Exercises — Pivot positions, free variables

KYA dhundhna hai: har nonzero row ki pehli nonzero entry — har ek ek leading 1 hai, aur har ek apne column mein akela baitha hai. Row 1 ki leading 1 column 1 mein hai. Row 2 ki column 2 mein hai. Row 3 ki column 4 mein hai.

• Pivot columns: $\{1,2,4\}$.
• Rank $r=3$ (teen pivots).
• Column 3 mein koi leading 1 nahi → $x_3$ ek maatra free variable hai.
• Check: $\#\text{free}=n-r=4-3=1$. ✓

• $A$: $n=2$ columns, ek leading 1 (row 1, col 1) toh $r=1$. $\#\text{free}=2-1=1$.
• $B$: $n=3$, diagonal par teen leading 1's toh $r=3$. $\#\text{free}=3-3=0$.
• $C$: $n=3$, columns 1 aur 2 mein leading 1's toh $r=2$; column 3 free hai. $\#\text{free}=3-2=1$.

Step 1 — pehle pivot ke neeche clear karo. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{cccc}1& 3& 1& 2\\ 0& 0& 2& 2\end{array}\right].$ Kyun? Hum column 1 mein leading 1 ke neeche zeros chahte hain taaki echelon staircase dikhaye. Step 2 — doosre pivot ko normalise karo. $R_2\leftarrow \frac{1}{2}{R}_2$: $x_3=1$ milta hai. Pivots columns 1 aur 3 mein hain; column 2 mein koi pivot nahi → $x_2$ free hai ($r=2$, $n=3$, toh $\#\text{free}=1$). Step 3 — back-substitute karo. Row 1: $x_1=2-3x_2-x_3=2-3x_2-1=1-3x_2$. $x_2=t$ ke saath: $\mathbf{x}=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+t\left[\begin{array}{c}-3\\ 1\\ 0\end{array}\right].$ Ek free variable ⇒ 3D mein ek line. ✓ (Dekho Null Space and Solution Sets of Ax=0 ki doosra vector null-space direction kyun hai.)

Step 1 — ek leading 1 lo. Rows swap karo taaki tidy row upar aaye, $R_1\leftrightarrow R_2$: $\left[\begin{array}{ccc}1& 3& 5\\ 2& 4& 8\end{array}\right].$ Step 2 — neeche clear karo. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{ccc}1& 3& 5\\ 0& -2& -2\end{array}\right]$. Step 3 — normalise karo. $R_2\leftarrow -\frac{1}{2}{R}_2\Rightarrow x_2=1$. Dono columns pivot columns hain → koi free variables nahi ($n-r=2-2=0$). Step 4 — back-substitute karo. $x_1=5-3x_2=5-3=2$. Unique solution $(2,1)$. ✓

$n=6$ variables, $r=3$ pivots. $\#\text{free}=n-r=6-3=3$. 3 free variables wala consistent system solutions ka ek 3-dimensional flat (ek "3-plane") rakhta hai. (Yeh exactly $\dim$ shifted null space ka hai; dekho Rank and the Rank-Nullity Theorem.)

Left matrix. Aakhri row poori zeros hai — ek harmless $0=0$, augmented column mein pivot nahi. Toh yeh consistent hai. Coefficient pivots columns 1 aur 3 mein hain; column 2 free hai ⇒ $\#\text{free}=1$. Solution set ek line hai. $x_2=t$ likhke: $x_1=4-2t$, $x_3=-1$, toh $\mathbf{x}=(4,0,-1)+t(-2,1,0)$. Right matrix. Bottom row $0=7$ padha jaata hai — augmented column ek pivot column ban jaata hai. Inconsistent → koi solution nahi. Free-variable count yahan irrelevant hai: freedom solution set ko sirf tab describe karta hai jab woh exist kare.

$r=3=n$, toh $A$ ka har column ek pivot column hai → koi free variables nahi. Kyunki $A$ mein har row mein bhi ek pivot hai ($r=3=m$), augmented column kabhi pivot column nahi ban sakta (koi all-zero coefficient row nahi hai jo nonzero right side ke upar baith sake). Toh system har $\mathbf{b}$ ke liye consistent hai, aur $0$ free variables ke saath solution unique hai — exactly ek point. (Yeh invertible case hai; dekho Existence and Uniqueness of Solutions.)

Ek plane = $2$ free variables. $n=4$ ke saath, hume chahiye $\#\text{free}=n-r=2$, toh $r=2$. Koi bhi RREF exactly do leading 1's ke saath banao, jaise $A=\left[\begin{array}{cccc}1& 0& 5& 3\\ 0& 1& -2& 4\\ 0& 0& 0& 0\end{array}\right].$ Columns 1, 2 mein pivots → $r=2$. Columns 3, 4 free → $\#\text{free}=4-2=2$. Ek homogeneous system hamesha consistent hota hai ($0$ kaam karta hai), toh solution set origin se guzarne wala ek sachcha 2-plane hai. ✓ (Iske do spanning directions null space ka basis banate hain — dekho Basis and Dimension.)

Columns independent hain $⟺$ har column pivot column hai $⟺r=2$ ($A\mathbf{x}=0$ mein koi free variables nahi; dekho Linear Independence). Reduce karo: $R_2\leftarrow R_2-2R_1$ deta hai $\left[\begin{array}{cc}1& 2\\ 0& k-4\end{array}\right]$.

• Agar $k-4\ne 0$ (jaise $k=0$): doosra pivot exist karta hai, $r=2$ → independent.
• Agar $k-4=0$, yani $k=4$: doosri row zero hai, sirf $r=1$, column 2 free hai → dependent. Answer: $k=4$ unhe dependent banata hai; $k=0$ (koi bhi $k\ne 4$) unhe independent rakhta hai.

Dono coefficient rows ko proportional banao lekin right-hand sides ko match mat karo. Jaise $\left[\begin{array}{cccc}1& 1& 1& 1\\ 2& 2& 2& 5\end{array}\right]\stackrel{R_2\leftarrow R_2-2R_1}{\to }\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& 0& 0& 3\end{array}\right].$ Bottom row $0=3$ padha jaata hai: augmented column ek pivot earn karta hai, toh system inconsistent hai. ✓

(a) Reduce karo. $R_2\leftarrow R_2-R_1$ aur $R_3\leftarrow R_3-2R_1$: $\left[\begin{array}{cccc}1& 2& 1& 1\\ 0& 0& 1& 2\\ 0& 0& 1& h-2\end{array}\right].$ Ab $R_3\leftarrow R_3-R_2$: $\left[\begin{array}{cccc}1& 2& 1& 1\\ 0& 0& 1& 2\\ 0& 0& 0& h-4\end{array}\right].$ (b) Consistency. Bottom row $0=h-4$ padha jaata hai. Yeh ek sach statement hai (koi augmented pivot nahi) exactly tab jab $h-4=0$, yani $h=4$. Kisi bhi $h\ne 4$ ke liye augmented column ek pivot le leta hai → inconsistent. (c) $h=4$ par. Coefficient pivots columns 1 aur 3 mein hain; column 2 free hai ($r=2$, $n=3$, $\#\text{free}=1$). $x_2=t$ ke saath back-substitute karo: row 2 deti hai $x_3=2$; row 1 deti hai $x_1=1-2x_2-x_3=1-2t-2=-1-2t$. Toh $\mathbf{x}=\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]+t\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right].$ (d) Shape. Ek free variable ⇒ 3D mein ek line, $(-1,0,2)$ se guzarti hui direction $(-2,1,0)$ mein.

Coefficient reduction same hai, pivots columns 1, 3 mein hain aur free column 2 hai — toh $\#\text{free}=1$ right-hand side se regardless (pivots sirf $A$ par depend karte hain). Constants zero set karke: $x_3=0$, $x_1=-2t$, $x_2=t$: ${\mathbf{x}}_{\text{hom}}=t\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right].$ Yeh origin se guzarne wali line hai same direction $(-2,1,0)$ mein. 5.1 ka inhomogeneous solution set yeh line hai jo particular point $(-1,0,2)$ se shift ki gayi hai — geometrically, do parallel lines. Woh "particular + null-space" split solution sets ka general structure hai (Null Space and Solution Sets of Ax=0).

Free variables non-pivot columns count karte hain ($n-r$), right-hand side sirf consistency control karta hai, aur pivot positions akele $A$ ki hain. Master formula aur woh do independent questions do jismein woh split hota hai ::: $\#\text{free}=n-r$; Question 1 (existence) $\mathbf{b}$/consistency par depend karta hai, Question 2 (solution set ka size) sirf $A$ ke pivots par depend karta hai.