Row echelon form and reduced row echelon form: Worked examples

4.5.10 · D3 · Maths › Linear Algebra (Full) › Row echelon form and reduced row echelon form

Shuru karne se pehle, vocabulary ke baare mein ek vaada. Ek matrix bas numbers ka ek grid hota hai. Jab hum equations ke system ko ek grid ke roop mein likhte hain, toh hum $x, y, z$ letters hata dete hain aur sirf unke coefficients rakhte hain, saath mein $=$ sign ke baad ke constants ek aakhri column mein ek bar se alag karke. Woh grid hi augmented matrix hai. Ek pivot kisi row mein left se right padhte hue pehla non-zero number hota hai. Ek pivot column woh column hota hai jisme ek pivot ho. Baaki sab kuch in teen shabdon se hi nikalta hai.

The scenario matrix

Koi bhi linear system, jab tum use RREF tak push karte ho, bilkul inhi outcome classes mein se ek mein land karta hai. Pivots ki sankhya versus unknowns ki sankhya — yahi sab decide karta hai.

Cell	Situation	RREF kaisi dikhti hai	Solution set
C1	pivots = unknowns, last column mein koi pivot nahi	left side par full identity block	bilkul ek point
C2	pivots < unknowns, last column mein koi pivot nahi	kisi column mein koi pivot nahi	infinitely many (free variables)
C3	pivot last column mein	ek row $0 = 1$ padhti hai	koi solution nahi (inconsistent)
C4	ek zero row aati hai	neeche $[00 \dots 0]$ hai	rank girta hai — C1/C2 se juda
C5	ek row swap chahiye (ek zero pivot slot block kar raha hai)	eliminate karne se pehle reorder karna zaroori	upar mein se koi bhi
C6	unknowns se zyada equations hain (tall)	aksar C3 ya C1 force hota hai	over-determined
C7	equations se zyada unknowns hain (wide)	hamesha kam se kam ek free variable	forced C2
C8	word problem → khud matrix banao	tumhari marzi	real answer
C9	exam twist: ek parameter $k$ cases split karta hai	pivot $k$ par depend karta hai	$k$ ke hisaab se branch
C10	degenerate: poori matrix zero hai	sab zero rows	har vector ise solve karta hai

Neeche ke examples har cell ko cover karte hain. Label dhyan se dekho.

Example 1 — Cell C1: ek clean point (square, easy case)

Forecast: teen equations, teen unknowns, kuch bhi degenerate nahi lagta — guess karo ek unique point. Aage padhne se pehle $x$ ke liye ek guess likh lo.

Augmented matrix: $121 1 - 1 2 11 - 1 632$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - R_{1}$ . Yeh step kyun? Top-left ka $1$ hamara pehla pivot hai. Sirf equation 1 mein $x$ ko isolate karne ke liye, hum neeche uske har $x$ ko khatam karte hain. $100 1 - 3 1 1 - 1 - 2 6 - 9 - 4$
$R_{2} \leftrightarrow R_{3}$ swap karo. Yeh step kyun? Doosra pivot column, column 2 hai. $+ 1$ ka pivot (puraane $R_{3}$ se) $- 3$ se zyada friendly hai; swap karne se pehle fractions se bacha ja sakta hai. (Yeh Cell C5 ka ek mini-preview hai.) $100 11 - 3 1 - 2 - 1 6 - 4 - 9$
$R_{3} \to R_{3} + 3 R_{2}$ . Yeh step kyun? Doosre pivot ke neeche ka $- 3$ khatam karo taaki $y$ equation 3 se chala jaye. $100110 1 - 2 - 7 6 - 4 - 21 (REF)$
$R_{3} \to - \frac{1}{7} R_{3}$ , row 3 mein leading $1$ milta hai. Phir upar clean karo: $R_{2} \to R_{2} + 2 R_{3}$ , $R_{1} \to R_{1} - R_{3}$ , phir $R_{1} \to R_{1} - R_{2}$ . Yeh step kyun? Yeh Gauss–Jordan hai: har pivot ko $1$ banao, phir har pivot ke upar ki entries zero karo taaki answer seedha padha ja sake. $100010001123 (RREF)$

Padh lo: $x = 1, y = 2, z = 3$ .

Verify: $1 + 2 + 3 = 6$ ✓; $2 (1) - 2 + 3 = 3$ ✓; $1 + 4 - 3 = 2$ ✓. Teen pivots, teen unknowns, last column mein koi pivot nahi → C1 confirmed.

Example 2 — Cell C2 & C7: ek free variable, answers ki poori line

Forecast: 3 unknowns par sirf 2 equations — wide (Cell C7). Teeno ko pin karne ke liye kaafi equations nahi hain. Guess: infinitely many solutions.

$[12241349]$

$R_{2} \to R_{2} - 2 R_{1}$ . Yeh step kyun? Pehle pivot ke neeche $x$ khatam karo. $[10201141] (REF)$
$R_{1} \to R_{1} - R_{2}$ . Yeh step kyun? Doosra pivot column 3 mein baitha hai (column 2 skip ho gaya — uski entry $0$ hai). RREF ke liye uske upar clear karo. $[10200131] (RREF)$

Pivots columns $1$ aur $3$ mein hain. Column $2$ mein koi pivot nahi ⇒ $y$ ek free variable hai. $y = t$ set karo: $x = 3 - 2 t, y = t, z = 1.$

Verify: $t = 1$ lo: $x = 1, y = 1, z = 1$ . Eq 1 check karo: $1 + 2 + 1 = 4$ ✓; eq 2: $2 + 4 + 3 = 9$ ✓. $t = 0$ lo: $x = 3, y = 0, z = 1$ : $3 + 0 + 1 = 4$ ✓; $6 + 0 + 3 = 9$ ✓. Do alag points dono kaam karte hain → ek line → C2/C7 confirmed. Neeche picture dekho.

Figure — Row echelon form and reduced row echelon form

Example 3 — Cell C3 & C6: inconsistent, koi solution nahi

Forecast: teen equations, do unknowns — tall (Cell C6). Do equations pehle se hi ek point fix kar deti hain; teesri disagree kar sakti hai. Guess: contradiction ke liye dekho.

$111 1 - 1 2 317$

$R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ . Yeh step kyun? Pehle pivot ke neeche $x$ khatam karo. $100 1 - 2 1 3 - 2 4$
$R_{2} \to - \frac{1}{2} R_{2}$ , phir $R_{3} \to R_{3} - R_{2}$ . Yeh step kyun? Doosre pivot ko clean $1$ banao, phir neeche khatam karo. $100110313$
Row 3 dekho: yeh $0 x + 0 y = 3$ , yaani $0 = 3$ padhti hai. Yeh kyun matter karta hai? Last (constants) column mein pivot hona algebra ka "impossible!" chillana hai. Koi bhi $x, y$ numbers $0$ ko $3$ barabar nahi bana sakte.

Verify: Sirf rows 1–2 se: $y = 1$ , $x = 2$ . Equation 3 mein test karo: $2 + 2 (1) = 4 \neq = 7$ . Teesri line wahan se miss karti hai jahan pehli do milti hain → koi solution nahi, C3/C6 confirmed.

Example 4 — Cell C5: woh row swap jo tum skip nahi kar sakte

Forecast: bilkul pehla coefficient $0$ hai. Tum $0$ ko pivot ke roop mein use nahi kar sakte (baad mein tum usse divide karte). Guess: pehle rows swap karni parengi, phir yeh ek normal one-point solve hai.

$012211112569$

$R_{1} \leftrightarrow R_{2}$ swap karo. Yeh step kyun? Pivot slot (row 1, column 1) mein $0$ hai. Ek aisi row upar swap karo jisme wahan non-zero entry ho taaki hamare paas ek real pivot ho. $102121112659$
$R_{3} \to R_{3} - 2 R_{1}$ . Yeh step kyun? Naye pivot ke neeche $x$ khatam karo. $100 12 - 1 110 65 - 3$
$R_{2} \to \frac{1}{2} R_{2}$ , phir $R_{3} \to R_{3} + R_{2}$ . Yeh step kyun? Doosre pivot ko $1$ banao, phir neeche clear karo. $100110 1 \frac{1}{2} \frac{1}{2} 6 \frac{5}{2} - \frac{1}{2}$
$R_{3} \to 2 R_{3}$ ( $z = - 1$ ), phir upar clean karo: $R_{2} \to R_{2} - \frac{1}{2} R_{3}$ , $R_{1} \to R_{1} - R_{3}$ , phir $R_{1} \to R_{1} - R_{2}$ . Yeh step kyun? Answer seedha padhne ke liye Gauss–Jordan finish. $100010001 43 - 1 (RREF)$

Padh lo: $x = 4, y = 3, z = - 1$ .

Verify: eq 1: $2 (3) + (- 1) = 5$ ✓; eq 2: $4 + 3 - 1 = 6$ ✓; eq 3: $2 (4) + 3 + 2 (- 1) = 9$ ✓. C5 confirmed — swap ke bina hum stuck ho jaate.

Example 5 — Cell C4 & C2: ek asli zero row aati hai

Forecast: equation 2 shak mein hai, lagti hai $2 \times$ equation 1 jaisi. Agar ek row doosri ki copy hai, toh elimination use poori tarah zeros mein badle degi. Guess: ek zero row aur ek free variable.

$121 12 - 1 121240$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - R_{1}$ . Yeh step kyun? Pehle pivot ke neeche $x$ khatam karo. $100 10 - 2 100 20 - 2$

Row 2 ban gayi $[0000]$ — ek zero row (Cell C4). Yeh koi information nahi carry karti (yeh $0 = 0$ hai, hamesha true). Ise neeche move karo.

$R_{2} \leftrightarrow R_{3}$ swap karo, phir $R_{2} \to - \frac{1}{2} R_{2}$ . Yeh step kyun? Zero rows neeche hoti hain; bachne wale pivot ko $1$ scale karo. $100110100210$
$R_{1} \to R_{1} - R_{2}$ . Yeh step kyun? RREF ke liye doosre pivot (column 2) ke upar clear karo. $100010100110 (RREF)$

Pivots columns $1, 2$ mein; column $3$ mein koi pivot nahi ⇒ $z$ free hai. $z = t$ set karo: $x = 1 - t, y = 1, z = t .$

Verify: $t = 0$ : $(1, 1, 0)$ : eq 1 $1 + 1 + 0 = 2$ ✓, eq 3 $1 - 1 + 0 = 0$ ✓. $t = 2$ : $(- 1, 1, 2)$ : eq 1 $- 1 + 1 + 2 = 2$ ✓, eq 3 $- 1 - 1 + 2 = 0$ ✓. Rank $= 2$ (do pivots, ek zero row) $< 3$ unknowns → free variable → C4/C2 confirmed.

Example 6 — Cell C10: degenerate all-zero right side (homogeneous)

Forecast: right side par sab $0$ hai. Toh $x = y = z = 0$ hamesha kaam karta hai (trivial solution). Bas sawaal yeh hai ki kya aur bhi solutions hain. Guess: haan, kyunki equation 2, equation 1 ka $2 \times$ hai — kaafi real equations nahi hain.

$[1224 - 1 - 2 00]$

$R_{2} \to R_{2} - 2 R_{1}$ . Yeh step kyun? Pivot ke neeche $x$ khatam karo. $[1020 - 1 0 00] (RREF)$

Doosri row poori tarah gaayab ho gayi. Ek pivot (column 1), toh $y$ aur $z$ dono free hain. $y = s, z = t$ ke saath: $x = - 2 s + t, y = s, z = t .$

Verify: $(s, t) = (0, 0)$ deta hai $(0, 0, 0)$ — trivial solution, kisi bhi homogeneous system ke liye guaranteed. $(s, t) = (1, 0)$ : $(- 2, 1, 0)$ : $- 2 + 2 - 0 = 0$ ✓. $(s, t) = (0, 1)$ : $(1, 0, 1)$ : $1 + 0 - 1 = 0$ ✓. Infinitely many → origin se jaati poori ek plane → C10 confirmed. Note karo: ek homogeneous system kabhi bhi inconsistent nahi ho sakta, kyunki all-zero vector hamesha ek solution hota hai.

Example 7 — Cell C8: ek word problem jo tumhe translate karna hai

Forecast: teen baskets, teen unknown prices — lagta hai Cell C1, ek clean answer. Pehle apple ki ek price guess karo.

Har basket ko ek equation mein translate karo, phir ek matrix row mein: $⎩ ⎨ ⎧ a + b + c = 10 2 a + b + 0 c = 12 a + 0 b + 3 c = 14 \Rightarrow 121110103101214$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - R_{1}$ . Yeh step kyun? Standard: pehle pivot ( $a$ ) ke neeche clear karo. $100 1 - 1 - 1 1 - 2 2 10 - 8 4$
$R_{2} \to - R_{2}$ , phir $R_{3} \to R_{3} + R_{2}$ . Yeh step kyun? Pivot $2$ ko positive $1$ banao, phir neeche clear karo. $10011012410812$
$R_{3} \to \frac{1}{4} R_{3}$ ( $c = 3$ ), phir upar clean karo: $R_{2} \to R_{2} - 2 R_{3}$ , $R_{1} \to R_{1} - R_{3}$ , phir $R_{1} \to R_{1} - R_{2}$ . Yeh step kyun? Prices seedhe padhne ke liye Gauss–Jordan. $100010001523 (RREF)$

Padh lo: $a = \$ 5,\ b = $2,\ c = $3$.

Verify (units ke saath): Basket A $= 5+2+3 = \$ 10 $✓ . B a s k e tB$ = 2(5)+2 = $12 $✓ . B a s k e tC$ = 5 + 3(3) = $14$ ✓. Sab prices positive aur sensible hain → C8 confirmed.

Example 8 — Cell C9: parameter $k$ wala exam twist

Forecast: coefficient $k$ matrix mein chhupa hai. Kahin ek pivot $k$ par depend karega; jab woh pivot $0$ ban jaye, toh behaviour flip ho jaata hai. Guess: ek special "bad" value of $k$ .

$[11 1 k 12]$

$R_{2} \to R_{2} - R_{1}$ . Yeh step kyun? Pivot ke neeche $x$ clear karo. $[10 1 k - 1 11]$
Pivot-2 slot mein entry $k - 1$ inspect karo. Yeh step kyun? Row 2 real doosra pivot deti hai ya nahi, yeh $k - 1$ ke non-zero hone par depend karta hai. Yahi branch point hai.

Case (a): $k \neq = 1$ . Toh $k - 1 \neq = 0$ ek genuine pivot hai. Do pivots, do unknowns, last column mein koi pivot nahi ⇒ unique solution. Solve karne par: $y = \frac{1}{k - 1}$ , aur $x = 1 - y = \frac{k - 2}{k - 1}$ .
Case (b/c): $k = 1$ . Row 2 ban jaati hai $[00 ∣ 1]$ , yaani $0 = 1$ ⇒ last column mein pivot ⇒ koi solution nahi (Cell C3). (Yahan koi aisa " $k$ " nahi hai jo infinitely many deta ho, kyunki jab $k = 1$ hota hai toh constant $1$ donon lines ko parallel lekin alag banata hai.)

Verify: $k = 3$ lo: formula deta hai $y = \frac{1}{2}$ , $x = \frac{1}{2}$ . Check karo: $x + y = 1$ ✓; $x + 3 y = \frac{1}{2} + \frac{3}{2} = 2$ ✓. $k = 0$ lo: $y = \frac{1}{- 1} = - 1$ , $x = \frac{- 2}{- 1} = 2$ . Check karo: $2 + (- 1) = 1$ ✓; $2 + 0 (- 1) = 2$ ✓. Aur $k = 1$ : equations $x + y = 1$ , $x + y = 2$ dono ek saath hold nahi kar sakti → koi solution nahi ✓. C9 confirmed.

Recall Which cell is which — self-test

"3 equations, 3 unknowns, RREF identity block hai" — yeh kaun sa cell hai? ::: C1 — unique solution. Bina pivot wale column ka matlab kya hai? ::: Ek free variable → infinitely many solutions (C2/C7/C10). Augmented last column mein pivot ka matlab kya hai? ::: Inconsistent, koi solution nahi (C3). Kabhi kabhi eliminate karne se pehle rows kyun swap karni padti hain? ::: Pivot slot mein $0$ pivot nahi ho sakta (tum zero se divide karte) — ek non-zero row upar swap karo (C5). Kya ek homogeneous system (right side par sab zeros) kabhi inconsistent ho sakta hai? ::: Nahi — all-zero vector hamesha ise solve karta hai (C10). Jis parameter value par pivot $0$ ban jaaye, wahan kya karo? ::: Us value ko alag case ki tarah analyse karo; general formula par bharosa mat karo (C9).

Connections

Har example yahan Gaussian Elimination se REF tak gaya, phir Gauss-Jordan Elimination se RREF tak.
Pivots ginana = Rank of a Matrix (Examples 5, 6 mein rank $<$ unknowns hai).
Outcome classes Solving Systems of Linear Equations ka dil hain.
Free variables (Examples 2, 5, 6) Free and Basic Variables mein formally define hote hain.
Sirf trivial solution wala homogeneous system columns ki Linear Independence se juda hai.
Row swaps aur scalings (Examples 1, 4) left side par act karne wali ek elementary matrix hain.
Constants ki jagah identity augment karne se Matrix Inverse via RREF milta hai.
Parent topic par wapas jao.