Pivot positions, free variables

4.5.11 · Maths › Linear Algebra (Full)

WHAT are pivots and free variables?

WHY this distinction exists: Row reduction ke baad, har woh equation jo abhi bhi "zinda" hai tumhe exactly ek leading variable solve karne deti hai baaki sab ke terms mein. Jo columns kabhi leading 1 nahi kamaa paate, woh kabhi solve nahi hote — woh parameters ki tarah rehte hain.

HOW to find them (the procedure)

1. $A$ (ya augmented $[A\mid \mathbf{b}]$) ko echelon form mein row-reduce karo.
2. Har nonzero row ki pehli nonzero entry ek pivot position mark karti hai → uski column ek pivot column hai.
3. Pivot columns ⇒ pivot variables. Non-pivot columns ⇒ free variables.
4. Count karo: agar $A$ $m\times n$ hai aur $r$ pivots hain, toh $\#\text{free variables}=n-r.$

WHY it controls the solution set

Worked Example 1 — one free variable (a line)

Solve karo $\left[\begin{array}{ccc}1& 2& 1\\ 2& 4& 3\end{array}\right]\mathbf{x}=\left[\begin{array}{c}1\\ 3\end{array}\right].$

Step 1. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{cccc}1& 2& 1& 1\\ 0& 0& 1& 1\end{array}\right].$ Yeh step kyun? Pehle pivot ke neeche zero karte hain taaki echelon structure expose ho sake.

Step 2. Pivots column 1 aur 3 mein hain → $x_1,x_3$ pivot variables hain; column 2 mein koi pivot nahi → $x_2$ free hai. Kyun? Column 2 ko kabhi leading entry nahi milti.

Step 3. Back-substitute karo. Row 2: $x_3=1$. Row 1: $x_1=1-2x_2-x_3=-2x_2$. Kyun? Har pivot variable ko free wale ke terms mein solve karo.

Result: $x_2=t$ ke saath, $\mathbf{x}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]+t\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right].$ $n-r=3-2=1$ free variable ⇒ ek line. ✓

Worked Example 2 — no free variables (unique solution)

$\left[\begin{array}{cc}1& 1\\ 0& 2\end{array}\right]\mathbf{x}=\left[\begin{array}{c}3\\ 4\end{array}\right].$

Step 1. $R_2\leftarrow \frac{1}{2}{R}_2$: $x_2=2$. Kyun? Leading entry ko 1 banao. Step 2. Dono columns pivot columns hain → koi free variable nahi. $n-r=2-2=0$. Step 3. $x_1=3-x_2=1$. Unique solution $(1,2)$. ✓

Worked Example 3 — inconsistent (free count irrelevant)

$\left[\begin{array}{ccc}1& -1& 2\\ 0& 0& 5\end{array}\right].$ Row 2 kehti hai $0=5$. Augmented column ek pivot column hai ⇒ koi solution nahi. Yeh kyun matter karta hai: free variables solution set ki dimension batate hain sirf tab jab solution exist karta ho. Pehle hamesha consistency check karo.

Common Mistakes (Steel-manned)

Flashcards

Connections

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

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions
• D5 · Question bank — concept traps