ರೇಖೀಯ ಸಮೀಕರಣಗಳ ವ್ಯವಸ್ಥೆಯ ಪರಿಹಾರಕ

Q: “ರೇಖೀಯ ಸಮೀಕರಣಗಳ ವ್ಯವಸ್ಥೆಯ ಪರಿಹಾರಕ” ಎಷ್ಟು ನಿಖರವಾಗಿದೆ?

ನಿಖರತೆ ನಿಮ್ಮ ಇನ್ಪುಟ್ಗಳು ಮತ್ತು ವಿಧಾನದ ಊಹೆಗಳನ್ನು ಅವಲಂಬಿಸಿದೆ. ದಶಮಾಂಶ ಗಣನೆ 50 ಸಾರ್ಥಕ ಅಂಕೆಗಳನ್ನು ಬಳಸುತ್ತದೆ. ಆದರೆ ಅಂದಾಜುಗಳು, ಸಂಖ್ಯಾತ್ಮಕ ವಿಧಾನಗಳು ಮತ್ತು ಮೂಲ ದತ್ತಾಂಶ ಕಡಿಮೆ ನಿಖರವಾಗಿರಬಹುದು; ಪ್ರದರ್ಶಿತ ಮೌಲ್ಯಗಳನ್ನು ರೌಂಡ್ ಮಾಡುವುದರಿಂದ ಈ ಮಿತಿಗಳು ನಿವಾರಣೆಯಾಗುವುದಿಲ್ಲ. ಸ್ವತಂತ್ರ ಮೂಲಗಳ ಪರಿಹಾರಗಳೊಂದಿಗೆ ಪರಿಶೀಲಿಸಿದ ಉದಾಹರಣೆಗಳು: 5. ಉದಾಹರಣೆಗೆ, “Wikipedia 3×3 example” ಅನ್ನು Wikipedia “Gaussian elimination” example; Python fractions: det = −1, ‖A‖∞ = 6, ‖A⁻¹‖∞ = 10 ಜೊತೆಗೆ ಪರಿಶೀಲಿಸಲಾಗುತ್ತದೆ.

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

ನವೀಕರಿಸಲಾಗಿದೆ ಸೆಪ್ಟೆಂಬರ್ 28, 2026ಪರಿಶೀಲಿಸಿದ ಉದಾಹರಣೆಗಳು: 5

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

ಲೆಕ್ಕಾಚಾರದ ವಿಧಾನ S

Augmented matrix
$\left[A \mid \mathbf b\right] = \left[\begin{array}{ccc|c}2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3\end{array}\right]$
Unknowns: x, y, z.
Column 1: make the pivot 1 and clear the rest of the column
$R_{1} \leftarrow \frac{1}{2}\,R_{1},\ R_{2} \leftarrow R_{2} + 3R_{1},\ R_{3} \leftarrow R_{3} + 2R_{1} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 2 & 1 & 5\end{array}\right]$
Column 2: make the pivot 1 and clear the rest of the column
$R_{2} \leftarrow 2\,R_{2},\ R_{1} \leftarrow R_{1} - \frac{1}{2}R_{2},\ R_{3} \leftarrow R_{3} - 2R_{2} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -1 & 1\end{array}\right]$
Column 3: make the pivot 1 and clear the rest of the column
$R_{3} \leftarrow -R_{3},\ R_{1} \leftarrow R_{1} + R_{3},\ R_{2} \leftarrow R_{2} - R_{3} \;\longrightarrow\; \left[\begin{array}{ccc|c}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1\end{array}\right]$
Read off the solution
$x = 2,\quad y = 3,\quad z = -1$

ರೇಖೀಯ ಸಮೀಕರಣಗಳ ವ್ಯವಸ್ಥೆಯ ಪರಿಹಾರಕ ಬಗ್ಗೆ

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.

ಪರಿಹರಿಸಿದ ಉದಾಹರಣೆಗಳು

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

ಪರಿಶೀಲನೆಯ ಮೂಲ: 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

ಪರಿಶೀಲನೆಯ ಮೂಲ: 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

ಪರಿಶೀಲನೆಯ ಮೂಲ: 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

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

ಪ್ರಶ್ನೆಗಳು

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.

“ರೇಖೀಯ ಸಮೀಕರಣಗಳ ವ್ಯವಸ್ಥೆಯ ಪರಿಹಾರಕ” ಎಷ್ಟು ನಿಖರವಾಗಿದೆ?

ನಿಖರತೆ ನಿಮ್ಮ ಇನ್‌ಪುಟ್‌ಗಳು ಮತ್ತು ವಿಧಾನದ ಊಹೆಗಳನ್ನು ಅವಲಂಬಿಸಿದೆ. ದಶಮಾಂಶ ಗಣನೆ 50 ಸಾರ್ಥಕ ಅಂಕೆಗಳನ್ನು ಬಳಸುತ್ತದೆ. ಆದರೆ ಅಂದಾಜುಗಳು, ಸಂಖ್ಯಾತ್ಮಕ ವಿಧಾನಗಳು ಮತ್ತು ಮೂಲ ದತ್ತಾಂಶ ಕಡಿಮೆ ನಿಖರವಾಗಿರಬಹುದು; ಪ್ರದರ್ಶಿತ ಮೌಲ್ಯಗಳನ್ನು ರೌಂಡ್ ಮಾಡುವುದರಿಂದ ಈ ಮಿತಿಗಳು ನಿವಾರಣೆಯಾಗುವುದಿಲ್ಲ. ಸ್ವತಂತ್ರ ಮೂಲಗಳ ಪರಿಹಾರಗಳೊಂದಿಗೆ ಪರಿಶೀಲಿಸಿದ ಉದಾಹರಣೆಗಳು: 5. ಉದಾಹರಣೆಗೆ, “Wikipedia 3×3 example” ಅನ್ನು Wikipedia “Gaussian elimination” example; Python fractions: det = −1, ‖A‖∞ = 6, ‖A⁻¹‖∞ = 10 ಜೊತೆಗೆ ಪರಿಶೀಲಿಸಲಾಗುತ್ತದೆ.

ಈ ವಿಧಾನದ ಮೂಲ ಯಾವುದು?

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

ಈ ಕ್ಯಾಲ್ಕುಲೇಟರ್ ಬಗ್ಗೆ

\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}

ಮೂಲಗಳು

ಆಧಾರಗಳೊಂದಿಗೆ ಪರಿಶೀಲಿಸಲಾಗಿದೆ

ಈ ಕ್ಯಾಲ್ಕುಲೇಟರ್‌ನಲ್ಲಿ ಸ್ವತಂತ್ರ ಆಧಾರಗಳಿಂದ ಉತ್ತರಗಳನ್ನು ಪಡೆದ 5 ಬಿಡಿಸಿದ ಉದಾಹರಣೆಗಳಿವೆ. ಇವು ಪರೀಕ್ಷಾ ಸಮೂಹದಲ್ಲಿ ನಡೆಯುತ್ತವೆ; ನೀವು ಇಲ್ಲಿಯೂ ನಡೆಸಬಹುದು.

Solution values

Augmented matrix

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

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

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

Read off the solution