem.html)