Systems of linear equations — matrix form Ax = b: Question bank

4.5.8 · D5 · Maths › Linear Algebra (Full) › Systems of linear equations — matrix form Ax = b

Notation reminder jo poore page mein use hogi (sab parent mein defined hain):

$A$ ek $m \times n$ coefficient matrix hai — $m$ rows (equations), $n$ columns (unknowns).
$x$ unknowns ka vector hai, $b$ right-hand side hai.
$[A ∣ b]$ augmented matrix hai: $A$ ke right side pe column $b$ chipka ke.
$rank$ independent rows/columns count karta hai; $Col (A)$ columns ka span hai; null space woh saare $x$ hain jiske liye $A x = 0$ ho.

True or false — justify

TF1. "Agar

A

m \times n

matrix hai jisme

m > n

(unknowns se zyada equations), to system ka koi solution nahi hoga."

False — solvability rank pe depend karti hai, row count pe nahi; agar extra rows dependent hain, to

b

phir bhi

Col (A)

mein ho sakta hai aur solution exist kar sakta hai. Dekho Rank of a matrix.

TF2. "

A x = 0

inconsistent ho sakta hai (koi solution nahi ho sakta)."

False —

x = 0

hamesha

A x = 0

satisfy karta hai, isliye ek homogeneous system kabhi inconsistent nahi hota; sirf sawaal ye hai ki nonzero solutions bhi exist karte hain ya nahi.

TF3. "Agar

det A = 0

A x = b

ka koi solution nahi."

False —

det A = 0

sirf unique solution ko khatam karta hai; system ke paas phir bhi infinitely many solutions ho sakte hain (agar

b \in Col (A)

) ya koi nahi (agar nahi hai). Dekho Determinant.

TF4. "Agar

A x = b

ke do alag solutions hain, to uske infinitely many solutions hain."

True — agar

A x_{1} = A x_{2} = b

A (x_{1} - x_{2}) = 0

, isliye ek nonzero null-space vector exist karta hai, aur

x_{1} + t (x_{1} - x_{2})

har real

t

ke liye ise solve karta hai.

TF5. "Ek square

A

ke liye,

A x = b

ka har every

b

ke liye unique solution hoga iff

A

invertible hai."

True — invertibility ka matlab hai columns saare

R^{n}

ko span karte hain aur independent hain, isliye har

b

exactly ek baar hit hota hai. Dekho Inverse of a matrix.

TF6. "Ek redundant equation add karna (kisi existing ki copy) solution set ko change kar sakta hai."

False — ek duplicate row

rank (A)

aur

rank ([A ∣ b])

dono ko unchanged chhod deti hai, isliye solution set identical rahta hai.

TF7. "Agar

A

ke columns linearly independent hain, to

A x = b

ka solution (jab exist kare) unique hoga."

True — independent columns ka matlab null space sirf

{0}

hai, isliye do alag inputs ek hi

b

pe map nahi kar sakte. Dekho Linear independence.

TF8. "

A x = b

solvable hone ka matlab hai

b

A

ki rows ka linear combination likha ja sakta hai."

False — iska matlab hai

b

A

ke columns ka combination hai (column view); rows bilkul alag space mein rehti hain.

TF9. "Agar

rank (A) = rank ([A ∣ b])

, to solution automatically unique hoga."

False — ye sirf existence guarantee karta hai; uniqueness ke liye additionally us common rank ka

n

(unknowns ki sankhya) ke barabar hona zaroori hai.

Spot the error

SE1. "

A x = b \Rightarrow x = b / A

, dono sides ko

A

se divide karke."

Matrix division hoti hi nahi; legal move hai left side se

A^{- 1}

multiply karna (sirf tab jab

A

square aur invertible ho), jisse ==

x = A^{- 1} b

== milta hai.

SE2. "

A x = b

, isliye

x = b A^{- 1}

Order galat hai — tumhe left se multiply karna hoga:

A^{- 1} A x = A^{- 1} b

x = A^{- 1} b

milta hai. Matrix products commute nahi karte, isliye

b A^{- 1}

yahan well-shaped bhi nahi hai.

SE3. "System mein 3 equations aur 3 unknowns hain, isliye iska unique solution hai."

Square shape akele kuch guarantee nahi karta; tumhe

det A \neq = 0

chahiye (equivalently full rank). Ek singular

3 \times 3

system ke infinitely many ya koi bhi solution nahi ho sakte.

SE4. "

A

2 \times 3

matrix hai, aur

det A = 0

, isliye koi unique solution nahi hai."

Determinant non-square matrices ke liye undefined hota hai; rank use karo. Yahan

n = 3

lekin rank

\leq 2 < n

hai, isliye unique solution kabhi bhi nahi ho sakta, chahe kuch bhi ho.

SE5. "Do solutions

x_{p}

aur

x_{p}^{'}

exist karte hain, isliye general solution

x_{p} + x_{p}^{'}

hai."

Galat structure — general solution

x_{p} + x_{h}

hai jahan

x_{h}

null space pe range karta hai; do particular solutions add karne se

A x = 2 b

milta hai,

b

nahi.

SE6. "rank

(A) = 2

, rank

([A ∣ b]) = 2

n = 4

, isliye

2

free parameters hain — yaani rank jitne."

Free parameters

n - r = 4 - 2 = 2

hain yahan coincidence se; rule hai ==

n - r

==, columns ki sankhya minus rank, rank itself nahi. (Jaise

r = 3, n = 4

1

free parameter milta hai,

3

nahi.)

SE7. "Kyunki

A x = b

consistent hai aur

A x = 0

ka sirf trivial solution hai, hum uniqueness ke baare mein kuch conclude nahi kar sakte."

Actually ye uniqueness force karta hai: trivial null space + existence ⇒ exactly ek solution, kyunki koi bhi do solutions null-space vector se differ karte.

Why questions

WHY1. Matrix multiplication "row-

i

dotted with

x

" ki tarah kyun define ki gayi hai na ki kisi aur rule se?

Kyunki wahi exact definition

(A x)_{i} = \sum_{j} a_{ij} x_{j}

banati hai jo equation

i

ko verbatim reproduce karti hai, isliye

A x = b

original system ko compress karta hai. Dekho Matrix multiplication.

WHY2. Rank — na ki equations ki sankhya — existence aur uniqueness kyun decide karta hai?

Rank independent constraints/directions ki sankhya count karta hai; redundant rows koi information add nahi karte, isliye sirf independent constraints unknowns ko pin down ya contradict kar sakte hain.

WHY3.

b

augmented rank ko

rank (A)

se upar raise kare to "no solution" kyun hota hai?

Iska matlab hai

b

ek aisi direction mein point karta hai jo columns reach nahi kar sakte (ek nayi independent direction), isliye

x

ki koi bhi weighting columns se

b

kabhi nahi bana sakti —

b \in / Col (A)

WHY4. General solution

x_{p} + x_{h}

kyun hota hai, sirf

x_{p}

kyun nahi?

Kyunki

A (x_{p} + x_{h}) = A x_{p} + A x_{h} = b + 0 = b

kisi bhi null-space vector

x_{h}

ke liye, isliye har particular solution poore null space se shift hota hai. Dekho Column space and null space.

WHY5.

x = A^{- 1} b

sirf square invertible

A

ke liye kyun kaam karta hai?

A^{- 1}

sirf tab exist karta hai jab

A

square ho aur independent columns hon; warna

A

ya information lose karta hai (undo nahi ho sakta) ya one-to-one map nahi hai, isliye koi inverse exist nahi karta.

WHY6. Hum

A x

ko "columns ka linear combination" kyun bol sakte hain instead of dot products compute karne ke?

Terms

a_{ij} x_{j}

ko column

j

ke hisaab se group karne se

x_{j}

times column

a_{(j)}

milta hai, isliye

A x = \sum_{j} x_{j} a_{(j)}

— weights literally unknowns hain.

WHY7. Nonzero null space hona uniqueness kyun destroy karta hai lekin existence nahi?

Ek nonzero

x_{h}

ko kisi bhi solution mein add karke doosra banaya ja sakta hai, uniqueness khatam ho jaati hai; lekin ye kuch nahi kehta ki

b

pehle jagah reachable hai ya nahi, jo ek alag (existence) sawaal hai.

Edge cases

EC1.

A x = b

kya describe karta hai jab

A

zero matrix hai aur

b = 0

Har vector

x

ise solve karta hai (constraint

0 = 0

vacuous hai), isliye solution set poora

R^{n}

hai — maximally-infinite case.

EC2. Kya hoga agar

A

zero matrix ho lekin

b \neq = 0

Koi solution nahi —

A x = 0

har

x

ke liye, isliye ye kabhi nonzero

b

ke barabar nahi ho sakta; yahan

rank (A) = 0 < rank ([A ∣ b]) = 1

EC3. Column view mein,

x = 0

(saare weights zero) hamesha kya produce karta hai?

Zero vector, kyunki

\sum_{j} 0 \cdot a_{(j)} = 0

— isliye

A x = 0

hamesha consistent hota hai.

EC4. Ek single equation

x + y + z = 2

(to

m = 1, n = 3

): kitne free parameters hain aur kyun?

n - r = 3 - 1 = 2

free parameters — 3D mein ek plane ek constraint hai (rank 1), jo do directions ko plane ke andar freely slide karne ke liye chhod deta hai.

EC5. Agar

m = n

lekin

A

ki do rows identical hain, to kya system square aur infinitely many solutions wala ho sakta hai?

Haan — identical rows rank ko

n

se neeche gira deti hain (isliye

det A = 0

), aur agar

b

dependence ke saath consistent hai, to square shape ke bawajood infinitely many solutions milte hain.

EC6.

A x = b

ke solution set ka kya hoga agar tum ek poori equation ko nonzero constant

c

se scale karo?

Kuch nahi —

[A ∣ b]

ki ek row ko

c \neq = 0

se multiply karna dono ranks aur solution set ko unchanged chhod deta hai; ye ek reversible row operation hai. Dekho Gaussian elimination.

EC7. Kya ek

n \times n

homogeneous system

A x = 0

ke exactly do solutions ho sakte hain (aur koi nahi)?

Nahi —

A x = 0

ke solutions ek subspace form karte hain, isliye ya to sirf

{0}

hai (ek) ya ek nonzero vector contain karta hai aur hence infinitely many (uske saare scalar multiples).

Connections

Systems of linear equations — matrix form Ax = b — woh parent jise ye bank drill karta hai.
Rank of a matrix — woh akela tool jo yahan zyaadatar traps resolve karta hai.
Column space and null space — existence ek mein rehti hai, uniqueness doosre mein.
Determinant — sirf square ke liye; " $det = 0 \Rightarrow$ no solution" trap ka source.
Inverse of a matrix — $x = A^{- 1} b$ ko legitimate karta hai aur koi doosra rearrangement nahi.
Linear independence — independent columns $⟺$ trivial null space $⟺$ uniqueness.
Gaussian elimination — woh row operations jo solution set kabhi change nahi karte.
Matrix multiplication — kyun $A x = b$ system hai, koi nayi rule nahi.