Gauss-Jordan Elimination Algorithm

2020-01-07
Jun Sok Huhh | 🏠lostineconomics.com

How to solve linear system

Ax=b A x = b

[A βˆ£ b]∈RmΓ—(n+1) \begin{bmatrix} A ~\vert ~ b \end{bmatrix} \in {\mathbb R}^{m \times (n +1)}

Row Echelon Form Matrix (REF)

Reduced Row Echelon Form Matrix (RREF)

[1βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—01βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—βˆ—0001βˆ—βˆ—βˆ—βˆ—βˆ—0000001βˆ—βˆ—00000001βˆ—000000000000000000] \begin{bmatrix} 1 & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 0 & 1 & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 0 & 0 & 0 & 1 & \ast & \ast & \ast & \ast & \ast \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & \ast & \ast \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & \ast \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

Matrix Operation

Solution

Inverse Matrix

AX=I AX = I

μ΄λ•Œ X=Aβˆ’1X = A^{-1}둜 λ‘˜ 수 μžˆλ‹€. μ‹μ˜ 양변에 λ™μ‹œμ— κΈ°λ³Έ ν–‰λ ¬ μ—°μ‚° rr λ²ˆμ„ μ „κ°œν•œλ‹€κ³  ν•˜μž.

ErErβˆ’1β‹―E1⏟(βˆ—)AX=ErErβˆ’1β‹―E1I \underbrace{E_r E_{r-1} \dotsb E_1}_{(*)} A X = E_r E_{r-1} \dotsb E_1 I

μ΄λ•Œ 이 μ—°μ‚°μ˜ κ²°κ³Ό μ‹μ˜ μ’Œλ³€μ΄ IXIX둜 바뀐닀고 ν•˜μž. μ΄λŠ” λ‹€μ‹œ μ•„λž˜μ™€ κ°™λ‹€.

ErErβˆ’1β‹―E1AX=ErErβˆ’1β‹―E1I=IX E_r E_{r-1} \dotsb E_1 A X = E_r E_{r-1} \dotsb E_1 I = IX

X=ErErβˆ’1β‹―E1X = E_r E_{r-1} \dotsb E_1이 되고, 이것이 곧 역행렬이닀.

 

Reference

https://en.wikipedia.org/wiki/Gaussian_elimination

🏠lostineconomics.com | Jun Sok Huhh