Phase 1 — Forward elimination → row echelon form (REF).
Goal: har leading entry (pivot) ke neeche sab kuch zero banana, left se right kaam karte hue.
Column j mein pivot choose karo (ek nonzero entry).
Uske neeche har row i ke liye, pivot row ka ajjaij times subtract karo. Yeh aij→0force karta hai.
Phase 2 — Back substitution.
Last equation ab ek unknown ke saath hai → use solve karo. Upar waali equation mein upward substitute karo, aur repeat karo.
(Optionally elimination upar continue karo aur pivots ko 1 tak scale karo → reduced row echelon form (RREF), woh identity-jaisi form jahan se tum solutions directly padh sakte ho. Yeh Gauss–Jordan hai.)
3 doston ko imagine karo jo ek secret number trio ke baare mein ek ek clue jaante hain. Akele clues messy hain. Toh tum combine karte ho clues — "mere clue ka double apne clue se le lo" — taaki ek ek unknown ko knock out karo. Tab tak karo jab tak ek clue sirf ek number mention na kare: use solve karo. Phir backwards chalo ise plug in karte hue. Gaussian elimination exactly yeh "ek cancel karo, phir backwards kaam karo" trick hai, ek grid mein neat tarike se ki gayi.
Infinitely many solutions (woh variable free hai).
Ek reduced row reads [0 0 ... 0 | c] jahan c ≠ 0. Iska kya matlab hai?
System inconsistent hai — koi solution nahi.
Partial pivoting kya hai aur ise kyun use karte hain?
Pivot column mein sabse bade magnitude wali entry ke saath row swap in karo; zero/tiny pivots se divide hone se bachta hai aur numerical error control karta hai.
REF aur RREF mein difference?
REF upper-triangular hai jisme pivots ke neeche zeros hain; RREF mein additionally pivots 1 ke barabar hain aur unke upar bhi zeros hain (identity-jaisa).
Gaussian elimination ka ek n×n system par approximate op count kya hai?
Lagbhag (2/3)n^3 floating-point operations.
Back-substitution bottom row se kyun karte hain?
REF ki bottom row mein sirf ek unknown hota hai, toh tum ise directly solve kar sakte ho aur upward substitute kar sakte ho.