Cramer's rule: Question bank — Hinglish deep dive

4.5.24 · D5 · Maths › Linear Algebra (Full) › Cramer's rule

Shuru karne se pehle, teen words jinhe hum rely karenge, har ek ke saath ek picture tumhare dimag mein pin karenge:

Sahi hai ya galat — justify karo

Neeche diya hua har claim ya to sahi hai ya galat. Reveal reason deta hai, jo asli point hai.

Cramer's rule kisi bhi square system ke liye kaam karta hai

Galat. Ise

det A \neq = 0

chahiye; warna box flat hai (columns dependent hain) aur tum zero se divide kar rahe hoge — koi unique recipe exist nahi karti.

Agar

det A \neq = 0

hai to Cramer's rule se mila solution ek hi solution hai

Sahi.

det A \neq = 0

ka matlab hai

A

invertible hai (dekho Invertible Matrix Theorem), isliye

A x = b

ka exactly ek solution hai, aur Cramer's rule usse name karta hai.

Cramer's rule ke liye

A

ka square hona zaroori hai

Sahi. Tumhe

n \times n

matrices ke

n

determinants chahiye; ek non-square

A

ka determinant hota hi nahi, isliye ratio undefined hai.

Target

b

ko double karne par har

x_{i}

double ho jaata hai

Sahi. Har

A_{i}

mein

b

ek column hai, aur

det

us column mein linear hai, isliye

det A_{i}

double ho jaata hai jabki

det A

unchanged rehta hai — har ratio double ho jaata hai.

Ek equation ko double karna (dono

A

aur

b

ki ek row) solution change kar deta hai

Galat. Ek poori equation ko scale karna ek equivalent system par ek valid row operation hai; yeh

det A_{i}

aur

det A

dono ko same factor se multiply karta hai, isliye har ratio

x_{i}

unchanged rehta hai.

det A_{i}

aur

det A

ka sign hamesha same hota hai

Galat. Unka ratio

x_{i}

negative ho sakta hai, isliye signs alag ho sakte hain. Sirf ratio system se fixed hai, individual signs nahi.

Cramer's rule computer par bade

50 \times 50

system ko solve karne ka sabse fast tarika hai

Galat. Ise

n + 1

determinants chahiye; honestly kiya jaaye to yeh Gaussian Elimination se kahin zyada slow aur numerically unstable hai. Cramer proofs aur tiny/symbolic systems ke liye shine karta hai.

Agar

b = 0

aur

det A \neq = 0

hai, to Cramer's rule

x = 0

deta hai

Sahi. Phir har

A_{i}

mein ek zero column hoga, isliye

det A_{i} = 0

, jo

x_{i} = 0/ det A = 0

deta hai — invertible

A

wale homogeneous system ka yahi ek solution hai.

Cramer's rule aur

A^{- 1} b

alag answers de sakte hain

Galat. Unhe agree karna hi hoga: Cramer literally

A^{- 1} = \frac{1}{d e t A} adj (A)

(dekho Matrix Inverse) hai jo

b

par apply hota hai, column by column unpack hoke.

Why questions

X_{i}

(identity jisme column

i

x

se replace kiya) banane par

det X_{i} = x_{i}

kyun hota hai?

Baaki saare columns basis vectors hain jo factor

1

contribute karte hain; expand karne par sirf diagonal entry

x_{i}

bachti hai. Yeh

det

ki multilinearity ka kaam hai.

Kyun

A

ka ek column

b

se replace hota hai, row nahi?

Kyunki

A x

A

ke columns ka combination hai; jab

X_{i}

mein column

i

mein

x

hota hai, to product

A X_{i}

exactly us column ko

A x = b

se swap kar deta hai.

Geometrically,

det A \neq = 0

condition kyun zaroori hai?

det A = 0

ka matlab hai column arrows ek lower dimension mein squash ho gaye hain (flat box). Tab

b

ya to reach nahi hota ya kai tarakon se reach hota hai — koi single recipe nahi, isliye koi unique

x_{i}

nahi.

Tiny

det A

answer ko numerically dangerous kyun banata hai?

Tum near-zero number se divide kar rahe ho, isliye

det A_{i}

det A

mein choti si errors hugely amplify ho jaati hain — solution ill-conditioned hai, yahi ek key reason hai ki practice mein Gaussian Elimination prefer karte hain.

Proof mein

det (M N) = det M det N

kyun aata hai?

A X_{i} = A_{i}

ke determinants lene ke liye

det (A X_{i})

det A \cdot det X_{i}

mein split karna padta hai. Multiplicativity exactly wahi split hai, matrix identity ko number equation

det A \cdot x_{i} = det A_{i}

mein convert karta hai.

Cramer's rule computation se zyada theory ke liye kyun suited hai?

Yeh har unknown ko ek clean closed-form ratio deta hai — facts prove karne ke liye perfect (jaise solutions ki entries mein smoothness). Lekin kai determinants compute karna costly hai, isliye real numbers par elimination se haarta hai.

Edge cases

b

A

ke columns mein se ek ke equal hai, maano

b = a_{1}

— Cramer kya deta hai?

Tab

x_{1} = 1

aur baaki saare

x_{i} = 0

i \neq = 1

ke liye,

A_{i}

mein do equal columns hain (

a_{1}

do baar aata hai) isliye

det A_{i} = 0

;

i = 1

ke liye,

A_{1} = A

deta hai

x_{1} = 1

A

identity matrix hai — solution kya hai?

A_{i}

identity ban jaata hai jisme column

i

b

se set hai, isliye

det A_{i} = b_{i}

aur

det A = 1

; thus

x_{i} = b_{i}

, yaani

x = b

, exactly jaisa

I x = b

demand karta hai.

1 \times 1

system

a x = b

— kya Cramer's rule tab bhi sense karta hai?

Haan:

det A = a

A_{1} = (b)

isliye

det A_{1} = b

, deta hai

x = b / a

— ordinary division, exactly tab valid jab

a \neq = 0

A

ke do columns identical hain — kya hota hai?

det A = 0

(repeated column ⇒ flat box), isliye Cramer's rule void hai. Geometrically do ingredients same hain, isliye

b

tak pahunchne ki recipe unique nahi hai (ya impossible hai).

System ka koi solution nahi hai (inconsistent) — kya Cramer detect karega?

Sirf indirectly: inconsistency ke liye

det A = 0

zaroori hai, isliye formula pehle se run karne se refuse karta hai. Cramer genuinely square inconsistent system ke liye kabhi fake answer nahi return karta, kyunki yeh division-by-zero gate par ruk jaata hai.

b

ki ek entry zero hai — kya yeh problem hai?

Nahi.

b

ki ek zero entry theek hai; sirf zero determinant

det A

rule ko todta hai. Column

b

mein freely zeros ho sakte hain.

Recall Har trap ki ek-line summary

In almost sabke peeche ek hi guardrail hai ::: Ek column ko $b$ se replace karo, aur rule sirf tab run karo jab $det A \neq = 0$ ho — baaki sab $det$ ke signed, multilinear, alternating volume hone se follow hota hai.

Connections

Cramer's rule — woh parent jise yeh bank stress-test karta hai.
Determinants — signed volume, har sign trap ka source.
Matrix Inverse — kyun Cramer aur $A^{- 1} b$ agree karte hain.
Cofactor Expansion — jahan alternating signs bite karte hain.
Gaussian Elimination — practical alternative jab $det A$ tiny ho ya $n$ bada ho.
Multilinear and Alternating Maps — abstract engine.
Invertible Matrix Theorem — $det A \neq = 0 ⟺$ unique solution.

Question bank — Cramer's rule