Pivot positions, free variables: Question bank — Hinglish deep dive

4.5.11 · D5 · Maths › Linear Algebra (Full) › Pivot positions, free variables

Shuru karne se pehle, teen words jo neeche har jagah dikhai dete hain, har ek parent note se liya gaya hai:

Pivot column = woh column jo ek leading entry (row ka pehla nonzero number) host karta hai jab matrix fully reduce ho jaati hai. Ek staircase imagine karo: har step ek column mein land karta hai, aur woh column ek pivot column hai.
Free variable = woh variable jiske column mein koi step land nahi karta — tum apni marzi se uski value choose kar sakte ho.
Rank $r$ = pivots ki sankhya = staircase mein stairs ki sankhya.
$n$ = columns ki sankhya = variables ki sankhya. Master count hai $free = n - r$ .

True or false — justify

TF1. Ek matrix jisme rows zyada hain columns se (

m > n

) mein kam se kam ek free variable zaroori hai.

False — "extra rows" reduce hone ke baad zero rows ban jaate hain, free variables nahi; agar saare

n

columns pivot kar lein (

r = n

) toh

n - r = 0

free variables hote hain. Free count hai

n - r

, kabhi bhi row count se nahi judta.

TF2. Agar ek system mein ek free variable hai, toh uske infinitely many solutions hain.

Sirf tab jab woh consistent ho. Ek inconsistent system (ek row

[0 \dots 0 ∣ c]

c \neq = 0

) ke zero solutions hote hain chahe free columns wahan baithe hon.

TF3. Usi

A

ke do alag echelon forms pivots ko alag-alag columns mein rakh sakte hain.

False — intermediate echelon forms alag hoti hain, lekin RREF unique hoti hai, isliye pivot positions (kaun se columns) sirf

A

se determine hote hain.

TF4. Ek square

n \times n

matrix mein hamesha exactly

n

pivots hote hain.

False — sirf ek invertible matrix mein hote hain. Ek singular square matrix mein

r < n

hota hai, isliye

n - r \geq 1

free variables aur

A x = 0

ke solutions ka ek poora flat hota hai.

TF5. Right-hand side

b

add karne se

A

ke kaun se columns pivot columns hain, yeh badal sakta hai.

A

ke columns ke liye False — woh pivots sirf

A

par depend karte hain.

b

jo kar sakta hai woh hai augmented column ko pivot column banana, jo inconsistency signal karta hai, koi naya free variable nahi.

TF6. Free variables ki sankhya echelon form mein zero rows ki sankhya ke barabar hoti hai.

False — free variables non-pivot columns count karte hain (

n - r

), jo ek column fact hai; zero rows ek row fact hain. Yeh sirf coincidence se milte hain.

TF7. Agar

A

ka full column rank hai (

r = n

), toh equation

A x = 0

ka sirf trivial solution hai.

True —

n - r = 0

free variables ka matlab hai kuch bhi roam nahi kar sakta, isliye

x = 0

forced hai. Yeh exactly column-independence condition hai.

TF8. Augmented column mein ek pivot ka matlab hai system ka ek unique solution hai.

False — augmented column mein pivot row

0 = c \neq = 0

hai; iska matlab hai koi solution nahi hai. Uniqueness

A

ke saare columns ke pivot hone ke baare mein hai, yeh ek alag baat hai.

TF9. Har free variable solution set mein ek dimension add karta hai.

Tab True jab consistent ho —

k

free variables ek

k

-dimensional flat dete hain: point (

k = 0

), line (

k = 1

), plane (

k = 2

), null-space dimension se match karta hai.

TF10. Agar

A

ka last column ek non-pivot column hai, toh

x_{n}

free hai.

True — "non-pivot column" hi definition hai free variable ke column ki. Position (pehla, aakhri, beech mein) irrelevant hai; sirf pivot-vs-not matter karta hai.

Why questions

WHY1. Har pivot exactly ek variable ke liye kyun solve karta hai, zyada nahi?

Ek leading entry ek column mein baith ti hai; reduced row us ek variable ko baaki ke terms mein express karta hai, exactly ek column aur ek equation use karke — yahi cheez ise ek rank-one contribution banati hai.

WHY2.

free = n - r

ek subtraction kyun hai aur kuch zyada complicated kyun nahi?

Har column ya toh ek pivot column hai (

r

unke) ya nahi; koi teesra option nahi hai. Isliye bacha hua count sirf total minus pivots hai — pure bookkeeping.

WHY3. Consistency ko free variables count karne se alag kyun check karna padta hai?

Count

n - r

solution flat ki shape describe karta hai, lekin ek flat empty bhi ho sakti hai. Consistency (koi augmented-column pivot nahi) decide karti hai ki woh flat populated hai ya nahi.

WHY4. Pivot positions unique kyun hote hain chahe echelon forms nahi hoti?

A

ki saari echelon forms aur reduce hokar ek aur sirf ek RREF tak pahunchti hain; pivot positions us final reduction mein unchanged survive karti hain, isliye woh

A

ki ek fixed property hain.

WHY5. Full column rank at most ek solution kyun guarantee karta hai?

r = n

force karta hai

n - r = 0

free variables, isliye vary karne ke liye kuch bachta nahi — koi bhi solution completely pinned ho jaata hai, uniqueness deta hai (existence ke liye consistency abhi bhi chahiye).

WHY6. Free variables ki sankhya null space ki dimension ke barabar kyun hoti hai?

Har free variable ek independent parameter hai jo tum dial kar sakte ho; ek ko

1

aur baaki ko

0

set karke null space ka ek basis vector banta hai, isliye free count = null-space dimension =

n - r

WHY7. Right-hand vector

b

kabhi bhi

A

ke andar ek naya pivot kyun nahi bana sakta?

[A ∣ b]

ka row reduction rows par operate karta hai;

A

ke pivot columns sirf

A

ki apni structure se set hote hain.

b

sirf extra column occupy karta hai, jahan woh at most ek inconsistency flag ban sakta hai.

Edge cases

EC1. Zero matrix

A = 0

(size

m \times n

A x = 0

ke liye kitne free variables hain?

Rank

r = 0

, isliye free

= n - 0 = n

— har variable free hai, aur solution set poora

R^{n}

hai.

EC2. Ek

1 \times 1

matrix

[0]

jisme

A x = [b]

: kya hota hai?

Agar

b = 0

hai, toh

x

free hai (koi bhi value kaam karta hai,

R^{1}

mein ek line); agar

b \neq = 0

hai toh single row padhti hai

0 = b

, inconsistent — koi solution nahi.

EC3. Ek wide matrix (

n > m

) jo consistent hai: kya uska kabhi unique solution ho sakta hai?

Nahi — at most

m

pivots hain lekin

n > m

columns hain, isliye

n - r \geq n - m \geq 1

free variables hamesha bache rehte hain; consistency ke saath phir infinitely many solutions hote hain, kabhi ek nahi.

EC4. Coefficient matrix ke roop mein ek identity matrix

I_{n}

: kitne free variables hain?

Har column pivot karta hai,

r = n

, isliye

n - r = 0

free variables — solution ek akela forced point hai.

EC5. Ek single equation

0 x_{1} + 0 x_{2} = 0

: solution set describe karo.

Dono columns non-pivot hain (rank

0

), isliye dono variables free hain — solution set poora plane

R^{2}

hai, ek

2

-dimensional flat.

EC6. Ek tall matrix (

m > n

) jisme

r = n

hai lekin ek nonzero augmented-column pivot hai: consistent hai ya nahi?

Inconsistent — full column rank hone ke bawajood (jo agar solvable hota toh uniqueness deta), augmented-column pivot ek

0 = c

row force karta hai, isliye zero solutions hain.

EC7. Ek homogeneous system

A x = 0

jisme

r < n

hai: kya woh inconsistent ho sakta hai?

Kabhi nahi —

x = 0

hamesha ise solve karta hai, isliye woh hamesha consistent hai;

r < n

ke saath iske infinitely many solutions hote hain jo origin se guzarti ek flat banate hain.

Recall One-line self-test

Woh ek sentence jo yahan ke zyaadatar traps ko dissolve kar deta hai ::: Free variables $= n - r$ non-pivot columns se aate hain, woh solution set ki dimension describe karte hain sirf consistency confirm hone ke baad, aur pivot positions sirf $A$ ke hote hain.

Connections

Pivot positions, free variables
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

Question bank — Pivot positions, free variables