Exercises — Row echelon form and reduced row echelon form

Rules ko ek checklist ki tarah padho. REF chahiye: (i) zero rows neeche, (ii) har pivot strictly upar wale se right mein, (iii) har pivot ke neeche zeros. RREF add karta hai: (iv) har pivot $=1$, (v) har pivot apne column mein akela (upar bhi zeros).

$A$: Pivots positions $(1,1),(2,2),(3,3)$ par hain — har ek aage right mein, koi zero rows nahi, neeche zeros hain. ✅ REF. Lekin pivots ke upar dekho: $2$ aur $3$ pivots ke upar non-zero entries hain. Toh (v) fail hota hai. $A$ sirf REF mein hai.

$B$: Zero row beech mein hai, neeche nahi → rule (i) turant fail ho jaata hai. $B$ neither hai.

$C$: Identity matrix. Har pivot $1$ hai, apne column mein akela hai, staircase perfect hai. ✅ RREF (aur isliye REF bhi).

Augmented matrix: $\left[\begin{array}{cccc}1& 1& 1& 6\\ 2& 3& 1& 11\\ 2& 1& 3& 14\end{array}\right]$ Step 1 — top-left pivot ke neeche walo ko khatam karo ($x$ ko equations 2,3 se bahar karo): $R_2\to R_2-2R_1$, $R_3\to R_3-2R_1$: $\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& -1& -1\\ 0& -1& 1& 2\end{array}\right]$ Step 2 — agla pivot $(2,2)$ par $1$ hai; uske neeche clear karo: $R_3\to R_3+R_2$: $\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& -1& -1\\ 0& 0& 0& 1\end{array}\right]$ Bottom row padhti hai $0=1$. Contradiction → system ka koi solution nahi (inconsistent). Ek pivot constants column mein land kiya — inconsistency ki algebraic "cheekh".

$\left[\begin{array}{cccc}2& 4& -2& 2\\ 1& 3& 1& 5\\ 3& 5& -5& 1\end{array}\right]$ Step 1 — top-left pivot ko friendly banao. $R_1\to \frac{1}{2}{R}_1$ scale karo: $\left[\begin{array}{cccc}1& 2& -1& 1\\ 1& 3& 1& 5\\ 3& 5& -5& 1\end{array}\right]$ Step 2 — neeche khatam karo: $R_2\to R_2-R_1$, $R_3\to R_3-3R_1$: $\left[\begin{array}{cccc}1& 2& -1& 1\\ 0& 1& 2& 4\\ 0& -1& -2& -2\end{array}\right]$ Step 3 — pivot $(2,2)$ ke neeche clear karo: $R_3\to R_3+R_2$: $\left[\begin{array}{cccc}1& 2& -1& 1\\ 0& 1& 2& 4\\ 0& 0& 0& 2\end{array}\right]$ Bottom row: $0=2$ → inconsistent, koi solution nahi. (Notice: is level ke do alag-alag dikhne wale systems dono inconsistent nikle — clue hamesha last column mein pivot hona hai.)

Step 1 — pivot $(1,1)$ ke neeche clear karo: $R_2\to R_2-2R_1$: $\left[\begin{array}{ccccc}1& 2& 0& 3& 1\\ 0& 0& 1& 2& 3\\ 0& 0& 1& 2& 3\end{array}\right]$ Step 2 — $R_2$ aur $R_3$ identical hain; column 3 mein pivot ke neeche clear karo: $R_3\to R_3-R_2$: $\left[\begin{array}{ccccc}1& 2& 0& 3& 1\\ 0& 0& 1& 2& 3\\ 0& 0& 0& 0& 0\end{array}\right](\text{REF})$ Pivots pehle se $1$ ke barabar hain aur row-2 pivot (column 3) ke upar column pehle se clean hai → yeh RREF bhi hai. Pivot columns: $1$ aur $3$. Non-pivot columns: $2$ aur $4$ → variables $y$ aur $w$ free hain. Equations padho:

• Row 1: $x+2y+3w=1\Rightarrow x=1-2y-3w$
• Row 2: $z+2w=3\Rightarrow z=3-2w$

Solution set ($y,w$ free ke saath): $(x,y,z,w)=(1-2y-3w,y,3-2w,w).$ Rank = pivots ki sankhya = ==$2$== (dekho Rank of a Matrix). $4$ variables aur rank $2$ ke saath, $4-2=2$ free variables hain → solutions ka poora 2-D plane solution space mein.

$\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 3& 2\\ 1& 1& k& k\end{array}\right]$ $R_2\to R_2-R_1$, $R_3\to R_3-R_1$: $\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& 1& 2& 1\\ 0& 0& k-1& k-1\end{array}\right]$ Decisive entry $(3,3)$ pivot slot hai: $k-1$, constant $k-1$ ke saath.

• Agar $k\ne 1$: pivot $k-1\ne 0$, toh teenon columns mein pivots hain → unique solution. Solve karo: $z=\frac{k-1}{k-1}=1$, phir $y+2z=1\Rightarrow y=-1$, phir $x+y+z=1\Rightarrow x=1$. Toh $(x,y,z)=(1,-1,1)$ har $k\ne 1$ ke liye.
• Agar $k=1$: bottom row $0=0$ ban jaati hai — ek zero row, koi contradiction nahi, aur $3$ variables ke liye sirf $2$ pivots → infinitely many solutions (ek free variable $z$). Row 2 se: $y+2z=1\Rightarrow y=1-2z$. Row 1 se: $x+y+z=1\Rightarrow x=1-y-z=1-(1-2z)-z=z$. Toh $(x,y,z)=(z,1-2z,z)$.
• $k$ ki koi bhi value no solution nahi deti, kyunki degenerate case ($k=1$) $0=0$ produce karta hai, kabhi $0=\text{nonzero}$ nahi. Toh case (c) empty hai.

Inhe ek matrix ke columns ki tarah rakh ke reduce karo (pivot columns ek independent set mark karte hain — dekho Linear Independence): $\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 6\\ 1& 3& 4\end{array}\right]$ $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$: $\left[\begin{array}{ccc}1& 2& 3\\ 0& 0& 0\\ 0& 1& 1\end{array}\right]$ $R_2\leftrightarrow R_3$ swap karo taaki zero row neeche jaaye: $\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]$ Pivots sirf columns $1$ aur $2$ mein hain → rank $=2$. Teen vectors lekin sirf $2$ pivots ⇒ woh linearly dependent hain. Concretely dependency isliye dikhti hai kyunki column 3 pivot column nahi hai: $v_3=v_1+v_2$ (check: $(1,2,1)+(2,4,3)=(3,6,4)$ ✅).

$2\times 2$ identity ke saath augment karo aur left block ko RREF tak drive karo; right block $A^{-1}$ ban jaata hai: $\left[\begin{array}{cccc}2& 1& 1& 0\\ 5& 3& 0& 1\end{array}\right]$ $R_1\to \frac{1}{2}{R}_1$: $\left[\begin{array}{cccc}1& \frac{1}{2}& \frac{1}{2}& 0\\ 5& 3& 0& 1\end{array}\right]$ $R_2\to R_2-5R_1$: $\left[\begin{array}{cccc}1& \frac{1}{2}& \frac{1}{2}& 0\\ 0& \frac{1}{2}& -\frac{5}{2}& 1\end{array}\right]$ $R_2\to 2R_2$: $\left[\begin{array}{cccc}1& \frac{1}{2}& \frac{1}{2}& 0\\ 0& 1& -5& 2\end{array}\right]$ $R_1\to R_1-\frac{1}{2}{R}_2$: $\left[\begin{array}{cccc}1& 0& 3& -1\\ 0& 1& -5& 2\end{array}\right]$ Toh $A^{-1}=\left[\begin{array}{cc}3& -1\\ -5& 2\end{array}\right]$. Check: $A{A}^{-1}=\left[\begin{array}{cc}2& 1\\ 5& 3\end{array}\right]\left[\begin{array}{cc}3& -1\\ -5& 2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ ✅.

Ek free variable ⇒ rank $2$, columns $1,3$ mein pivots (toh column $2$, yaani $y$, free hai). Ek clean RREF jo yeh meet karta hai: $\left[\begin{array}{cccc}1& 2& 0& 5\\ 0& 0& 1& 3\\ 0& 0& 0& 0\end{array}\right]$

• Pivots: columns $1,3$ ✅; column $2$ pivot-free → $y$ free ✅.
• Last (constant) column mein koi pivot nahi → consistent ✅; rank $2<3$ variables → one-parameter family ✅. Solution: row 1 se, $x=5-2y$; row 2 se, $z=3$; $y$ free: $(x,y,z)=(5-2y,y,3).$ Ek valid original system (un-reduce karo): $x+2y=5,z=3,0=0$ — ya ise dress up karo as $x+2y=5,2x+4y+z=13,z=3$.

Setup. Ek square matrix ka RREF ya toh $I_n$ hota hai (har column mein ek pivot) ya usme kam se kam ek pivot-free column hota hai (aur isliye kam se kam ek zero row, kyunki $n$ pivots fit nahi ho sakte agar koi column skip ho). Yeh ek square matrix ke liye sirf do possibilities hain.

($\Leftarrow$) RREF $=I_n\Rightarrow$ invertible. Har row operation ek elementary matrix $E_k$ se left-multiplication ke barabar hai, aur har $E_k$ invertible hai. Agar $E_m\cdots E_{1}A=I_n$, toh $A=(E_m\cdots E_1{)}^{-1}=E_1^{-1}\cdots E_m^{-1}$, invertibles ka product → $A$ invertible, $A^{-1}=E_m\cdots E_1$ ke saath.

($\Rightarrow$) invertible $\Rightarrow$ RREF $=I_n$. Suppose nahi: RREF $R$ mein ek zero row hai. Toh $\det R=0$, lekin $R=E_m\cdots E_{1}A$ saare invertible factors ke saath, toh $\det R=(\prod \det E_k)\det A$. Kyunki har $\det E_k\ne 0$ aur (assumption se) $\det A\ne 0$, product $\det R\ne 0$ — contradiction zero row ke saath jo $\det R=0$ force karta hai. Isliye RREF mein koi zero row nahi → har column mein pivot → RREF $=I_n$. $■$

Neeche har line Prompt ::: Answer format mein likhi hai — ::: ke baad waala part cover karo aur khud test karo, phir reveal karo. Maan lo $r$ = coefficient block ka rank, $r^{\prime }$ = augmented matrix ka rank, $n$ = unknowns ki sankhya. Agar $r^{\prime }>r$ (last column mein pivot) ::: koi solution nahi (inconsistent) Agar $r^{\prime }=r=n$ ::: unique solution Agar $r^{\prime }=r<n$ ::: infinitely many, $n-r$ free variables ke saath