Worked examples — Pivot positions, free variables

4.5.11 · D3 · Maths › Linear Algebra (Full) › Pivot positions, free variables

Shuru karne se pehle, kuch reminders plain words mein:

• $A$ numbers ki ek grid hai ($m$ rows, $n$ columns). Har column ek unknown ke liye khada hai $x_1,\dots ,x_n$.
• $\mathbf{b}$ right-hand side hai: numbers ka woh column jiske barabar equations ko hona chahiye. Toh $A\mathbf{x}=\mathbf{b}$ ka matlab hai "har row of $A$, unknowns ke saath dot karo, $\mathbf{b}$ ki matching entry milti hai."
• Augmented matrix $[A\mid \mathbf{b}]$ ek extra column ke roop mein $\mathbf{b}$ ko chipka deta hai, ek vertical bar se alag karke. Iska aakhiri column (bar ke daayein) woh hai jise hum "$\mathbf{b}$-column" kehte hain. Yeh sirf bookkeeping hai: bar humein yaad dilata hai ki woh column answers carry karta hai, koi unknown nahi.

• Row-reduce = woh moves apply karo jab tak har row ka pehla nonzero number upar wali row se daayein na ho (staircase shape). Woh staircase form echelon form kehlaati hai.
• Pivot = reduction ke baad nonzero row ki leading (pehli nonzero) entry. Pivots ki sankhya rank $r$ hai (dekho Rank and the Rank-Nullity Theorem).

Yeh poora page Pivot positions, free variables ka companion hai.

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The scenario matrix

Har linear system exactly inhi cells mein se ek mein aata hai. Do independent switches outcome decide karte hain: (A) Kya system consistent hai (koi bhi solution exist karta hai)? (B) Kitne free variables hain $=n-r$?

#	Consistency	Free vars $n-r$	Solution set ki shape	Special feature	Example
C1	Consistent	$0$	single point (unique)	square, full rank	Ex 1
C2	Consistent	$1$	ek line	ek free column	Ex 2
C3	Consistent	$2$	ek plane	do free columns	Ex 3
C4	Inconsistent	(irrelevant)	empty (no solution)	$\mathbf{b}$-column ek pivot hai	Ex 4
C5	Consistent	$n$ (sab free)	poora space	$A=0$, degenerate	Ex 5
C6	Consistent	$1$ (yahan $n=2,r=1$)	ek line	rows zyada cols se, zero rows aate hain	Ex 6
C7	Consistent	$0$	unique point	rows zyada cols se, phir bhi full column rank	Ex 7
C8	Consistent	$1$	ek line	word problem (real units)	Ex 8
C9	either	$0$ ya contradiction	point ya empty	exam twist: parameter $k$ case ko flip karta hai	Ex 9

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Ex 1 — C1: unique solution (zero free variables)

Forecast: $2$ columns; rows $(2,1)$ aur $(1,3)$ proportional nahi hain, toh dono columns pivot kamaayenge → $r=2=n$ → $n-r=0$, ek single point.

1. Leading $1$ upar laane ke liye swap karo: $R_1\leftrightarrow R_2$ → $\left[\begin{array}{ccc}1& 3& 5\\ 2& 1& 5\end{array}\right].$ Yeh step kyun? Ek leading $1$ elimination arithmetic ko clean banata hai.
2. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{ccc}1& 3& 5\\ 0& -5& -5\end{array}\right].$ Kyun? Pehle pivot ke neeche entry kill karo taaki staircase expose ho.
3. $R_2\leftarrow -\frac{1}{5}{R}_2$: deta hai $x_2=1$. Kyun? Pivot ko $1$ pe scale karo (RREF ki taraf ek kadam) taaki $x_2$ seedha padha ja sake.
4. Back-substitute $x_1=5-3x_2=5-3=2$. Kyun? Row 1 apna pivot variable solve karta hai ab-jaani value use karke.

Result: columns 1 aur 2 mein pivots ⇒ koi free columns nahi, $n-r=2-2=0$. Unique solution $(x_1,x_2)=(2,1)$.

Verify: $2(2)+1(1)=5$ ✓ aur $1(2)+3(1)=5$ ✓. Dono equations satisfied.

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Ex 2 — C2: one free variable (a line)

Forecast: $n=3$ unknowns lekin sirf $m=2$ rows, toh $r\le 2$ aur $n-r\ge 1$. Kyun? Har pivot apni row mein rehta hai, toh pivots rows ki sankhya se zyada nahi ho sakte; kam se kam ek non-pivot column ke saath, ek line expect karo.

1. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 0& 2& 2\end{array}\right].$ Kyun? Pehle pivot ke neeche zero karo.
2. $R_2\leftarrow \frac{1}{2}{R}_2$: $\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 0& 1& 1\end{array}\right].$ Kyun? Doosre pivot ko leading $1$ banao.
3. Pivots columns 1 aur 3 mein hain; column 2 mein koi pivot nahi → $x_2$ free hai. Set $x_2=t$. Kyun? Har non-pivot column by definition ek free parameter hai.
4. Row 2: $x_3=1$. Row 1: $x_1=4-2x_2-3x_3=4-2t-3=1-2t$. Kyun? Har pivot variable ko free wale ke terms mein solve karo.

Result: $n-r=3-2=1$ free variable ⇒ ek line: $\mathbf{x}=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+t\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right].$

Neeche wali figure is line ko $x_1$–$x_2$ plane mein draw karti hai (pivot value $x_3=1$ poori line mein fixed hai). Coral dot particular solution $(1,0,1)$ hai; mint arrow direction $(-2,1,0)$ hai jise free variable $t$ sweep karta hai — yeh direction exactly null-space vector hai. Butter dots $t=-1$ aur $t=1$ pe sample solutions hain; lavender line pe har point system solve karta hai.

Figure Ex 2: ek free variable ek line trace karta hai — coral = particular solution, mint arrow = free-variable direction, butter dots = sample solutions.

Verify $t=0$ pe: $x_1+2x_2+3x_3=1+0+3=4$ ✓; $2+0+8=10$ ✓. $t=1$ pe: $\mathbf{x}=(-1,1,1)$: $-1+2+3=4$ ✓, $-2+4+8=10$ ✓.

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Ex 3 — C3: two free variables (a plane)

Forecast: $n=4$ unknowns aur clearly $2$ staircase pivots, toh $n-r=4-2=2$. Kyun? Do rows se zyada se zyada do pivots milte hain aur dono clearly present hain, do non-pivot columns chodke → ek plane.

1. Pivots padho: column 1 (row 1) aur column 3 (row 2) mein leading entries. Kyun? Yeh har row ki pehli nonzero entries hain.
2. Non-pivot columns 2 aur 4 hain → $x_2=s$, $x_4=t$ free. Kyun? Wahan koi leading entry nahi rehti.
3. Row 2: $x_3=6-4x_4=6-4t$. Row 1: $x_1=5-2x_2-3x_4=5-2s-3t$. Kyun? Har pivot variable ko dono free waalon ke against solve karo.

Result: $n-r=4-2=2$ ⇒ ek plane: $\mathbf{x}=\left[\begin{array}{c}5\\ 0\\ 6\\ 0\end{array}\right]+s\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right]+t\left[\begin{array}{c}-3\\ 0\\ -4\\ 1\end{array}\right].$

Verify $(s,t)=(0,0)$ pe: eq1 $=5$ ✓, eq2 $=6$ ✓. $(s,t)=(1,1)$ pe: $\mathbf{x}=(0,1,2,1)$ → eq1 $0+2+0+3=5$ ✓, eq2 $2+4=6$ ✓.

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Ex 4 — C4: inconsistent (no solution, free count irrelevant)

Forecast: $A$ ki row 2 exactly $2\times$ row 1 hai, lekin $\mathbf{b}$ ki entries $1,5$ usi ratio mein nahi hain. Trouble kyun suspect karein? $A$ mein proportional rows proportional right-hand sides maangti hain; mismatch ek contradiction force karta hai ($0=$ nonzero).

1. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 3\end{array}\right].$ Kyun? Pehle pivot ke neeche eliminate karo.
2. Row 2 padhta hai $0x_1+0x_2=3$, yaani $0=3$. Yeh sab kuch kyun khatam kar deta hai? Augmented column ab ek leading entry hold karta hai — $\mathbf{b}$-column khud ek pivot column hai, inconsistency ka flag.

Result: koi solution nahi. Dhyaan do column 2 mein $A$ mein koi pivot nahi hai, toh naively "ek free variable hai" — lekin zero solutions ke saath, woh freedom kuch bhi describe nahi karti. Consistency pehle check hoti hai.

Verify: $x_1+x_2=1$ force karta hai $2x_1+2x_2=2\ne 5$. Contradiction confirmed — koi bhi $(x_1,x_2)$ kaam nahi karta.

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Ex 5 — C5: the all-zero matrix (every variable free)

Forecast: matrix bilkul koi constraint impose nahi karta, toh koi column pivot nahi rakh sakta → $r=0$ → $n-r=2$. Kyun? Rank $0$ ke saath, har column non-pivot hai, toh har variable free hai — poora plane.

1. Zero matrix ka RREF zero matrix hi hai: koi nonzero rows nahi ⇒ koi pivots nahi, toh $r=0$. Kyun? Pivot ek row ki pehli nonzero entry hoti hai; koi hai hi nahi.
2. Dono columns non-pivot hain ⇒ $x_1=s$, $x_2=t$ dono free. Kyun? Har column mein leading entry nahi hai.

Result: $n-r=2-0=2$ ⇒ solution set hai poora ${\mathbb{R}}^2$: $\mathbf{x}=s\left[\begin{array}{c}1\\ 0\end{array}\right]+t\left[\begin{array}{c}0\\ 1\end{array}\right],s,t\in \mathbb{R}.$ Yeh $n-r$ ki extreme hai: rank zero maximum freedom deta hai.

Verify: koi bhi $(s,t)$ dono rows mein $0=0$ deta hai ✓. Degenerate limiting case handle kiya.

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Ex 6 — C6: columns se zyada rows, ek redundant (zero) row ke saath

Forecast: $A$ ki teeno rows $(1,2)$ ke multiples hain, toh sirf $r=1$ pivot bachta hai → $n-r=2-1=1$, ek line. Kyun? Proportional rows ek single independent equation mein collapse ho jaate hain; baaki zero rows ban jaate hain, ek free column chodke.

1. $R_2\leftarrow R_2-2R_1$, $R_3\leftarrow R_3-3R_1$: $\left[\begin{array}{ccc}1& 2& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right].$ Kyun? Pehle pivot ke neeche eliminate karo; do lower rows collapse ho jaate hain.
2. Ek pivot (column 1); column 2 free hai, $x_2=t$. Do zero rows hain lekin sirf ek free variable. Kyun myth fail karta hai: free variables non-pivot columns count karte hain $=n-r=2-1=1$, zero rows nahi.
3. Row 1: $x_1=1-2x_2=1-2t$. Kyun? Single pivot variable ko free wale ke against solve karo.

Result: ek line $\mathbf{x}=\left[\begin{array}{c}1\\ 0\end{array}\right]+t\left[\begin{array}{c}-2\\ 1\end{array}\right]$. Do zero rows ≠ do free variables.

Verify $t=0$ pe: teeno original equations $1,2,3$ dete hain $\mathbf{b}$ se match karte hue ✓. $t=1$ pe: $\mathbf{x}=(-1,1)$ → $-1+2=1$, $-2+4=2$, $-3+6=3$ ✓.

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Ex 7 — C7: columns se zyada rows, full column rank (unique)

Forecast: pehli do rows independent hain, toh $r=2=n$ → $n-r=0$, ek unique point — even though rows columns se zyada hain. Kyun? Full column rank har variable ko pin karta hai; teesri row phir ek redundant combination honi chahiye (aur use consistent hona chahiye).

1. $R_3\leftarrow R_3-R_1-R_2$: $\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 0\end{array}\right].$ Kyun? Row 3 ko upar ke do pivots ke against cancel karo.
2. Columns 1 aur 2 mein pivots ⇒ full column rank, koi free variables nahi ($n-r=2-2=0$). Extra row zero ban gayi aur, crucially, uska $\mathbf{b}$-side bhi $0$ hai → consistent. $\mathbf{b}$-side check kyun karein? Wahan nonzero (jaise Ex 4) ise kill kar deta.

Result: unique solution $x_1=2$, $x_2=3$.

Verify: $1(2)+0=2$ ✓; $0+1(3)=3$ ✓; $1(2)+1(3)=5$ ✓. Teeno rows agree karti hain.

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Ex 8 — C8: word problem (real units, one free variable)

Forecast: $n=3$ unknowns lekin sirf $m=2$ constraints, toh $r\le 2$ aur $n-r\ge 1$ → recipes ki ek line. Kyun? Do equations zyada se zyada do variables pin kar sakti hain; teesra free rehta hai (phir real-world $\ge 0$ conditions line ko ek segment mein clip karti hain).

1. Model banao: $A+O+G=10,2A+3O+5G=34.$ Matrix $\left[\begin{array}{cccc}1& 1& 1& 10\\ 2& 3& 5& 34\end{array}\right].$ Kyun? Har unknown ek column hai; har rule ek row hai.
2. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{cccc}1& 1& 1& 10\\ 0& 1& 3& 14\end{array}\right].$ Kyun? Pehle pivot ke neeche eliminate karo.
3. Columns 1, 2 mein pivots; column 3 (grape) free hai, $G=t$. Kyun? Column 3 mein koi leading entry nahi.
4. Row 2: $O=14-3G=14-3t$. Row 1: $A=10-O-G=10-(14-3t)-t=-4+2t$. Kyun? Pivots ko free variable ke against solve karo.

Result: $\mathbf{x}=(A,O,G)=(-4+2t,14-3t,t)$, $n-r=3-2=1$ ⇒ recipes ki ek line. Physically humein $A,O,G\ge 0$ bhi chahiye: $A\ge 0\Rightarrow t\ge 2$; $O\ge 0\Rightarrow t\le \frac{14}{3}\approx 4.67$; toh $2\le t\le \frac{14}{3}$ L grape. Maths ek line deta hai; units ise ek segment mein clip karti hain.

Verify $t=3$ pe: $(A,O,G)=(2,5,3)$ → volume $2+5+3=10$ ✓, sugar $2(2)+3(5)+5(3)=4+15+15=34$ ✓, sab $\ge 0$ ✓.

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Ex 9 — C9: exam twist (ek parameter case flip karta hai)

Forecast: eliminate karne par $k-6$ doosre pivot slot mein aata hai, toh $k=6$ ek dangerous value hai. Kyun? Jab pivot entry $0$ ho jaata hai toh rank drop hota hai; phir yeh "no solution" deta hai ya "infinitely many", yeh $\mathbf{b}$-column par hinge karta hai.

1. $R_2\leftarrow R_2-3R_1$: $\left[\begin{array}{ccc}1& 2& 1\\ 0& k-6& 2\end{array}\right].$ Kyun? Pivot ke neeche eliminate karo; $(2,2)$ entry $k-6$ ban jaati hai.
2. Case $k\ne 6$: entry $k-6\ne 0$ → column 2 mein bhi pivot. Dono columns pivoted ⇒ $n-r=0$ ⇒ (i) unique solution. Kyun? Full column rank, consistent.
3. Case $k=6$: row 2 ban jaata hai $\left[00\mid 2\right]$, yaani $0=2$ → $\mathbf{b}$-column ek pivot hai ⇒ (ii) no solution. Kyun? Contradiction, jaise Ex 4.
4. Kya (iii) kabhi reachable hai? Infinitely many ke liye ek free column aur consistency chahiye. Yahan $k=6$ hi akela rank-drop value hai, lekin yeh $\mathbf{b}$ ko inconsistent banata hai, toh is $\mathbf{b}$ ke liye (iii) kabhi nahi hota. Yeh note kyun karein? Exam bait hai yeh assume karna ki "rank drop ⇒ infinitely many" — galat hai jab tak $\mathbf{b}$-side cooperate na kare.

Result: sab $k\ne 6$ ke liye unique; $k=6$ pe no solution; kabhi infinitely many nahi.

Verify unique case solve karo: eliminate deta hai $x_2=\frac{2}{k-6}$, $x_1=1-2x_2$. $k=8$ pe: $x_2=1$, $x_1=-1$ → check $1(-1)+2(1)=1$ ✓, $3(-1)+8(1)=5$ ✓. $k=6$ pe: $0=2$ contradiction ✓.

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Recall Main kis cell mein hun? (do sawaal)

Pehle: kya $\mathbf{b}$-column ek pivot column hai? ::: Agar haan → inconsistent (C4/Ex 9 case ii), ruko. Agar nahi → consistent, continue karo. Phir: kitne non-pivot columns hain? ::: Woh count hai $n-r$: $0$→point, $1$→line, $2$→plane, $n$→poora space.

Recall Woh trap jo teeno galtiyan share karti hain

"Zero rows = free vars", "free var ⇒ infinite solutions", aur "rank drop ⇒ infinite" sab kyun fail karte hain? ::: Har ek ek alag switch ko ignore karta hai — free variables non-pivot columns count karte hain (zero rows nahi), aur kisi bhi solution count ke liye pehle $\mathbf{b}$-column ke through consistency check honi chahiye.

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Connections

• Pivot positions, free variables
• Row Reduction and Echelon Forms
• Rank and the Rank-Nullity Theorem
• Null Space and Solution Sets of Ax=0
• Existence and Uniqueness of Solutions
• Basis and Dimension
• Linear Independence

Case Map (pehle yeh padho, phir diagram)

Plain words mein: kisi bhi system ko classify karne ke liye, do checks karo order mein.

1. Reduce karo augmented matrix ko echelon form mein.
2. Consistency check — agar $\mathbf{b}$-column ek pivot hold karta hai (ek row $[0\cdots 0\mid c]$ jahan $c\ne 0$), ruko: no solution (cell C4).
3. Agar consistent hai, non-pivot columns count karo $=n-r$: $0$ free → single point (C1/C7); $1$ free → line (C2/C6/C8); $2$ free → plane (C3); sab $n$ free → poora space (C5).

Neeche wala diagram same logic ko arrows ki tarah draw karta hai.