Row echelon form and reduced row echelon form

4.5.10 · Maths › Linear Algebra (Full)

YEH forms KYA hain?

Staircase KYON kaam karti hai? (first-principles derivation)

Sirf teen elementary row operations use karne ki ijazat hai, kyunki har ek solution set ko preserve karti hai:

REF tak KAISE pahunchein — Gaussian elimination

1. Leftmost column dhoondo jisme non-zero entry ho → pivot column.
2. Kisi non-zero entry ko swap karke pivot position par lao ($0$ se divide karne se bachne ke liye).
3. Replace use karo taaki pivot ke neeche sab kuch zero ho jaaye.
4. Us row+column ko ignore karo, neeche-daayein submatrix par recurse karo.

RREF tak KAISE pahunchein — Gauss–Jordan elimination

Upar sab karo, phir bottom pivot se upar ki taraf jaate hue: 5. Har pivot row ko scale karo taaki pivot $1$ ban jaaye. 6. Replace use karo taaki har pivot ke upar bhi entries zero ho jaayein.

Worked Example 1 — REF phir RREF tak

Solve karo $\{\begin{array}{l}x+2y+z=6\\ 2x+5y+3z=15\\ x+3y+3z=11\end{array}$

Augmented matrix: $\left[\begin{array}{cccc}1& 2& 1& 6\\ 2& 5& 3& 15\\ 1& 3& 3& 11\end{array}\right]$

Step A: $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$. Yeh step kyun? Pivot top-left ka $1$ hai; hum uske neeche wali entries ko khatam karte hain taaki $x$, equations 2 aur 3 se chala jaaye. $\left[\begin{array}{cccc}1& 2& 1& 6\\ 0& 1& 1& 3\\ 0& 1& 2& 5\end{array}\right]$

Step B: $R_3\to R_3-R_2$. Yeh step kyun? Agla pivot row 2, column 2 mein $1$ hai. Uske neeche wali entry khatam karo taaki $y$ equation 3 se chala jaaye. $\left[\begin{array}{cccc}1& 2& 1& 6\\ 0& 1& 1& 3\\ 0& 0& 1& 2\end{array}\right]\text{(REF — staircase done!)}$

REF se, back-substitute karo: $z=2$, phir $y+z=3\Rightarrow y=1$, phir $x+2y+z=6\Rightarrow x=2$.

Ab RREF ki taraf (pivots ke upar clean karo, bottom-up):

Step C: $R_2\to R_2-R_3$, $R_1\to R_1-R_3$. Kyun? Row 3 ke pivot ki column clear karo. $\left[\begin{array}{cccc}1& 2& 0& 4\\ 0& 1& 0& 1\\ 0& 0& 1& 2\end{array}\right]$

Step D: $R_1\to R_1-2R_2$. Kyun? Row 2 ke pivot ke upar clear karo. $\left[\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 1& 2\end{array}\right]\text{(RREF)}$ Seedha padho: $x=2,y=1,z=2$. ✅

Worked Example 2 — free variables aur infinite solutions

$\left[\begin{array}{cccc}1& 2& 3& 4\\ 2& 4& 7& 9\end{array}\right]$

Step A: $R_2\to R_2-2R_1$. Kyun? Leading $1$ ke neeche khatam karo. $\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 0& 1& 1\end{array}\right](\text{REF})$

Step B: $R_1\to R_1-3R_2$ (column 3 mein pivot ke upar clean karo). $\left[\begin{array}{cccc}1& 2& 0& 1\\ 0& 0& 1& 1\end{array}\right](\text{RREF})$

Pivot columns: $1,3$. Column $2$ mein koi pivot nahi ⇒ $y$ ek free variable hai. $x=1-2y$, $z=1$, $y$ free. Infinitely many solutions hain, ek poori line. Kyun? Variables se kam pivots hain ⇒ bachi hui freedom.

Worked Example 3 — inconsistent system

$\left[\begin{array}{ccc}1& 1& 2\\ 2& 2& 5\end{array}\right]\stackrel{R_2\to R_2-2R_1}{\to }\left[\begin{array}{ccc}1& 1& 2\\ 0& 0& 1\end{array}\right]$ Row 2 padhi jaati hai $0=1$. Augmented (last) column mein ek pivot ⇒ no solution. Yeh kyun matter karta hai: constants column mein pivot algebraically yeh cheekh kar kehta hai "contradiction!".

Flashcards

Connections

• Gaussian Elimination and Gauss-Jordan Elimination
• Rank of a Matrix (= number of pivots)
• Solving Systems of Linear Equations
• Free and Basic Variables
• Matrix Inverse via RREF (augment with $I$)
• Linear Independence (pivot columns are independent)
• Elementary Matrices (each row op = left-multiplication)

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions ke saath
• D5 · Question bank — concept traps