Gaussian elimination — forward elimination, back substitution: Worked examples

4.5.9 · D3 · Maths › Linear Algebra (Full) › Gaussian elimination — forward elimination, back substitutio

Yeh parent note ka workbench child hai. Parent ne tumhe rules bataye; yahan hum har woh haath khelenge jo yeh game deal kar sakti hai. Ek bhi example se pehle, hum un sab situations ka ek matrix banate hain jo elimination mein aa sakti hain — phir hum unhe ek-ek karke dhundhte hain.

Definition Index letters jo hum har jagah use karte hain (

$m$ , $n$ , $i$ , $k$ ) Inhe ek baar fix karo taaki koi bhi symbol kabhi surprise na kare:

$m$ = number of equations (coefficient grid ki rows ki sankhya).
$n$ = number of unknowns (coefficient grid ki columns ki sankhya, bar se pehle).
$i$ aur $k$ sirf row/column counters hain — pointers jo grid pe chalte hain. Hum $a_{ik}$ likhte hain "row $i$ , column $k$ mein baithe number" ke liye.
Ek pivot diagonal pe position (row $k$ , column $k$ ) pe rehta hai; hum us number ko $a_{k k}$ bolte hain — wahi counter $k$ do baar use karna matlab "row-number same hai column-number ke barabar", yaani ek diagonal spot. Toh " $a_{k k} = 0$ " ka seedha matlab hai "jo diagonal number main pivot karna chahta tha woh zero hai", aur " $m \neq = n$ " ka matlab hai "grid rectangle hai, square nahi."

Scenario matrix

Har linear system jo tum elimination ko dete ho, inhi cells mein se kisi ek mein land karta hai. Last column us example ka naam deta hai jo use nail karta hai.

#	Case class	Kya special hai	Outcome jo dikhega	Example
C1	Clean square system	har pivot as-is nonzero	unique solution	Ex 1
C2	Zero pivot, swap fix karta hai	diagonal pivot $a_{k k} = 0$ par koi lower row nonzero hai	swap ke baad unique solution	Ex 2
C3	Tiny pivot (stability)	pivot $\approx 0$ , division round-off bigaad deta hai	accuracy ke liye swap	Ex 3
C4	Inconsistent	koi row $0 \dots 0 ∣ d, d \neq = 0$ ban jaati hai	koi solution nahi	Ex 4
C5	Free variable / infinite	koi row $0 \dots 0 ∣ 0$ ban jaati hai	infinitely many	Ex 5
C6	Negative & fractional pivots	har jagah signs aur fractions	unique, sign discipline	Ex 6
C7	Rectangular ( $m \neq = n$ : rows $\neq =$ columns)	over- ( $m > n$ ) aur under-determined ( $m < n$ )	rank batata hai sab	Ex 7a, 7b
C8	Word problem (real units)	translate → solve → interpret	numeric answer with units	Ex 8
C9	Exam twist (unknown parameter)	matrix mein ek letter $k$	har outcome ke liye $k$ par condition	Ex 9

Hum har cell cover karenge. Cells C4/C5/C9 Solving Linear Systems — Consistency se connect hain; C7 Rank of a Matrix se; C3 Pivoting and Numerical Stability se.

Figure — Gaussian elimination — forward elimination, back substitution

Intuition Upar wali compass figure ko kaise padhein

Yeh single picture woh skeleton hai jis par neeche ke saare examples tike hain. Ise teen passes mein padho:

$*$ aur $0$ ka grid. Har $*$ ek still-live coefficient hai; har pink $0$ ek aisi entry hai jo humne deliberately elimination ke dauran banai. Dhyan do ki saare pink zeros diagonal ke neeche hain — zeros ka woh lower-left triangle hi "upper-triangular $U$ " ka literal matlab hai.
Yellow DOWN arrow (left). Forward elimination columns ko left-to-right process karta hai, aur kisi column ke andar top-to-bottom kaam karta hai, woh pink zeros plant karne ke liye pivot-rows subtract karta hai. Har Ex 1–9 is arrow ko travel karke shuru hota hai.
Blue UP arrow (right). Jab $U$ ban jaata hai, back substitution ulti direction mein chalta hai: sabse pehle bottom row (sirf ek unknown, apne blue $b$ ke saath) solve hoti hai, phir hum upar chadhte hain, har step exactly ek nayi variable expose karta hai. Jab koi example "back-substitute (bottom-up)" kehta hai, toh woh yahi blue arrow walk kar raha hai. Yeh image yaad rakho: neeche jaate hue pink zeros banao, phir upar jaate hue variables padho.

Intuition Main KAISE PATA karun ki kab eliminating band karoon aur kab padhna shuru karoon? (ek saanp mein strategy)

Poore method mein exactly do phases hain aur boundary par ek decision:

Neeche jao zeros plant karte hue jab tak kisi column mein pivot ke neeche kuch bhi kill karne ko na bache — grid ab staircase (echelon) form mein hai.
Climb karne se pehle, har finished row inspect karo. Jo row $0 \dots 0 ∣ d$ padhti ho jahan $d \neq = 0$ ho woh ek contradiction hai — ruko, system ka koi solution nahi. Jo row sab zeros $0 \dots 0 ∣ 0$ padhti ho woh koi information carry nahi karti — yeh ek free variable signal karti hai (infinitely many solutions). Jo bhi column kabhi pivot ka host nahi bana woh free variable ka column hai.
Sirf tab jab finished rows consistent hain tum chadhte ho (back-substitute karte ho). Rectangular systems ke liye yeh inspection step optional garnish nahi hai — yahi batata hai ki ek answer expect karo, koi nahi, ya poora family. Pivots gino: pivots $=$ rank; ise $n$ (unknowns) se compare karo taaki substitute karne se pehle hi outcome predict kar sako.

Ex 1 — Clean square system (cell C1)

Forecast: har pivot already nonzero hai, toh humein bina kisi swap ke seedha neeche march karna chahiye aur ek unique triple pe land karna chahiye. Guess karo ki numbers "nice" hain ya nahi (hain).

Augmented matrix: $121134 17 - 1 6254$

Pivot $a_{11} = 1$ ; column 1 clear karo. $m_{21} = 2/1 = 2$ toh $R_{2} \to R_{2} - 2 R_{1}$ ; $m_{31} = 1/1 = 1$ toh $R_{3} \to R_{3} - R_{1}$ . Yeh step kyon? Hum chahte hain ki pivot ke neeche wala $2$ aur $1$ zero ho jaayein — multiplier exactly "kitni pivot-rows unhe cancel karein" hai. $100113 15 - 2 613 - 2$
Pivot $a_{22} = 1$ ; neeche clear karo. $m_{32} = 3/1 = 3$ toh $R_{3} \to R_{3} - 3 R_{2}$ . Yeh step kyon? Column 2 mein wale $3$ ko khatam karo taaki row 3 mein sirf last variable bache. $100110 15 - 17 613 - 41$
Back-substitute (bottom-up). Bottom-up kyon? Sirf last row mein ek hi unknown hai.
- Row 3: $- 17 z = - 41 \Rightarrow z = \frac{41}{17}$ .
- Row 2: $y + 5 z = 13 \Rightarrow y = 13 - 5 \cdot \frac{41}{17} = \frac{221 - 205}{17} = \frac{16}{17}$ .
- Row 1: $x + y + z = 6 \Rightarrow x = 6 - \frac{16}{17} - \frac{41}{17} = \frac{102 - 57}{17} = \frac{45}{17}$ .

Answer: $x = \frac{45}{17}, y = \frac{16}{17}, z = \frac{41}{17}$ .

Worked example Verify

Row 3 original: $x + 4 y - z = \frac{45}{17} + \frac{64}{17} - \frac{41}{17} = \frac{68}{17} = 4$ ✓.

Ex 2 — Zero pivot, swap bachata hai (cell C2)

Forecast: position $(1, 1)$ par $0$ hai — tum isse divide nahi kar sakte. Predict karo ki ek row swap forced hai (optional nahi), aur us ke baad hume unique solution milega.

Swap $R_{1} \leftrightarrow R_{2}$ . Yeh step kyon? Pivot nonzero hona chahiye; row 2 column 1 mein ek clean $1$ offer karta hai. $103211120647$
Pivot $a_{11} = 1$ ; column 1 clear karo. Sirf $R_{3}$ ko kaam chahiye: $m_{31} = 3$ , $R_{3} \to R_{3} - 3 R_{1}$ . Yeh step kyon? Pivot ke neeche wale $3$ ko zero karo. $100 21 - 5 12 - 3 64 - 11$
Pivot $a_{22} = 1$ ; neeche clear karo. $m_{32} = - 5/1 = - 5$ toh $R_{3} \to R_{3} - (- 5) R_{2} = R_{3} + 5 R_{2}$ . Yeh step kyon? Ek negative multiplier ka matlab hai hum pivot row add karte hain — " $- m$ " template sign automatically handle karta hai. $100210127649$
Back-substitute karo. Row 3: $7 z = 9 \Rightarrow z = \frac{9}{7}$ . Row 2: $y + 2 z = 4 \Rightarrow y = 4 - \frac{18}{7} = \frac{10}{7}$ . Row 1: $x + 2 y + z = 6 \Rightarrow x = 6 - \frac{20}{7} - \frac{9}{7} = 6 - \frac{29}{7} = \frac{13}{7}$ .

Answer: $x = \frac{13}{7}, y = \frac{10}{7}, z = \frac{9}{7}$ .

Worked example Verify

Row 1 original ( $0 \cdot x + y + 2 z = 4$ ): $\frac{10}{7} + \frac{18}{7} = \frac{28}{7} = 4$ ✓. Row 3 original: $3 \cdot \frac{13}{7} + \frac{10}{7} = \frac{49}{7} = 7$ ✓.

Ex 3 — Tiny pivot, accuracy ke liye swap (cell C3)

Forecast: pivot $0.001$ legal hai (nonzero) par bahut chhota hai. Isse divide karne par multiplier huge ho jaata hai aur round-off answer kharaab kar deta hai. Predict karo: partial pivoting (bade entry ko swap karo) fix hai — Pivoting and Numerical Stability dekho.

Naive way (koi swap nahi). $m_{21} = 1/0.001 = 1000$ , $R_{2} \to R_{2} - 1000 R_{1}$ :
- new $a_{22} = 1 - 1000 \cdot 1 = - 999$
- new RHS $= 2 - 1000 \cdot 1 = - 998$ . 3 sig figs tak round karo: $- 999 y = - 998 \Rightarrow y \approx 0.999$ . Phir $0.001 x + 0.999 = 1 \Rightarrow x = \frac{0.001}{0.001} = 1.00$ . Yahaan abhi bhi theek hai, par jab zyada digits khote hain toh chhota pivot error badhaata hai.
Stable way (pehle swap karo). Yeh step kyon? Column-1 ki sabse badi entry ko upar rakho taaki multiplier $\leq 1$ ho aur round-off blow up na kar sake. $[1 0.001 1121]$ $m_{21} = 0.001/1 = 0.001$ , $R_{2} \to R_{2} - 0.001 R_{1}$ : $a_{22} = 1 - 0.001 = 0.999$ , RHS $= 1 - 0.002 = 0.998$ . Back-sub: $0.999 y = 0.998 \Rightarrow y \approx 0.999$ ; $x = 2 - y \approx 1.00$ .

Answer (true value): $x = \frac{1000}{999}, y = \frac{998}{999}$ . Exact system: $0.001 x + y = 1, x + y = 2$ . Subtract: $(1 - 0.001) x = 1 \Rightarrow x = \frac{1}{0.999} = \frac{1000}{999}$ , aur $y = 2 - x = \frac{998}{999}$ .

Worked example Verify

$x = \frac{1000}{999}, y = \frac{998}{999}$ : eqn 1 $0.001 x + y = \frac{1}{999} + \frac{998}{999} = \frac{999}{999} = 1$ ✓; eqn 2 $x + y = \frac{1998}{999} = 2$ ✓.

Ex 4 — Inconsistent system, koi solution nahi (cell C4)

Forecast: row 2 ka left column exactly row 1 ka $- 2$ times hai — coefficients proportional hain par RHS nahi. Predict karo ek row $00 ∣ d$ form ki milegi jahan $d \neq = 0$ : koi solution nahi.

Column 1 clear karo. $m_{21} = - 2/1 = - 2$ , $R_{2} \to R_{2} - (- 2) R_{1} = R_{2} + 2 R_{1}$ . Yeh step kyon? Standard elimination; negative multiplier ka matlab add karo. $[10 - 2 0 37]$
Bottom row padho. Yeh kehti hai $0 x + 0 y = 7$ , yaani $0 = 7$ — impossible. Yeh contradiction kyun reveal karta hai? Elimination $0 =$ nonzero tabhi banata hai jab equations genuinely disagree karti hain.

Answer: Inconsistent — koi solution nahi. (Do parallel lines, kabhi nahi miltiyan.)

Worked example Verify

Elimination ke baad RHS: $1 + 2 \cdot 3 = 7 \neq = 0$ , inconsistency confirm karta hai. Solving Linear Systems — Consistency dekho.

Ex 5 — Free variable, infinitely many solutions (cell C5)

Forecast: Ex 4 wala hi left side, par ab RHS bhi row 1 ka $- 2 \times$ hai. Predict karo ek full zero row $00 ∣ 0$ : infinitely many solutions with one free variable.

Column 1 clear karo. $m_{21} = - 2$ , $R_{2} \to R_{2} + 2 R_{1}$ . $[10 - 2 0 30]$
Interpret karo. Row 2 hai $0 = 0$ — hamesha true, koi info carry nahi karta. Free variable kyon? Sirf ek real equation $x - 2 y = 3$ do unknowns ke liye; ek variable free hai.
Parametrise karo. $y = t$ lo (free). Phir $x = 3 + 2 t$ .

Answer: infinitely many: $(x, y) = (3 + 2 t, t)$ kisi bhi real $t$ ke liye.

Worked example Verify

$t = 1$ lo: $(x, y) = (5, 1)$ . Eqn 1: $5 - 2 = 3$ ✓. Eqn 2: $- 10 + 4 = - 6$ ✓. $t = 0$ lo: $(3, 0)$ , eqn 1 $= 3$ ✓.

Ex 6 — Negative aur fractional pivots (cell C6)

Forecast: pehla pivot negative hai. Predict karo fractions aayenge par ek clean unique answer milega agar tum " $R_{i} \to R_{i} - m R_{k}$ " template religiously follow karo.

Pivot $a_{11} = - 2$ ; column 1 clear karo. $m_{21} = 4/ (- 2) = - 2$ , $R_{2} \to R_{2} + 2 R_{1}$ ; $m_{31} = - 6/ (- 2) = 3$ , $R_{3} \to R_{3} - 3 R_{1}$ . Negative pivot theek kyon hai? Multiplier formula ne kabhi positivity assume nahi ki — yeh bas cancel karta hai. $- 2 00 112 38 - 10 920 - 30$
Pivot $a_{22} = 1$ ; neeche clear karo. $m_{32} = 2/1 = 2$ , $R_{3} \to R_{3} - 2 R_{2}$ . $- 2 00 110 38 - 26 920 - 70$
Back-substitute karo. Row 3: $- 26 z = - 70 \Rightarrow z = \frac{70}{26} = \frac{35}{13}$ . Row 2: $y + 8 z = 20 \Rightarrow y = 20 - \frac{280}{13} = \frac{260 - 280}{13} = - \frac{20}{13}$ . Row 1: $- 2 x + y + 3 z = 9 \Rightarrow - 2 x = 9 + \frac{20}{13} - \frac{105}{13} = 9 - \frac{85}{13} = \frac{32}{13} \Rightarrow x = - \frac{16}{13}$ .

Answer: $x = - \frac{16}{13}, y = - \frac{20}{13}, z = \frac{35}{13}$ .

Worked example Verify

Row 2 original: $4 x - y + 2 z = - \frac{64}{13} + \frac{20}{13} + \frac{70}{13} = \frac{26}{13} = 2$ ✓.

Ex 7a — Rectangular: over-determined, $m > n$ (cell C7)

Forecast: yahaan $m = 3$ equations aur $n = 2$ unknowns hain, toh $m > n$ . Aam taur par over-determined systems ka koi solution nahi hota — jab tak teesri equation pehli do ki hidden combination na ho. Predict karo hume eliminating finish karni hogi, phir last row inspect karni hogi (upar wale strategy box ke hisaab se): agar woh $0 = 0$ mein collapse ho jaaye, toh extra equation redundant thi aur ek unique solution bachta hai. (Yahi Rank of a Matrix action mein hai.)

Column 1 clear karo. $R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - 2 R_{1}$ . $100 1 - 2 - 1 2 - 2 - 1$
Column 2 clear karo. Pivot $a_{22} = - 2$ , $m_{32} = - 1/ - 2 = \frac{1}{2}$ , $R_{3} \to R_{3} - \frac{1}{2} R_{2}$ . Yeh step kyon? Test karo ki row 3 nayi information deta hai ya collapse ho jaata hai. $100 1 - 2 0 2 - 2 0$
Inspect karo, phir interpret karo. Bottom row $0 = 0$ : consistent aur koi info carry nahi karta, toh teesri equation dependent thi. Do unknowns ke liye do pivots → rank $= n = 2$ → unique solution. Back-sub: $- 2 y = - 2 \Rightarrow y = 1$ ; $x + y = 2 \Rightarrow x = 1$ .

Answer: $x = 1, y = 1$ (aur yeh teeno originals satisfy karta hai).

Worked example Verify

Eqn 2: $x - y = 1 - 1 = 0$ ✓. Eqn 3: $2 x + y = 2 + 1 = 3$ ✓. Rank $= 2 =$ unknowns ki sankhya → unique.

Ex 7b — Rectangular: under-determined, $m < n$ (cell C7)

Forecast: yahaan $m = 2$ equations aur $n = 3$ unknowns hain, toh $m < n$ . Variables zyada hain jitne pivots pin down kar sakein, toh predict karo kam se kam ek free variable hoga aur isliye infinitely many solutions milenge (assuming koi contradiction nahi). Rank yahi batata hai: Rank of a Matrix dekho.

Column 1 clear karo. $m_{21} = 2/1 = 2$ , $R_{2} \to R_{2} - 2 R_{1}$ . Yeh step kyon? Pehle pivot ke neeche wala $2$ zero karo, as always. $[10211143]$
Pivots gino (stop-and-inspect step). Pivots columns 1 aur 2 mein hain (values $1$ aur $1$ ). Column 3 mein koi pivot nahi hai. Koi row $0 \dots 0 ∣ d$ nahi hai, toh system consistent hai; rank $= 2 < n = 3$ . Yeh kyon matter karta hai? Bina pivot wala column ek free variable correspond karta hai — kuch bhi uski value force nahi karta, toh hum ek puri family expect karte hain, single point nahi.
Free variable ko parametrise karo. Pivot-less column, column 3 hai, jiska unknown $z$ hai, toh hum $z = t$ (free) set karte hain aur ise sab real numbers pe range karne dete hain. $z$ kyon aur $x$ ya $y$ kyon nahi? $x$ aur $y$ dono ke paas pivot hai (columns 1 aur 2), toh unki values $z$ choose karne ke baad forced hain; sirf pivot-less variable genuinely free hai.
- Row 2: $y + z = 3 \Rightarrow y = 3 - t$ . Yeh substitution kyon? $z = t$ jaanne ke baad, row 2 mein sirf ek remaining unknown $y$ hai.
- Row 1: $x + 2 y + z = 4 \Rightarrow x = 4 - 2 (3 - t) - t = 4 - 6 + 2 t - t = - 2 + t$ . Last kyon? Upar chadhte hue, row 1 mein ab sirf $x$ unknown hai jab $y$ aur $z$ haath mein hain.

Answer: infinitely many: $(x, y, z) = (- 2 + t, 3 - t, t)$ kisi bhi real $t$ ke liye. (Rank $= 2 < 3 =$ unknowns, toh exactly ek free variable — 3-D space mein solutions ki ek line.)

Worked example Verify

$t = 2$ lo: $(0, 1, 2)$ . Eqn 1: $0 + 2 + 2 = 4$ ✓. Eqn 2: $0 + 5 + 6 = 11$ ✓. $t = 0$ lo: $(- 2, 3, 0)$ . Eqn 1: $- 2 + 6 + 0 = 4$ ✓; eqn 2: $- 4 + 15 + 0 = 11$ ✓.

Ex 8 — Word problem with real units (cell C8)

Forecast: teen sentences ko teen equations mein translate karo, phir yeh ek plain $3 \times 3$ hai. Guess karo ki counts whole numbers aate hain ya nahi (aane chahiye — aadha coin nahi ho sakta!).

$n, d, q$ ko counts maano. $n + d + q 5 n + 10 d + 25 q d = 12 = 170 = n + q \Rightarrow - n + d - q = 0 (count) (cents) (dimes rule)$

$15 - 1 1101 125 - 1 121700$

Column 1 clear karo. $m_{21} = 5/1 = 5$ , $R_{2} \to R_{2} - 5 R_{1}$ ; $m_{31} = - 1/1 = - 1$ , $R_{3} \to R_{3} + R_{1}$ . Yeh step kyon? Pehle pivot ke neeche $5$ aur $- 1$ zero karo taaki rows 2–3 variable $n$ kho dein. $10015212001211012$
Column 2 clear karo. Pivot $a_{22} = 5$ , $m_{32} = 2/5$ , $R_{3} \to R_{3} - \frac{2}{5} R_{2}$ . Yeh step kyon? Column 2 mein $2$ ko khatam karo taaki row 3 mein sirf last variable $q$ bache. $100150 120 - 8 12110 - 32$
Back-substitute (bottom-up) karo. Bottom-up kyon? Last row mein sirf ek unknown $q$ hai. $- 8 q = - 32 \Rightarrow q = 4$ . Phir $5 d + 20 \cdot 4 = 110 \Rightarrow 5 d = 30 \Rightarrow d = 6$ . Phir $n + 6 + 4 = 12 \Rightarrow n = 2$ .

Answer: $n = 2$ nickels, $d = 6$ dimes, $q = 4$ quarters.

Worked example Verify (with units)

Count: $2 + 6 + 4 = 12$ coins ✓. Value: $2 (5) + 6 (10) + 4 (25) = 10 + 60 + 100 = 170$ ¢ $=\$ 1.70 $✓ . D im esr u l e :$ d=6,\ n+q=6$ ✓.

Ex 9 — Exam twist: ek parameter $k$ (cell C9)

Forecast: column 2 mein pivot mein $k$ hoga. Predict karo $k$ ki ek special value hogi jo us pivot ko $0$ banaa de; phir RHS decide karega "koi solution nahi" aur "infinitely many" ke beech. Yeh ek Solving Linear Systems — Consistency classic hai.

Column 1 clear karo. $m_{21} = 2/1 = 2$ , $R_{2} \to R_{2} - 2 R_{1}$ . Yeh step kyon? Saari $k$ -dependence ko ek entry mein isolate karo taaki fork clearly dikhe. $[10 2 k - 4 32]$
Pivot $k - 4$ par case-split karo. Yahaan split kyon? Agla step $k - 4$ se divide karta hai; behaviour exactly tab badalta hai jab woh pivot zero ho.
- Agar $k \neq = 4$ : pivot $\neq = 0$ , toh $y = \frac{2}{k - 4}$ , phir $x = 3 - 2 y = 3 - \frac{4}{k - 4}$ . Unique solution.
- Agar $k = 4$ : row 2 ban jaati hai $00 ∣ 2$ , yaani $0 = 2$ . Inconsistent — koi solution nahi.
- Infinitely many? Row 2 $= 00 ∣ 0$ chahiye. Uske liye $k - 4 = 0$ aur RHS $= 0$ chahiye; par RHS $2 \neq = 0$ hai, toh yeh kabhi nahi hota yahaan.

Answer: sabhi $k \neq = 4$ ke liye unique; $k = 4$ par koi solution nahi; $k$ ki kisi bhi value ke liye infinitely many nahi.

Worked example Verify

$k = 6$ : $y = \frac{2}{2} = 1, x = 3 - 2 = 1$ . Eqn 2 check karo: $2 (1) + 6 (1) = 8$ ✓. $k = 4$ : row deta hai $0 = 2$ , inconsistent ✓.

Recall checkpoint

Connections

Solving Linear Systems — Consistency — cells C4, C5, C9 exactly uske teen outcomes hain.
Rank of a Matrix — pivots ki sankhya (Ex 7a, 7b) = rank; rank vs. unknowns uniqueness decide karta hai.
Pivoting and Numerical Stability — "sabse bade par swap kyon karo" Ex 3 se.
Row Echelon Form / Reduced Row Echelon Form — woh staircase shapes jo yeh examples produce karte hain.
LU Decomposition — jo multipliers humne compute kiye woh $L$ ki entries hain.
Determinant — ek full zero row (Ex 5, Ex 7a) zero determinant signal karta hai.