Solving systems using matrix inversion: Question bank

2.6.12 · D5 · Maths › Matrices & Determinants — Introduction › Solving systems using matrix inversion

Yeh page misconceptions ko dhundhta hai, arithmetic ko nahi. Har item ek one-line reveal hai: prompt padho, apna jawab zor se bolo, phir check karo. Agar tumhara reason reveal se alag hai — chahe verdict same ho — toh wahi gap lesson hai.

Parent: matrix inversion method. Prerequisite gates: Determinants, Adjoint and Inverse of a Matrix, Consistency of Linear Systems.

Intuition Neeche diye har trap ke peeche ek hi idea — aur uski picture

$A$ ek machine hai jo ek input vector $X$ ko output $B$ mein badal deti hai. Solve karne ka matlab hai machine ko ulta chalana. Ulta tabhi kaam karta hai jab machine kabhi do alag inputs ko same output par squash na kare — aur yahi squashing hai jo $== det A = 0 ==$ detect karta hai. Figure dekho: $A$ chhote unit square ko (do basis arrows se bana) ek parallelogram mein badal deta hai. $det A$ us parallelogram ka signed area hai. Jab machine square ko ek line par flatten kar deti hai, area $0$ ho jaata hai — yahi $det A = 0$ hai, aur machine ab reverse nahi ho sakti.

Is page par almost har trap ya toh us area-gate ko check na karne ki disguised failure hai, ya yeh bhool jana ki matrix multiplication ka side hota hai (left vs right matter karta hai).

True ya false — justify karo

Yahan har statement ya toh subtly true hai ya subtly false. Reveal reason deta hai, yahi point hai.

Equation

A X = B

sirf original system of equations ko rewrite karna hai, koi naya object nahi.

True —

A

ki row

i

ko column

X

se multiply karne par equation

i

ka left side milta hai, aur use

b_{i}

ke equal karne par exactly equation

i

milti hai. Yeh repackaging hai, approximation nahi.

Agar

A^{- 1}

exist karta hai, toh

X = A^{- 1} B

ek valid solution hai lekin aur bhi ho sakte hain.

False — jab

A^{- 1}

exist karta hai toh solution unique hota hai. Agar

A X_{1} = B

aur

A X_{2} = B

, toh left-multiply karo

A^{- 1}

se aur

X_{1} = X_{2}

milega, toh koi doosra solution chhup nahi sakta.

det A \neq = 0

guarantee karta hai ki solution exist karta hai;

det A = 0

guarantee karta hai ki koi solution nahi hai.

Pehla aadha true hai, doosra false hai.

det A = 0

ka matlab hai no unique solution — phir bhi infinitely many ho sakte hain, iske upar depend karta hai ki

(adj A) B = O

hai ya nahi (zero column).

2 \times 2

matrix ke liye, "diagonal swap karo aur off-diagonal negate karo" seedha inverse deta hai.

False — woh operation adjugate deta hai, inverse nahi. Tumhe phir bhi

det A = a d - b c

se divide karna hoga; division skip karne par ek scalar factor se galat answer aata hai.

A X = B

aur

X A = B

same system describe karte hain.

False — matrix multiplication commutative nahi hai, aur column vector

X

ke liye product

X A

aksar defined bhi nahi hota. Order aur side dono ka matlab hota hai.

Agar ek system mein infinitely many solutions hain, toh uske coefficient matrix ka

det A = 0

hona zaroori hai.

True — non-zero determinant ek unique solution force karta hai, toh infinitely many solutions tabhi ho sakte hain jab inverse exist na kare, yani

det A = 0

Identity

A adj (A) = (det A) I

tab bhi hold karta hai jab

det A = 0

ho.

True — yeh ek algebraic identity hai jo har square matrix ke liye valid hai. Jab

det A = 0

ho toh yeh simply kehta hai

A adj (A) = O

(zero matrix), jo consistency test ko power deta hai.

Do alag linear systems same coefficient matrix

A

share kar sakte hain lekin alag solvability ho sakti hai.

True — same

A

ka matlab same

det A

hai, toh dono unique honge ya dono degenerate; lekin agar

det A = 0

hai, toh

B

ki value decide karti hai "no solution" hai ya "infinitely many". Degenerate case mein solvability sirf

A

se determine nahi hoti.

Error dhundho

Har line mein ek claimed step ya fact hai. Dhundho kya galat hai aur kyun.

A X = B

se, dono sides ko

A

se divide karo aur

X = B / A

milega."

Matrix division hota hi nahi. Legal move hai left-multiply by

A^{- 1}

, jisse

X = A^{- 1} B

milta hai; notation

B / A

meaningless hai kyunki yeh chhupa deta hai ki inverse kis side par baithta hai.

A X = B

se, hum

A^{- 1}

se left-multiply karte hain aur

X = B A^{- 1}

milta hai."

Verdict "left-multiply" sahi hai lekin result galat hai:

B

ko left-multiply karne par

A^{- 1} B

milta hai, inverse left par hota hai.

B A^{- 1}

right-multiplication hoga aur dimensionally valid bhi nahi ho sakta.

det A = 0

, toh main

A^{- 1} = \frac{1}{0} adj (A)

compute karke carefully aage badhta hun."

Zero se divide nahi kar sakte —

A^{- 1}

yahan exist nahi karta. Method khatam; tumhe

(adj A) B

use karke consistency analysis par switch karna hoga.

adj (A)

cofactors

C_{ij}

ka matrix hai jo unki natural positions mein rakhe gaye hain."

Transpose missing hai:

adj (A)_{ij} = C_{j i}

. Adjugate cofactor matrix ka transpose hota hai; ise bhoolna silently diagonal ke bahar har entry corrupt kar deta hai.

2 \times 2

[a c b d]

ke liye, inverse hai

\frac{1}{a d - b c} [a - c - b d]

Diagonal entries swap honi chahiye: hona chahiye

[d - c - b a]

a

aur

d

ko jagah par rakhne par check

A A^{- 1} = I

fail ho jaata hai.

"Maine

X = A^{- 1} B

find kar liya, toh substitute back karne ki zaroorat nahi — algebra exact hai."

Algebra tabhi exact hai jab tumne

A^{- 1}

sahi compute kiya ho; adj ya det mein ek bhi sign slip tab tak invisible hai jab tak substitute nahi karte. Check

A X = B

un slips ko saste mein pakad leta hai.

det A = 0

aur

(adj A) B = O

, isliye system mein definitely infinitely many solutions hain."

Reveal ek maybe hai:

(adj A) B = O

"no solution" verdict hatata hai lekin khud se existence prove nahi karta — tumhe substitution se check karna hoga, kyunki kuch configurations mein phir bhi inconsistent ho sakta hai.

Kyun consistency test zaroori hai lekin sufficient nahi — ek visual walk-through

Intuition Step 2 — yeh sufficient kyun nahi hai (do parallel lines vs ek line ki picture)

Figure singular case ko do straight lines ke roop mein dikhata hai (har equation ek line hai). Jab $det A = 0$ ho toh teen cheezein ho sakti hain:

Same line do baar draw ki gayi (green): us par har point dono equations solve karta hai → infinitely many solutions.
Do parallel lekin distinct lines (red): woh kabhi nahi miltiein → no solution. Dono singular pictures pehle ki "not-obviously-inconsistent" screen vector level par pass kar sakti hain, toh $(adj A) B = O$ pass karna guaranteed-inconsistent case hatata hai lekin "same line" ko subtly inconsistent case se distinguish nahi kar sakta. Yahi gap exactly hai kyun tumhe substitute back and check karna hota hai — test necessary hai, sufficient nahi.

Why questions

Reason batao, sirf fact nahi.

Hum

A^{- 1}

se right-multiply ki jagah left-multiply kyun karte hain?

Kyunki

A

A X = B

mein

X

ke left par baitha hai, aur sirf left-multiplication hi

A^{- 1} A = I

X

ke adjacent produce karta hai; right-multiplying inverse ko

A

ke galat side par stranded kar deta.

det A \neq = 0

unique solution ke liye gate ka kaam kyun karta hai?

Kyunki

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

det A

se divide karna hota hai; geometrically

det A \neq = 0

ka matlab hai unit square ek non-zero area ke parallelogram mein map hota hai (space flatten nahi hoti), toh distinct inputs distinct rehte hain aur reverse ho sakte hain.

Ek "flattening" machine (

det A = 0

) uniquely undo kyun nahi ki ja sakti?

Flattening alag-alag inputs ki poori line (ya plane) ko same output par map kar deti hai, toh output dekhkar tum nahi bata sakte kaun sa input tha — reverse machine ka koi single jawab nahi hota.

Identity

A adj (A) = (det A) I

consistency test kyun explain karta hai?

A X = B

adj (A)

se multiply karne par

(det A) X = (adj A) B

milta hai; jab

det A = 0

ho toh left side

O

hai, jo

(adj A) B = O

force karta hai, isliye us equality ka koi bhi violation prove karta hai ki koi

X

exist nahi karta.

Matrix inversion ko elimination se "mistake-proof" kyun kaha jaata hai?

Steps ek fixed mechanical recipe hain — det, adjugate, multiply — koi branching human choices nahi ki kaun si row subtract karo, isliye same input hamesha same procedure yield karta hai aur ek built-in check

A X = B

bhi hota hai.

X = A^{- 1} B

ke liye

A

square hona kyun zaroori hai?

Sirf square matrices ka ek two-sided inverse ho sakta hai jo

A^{- 1} A = A A^{- 1} = I

satisfy kare; non-square

A

equations aur unknowns ki alag-alag sankhya describe karta hai aur iss tarah uniquely reverse nahi ho sakta.

det A

se divide karna, geometrically, scaling ke roop mein kyun appear hota hai?

det A

woh factor hai jisse

A

area/volume stretch karta hai; transformation reverse karne ke liye inverse ko usi factor se shrink karna hota hai, jo precisely

\frac{1}{d e t A}

aage hota hai.

Edge cases

Woh boundary aur degenerate scenarios jo method invite karta hai. Figure do homogeneous cases ke liye concrete pictures deta hai taaki statements sirf words na rahein.

Figure — Solving systems using matrix inversion

B = O

(sab constants zero, zero column) aur

det A \neq = 0

X

kya hai?

Unique solution hai

X = A^{- 1} O = O

— trivial solution. Invertible coefficient matrix wala ek homogeneous system har variable ko zero force karta hai (figure ka left panel: do lines sirf origin par cross karti hain).

B = O

aur

det A = 0

: kya badalta hai?

(adj A) B = (adj A) O = O

automatically, toh "no solution" branch impossible hai; ek homogeneous singular system ke paas hamesha infinitely many solutions hote hain — origin se guzarti poori overlapping line (figure ka right panel).

Jab

A = I

, identity machine ho, toh

A^{- 1}

kya hai?

A^{- 1} = I

khud, kyunki

I \cdot I = I

; jo machine kuch nahi badlati usse undo karna kuch nahi karne jaisa hai, aur phir

X = I B = B

Agar

A

ki do rows identical hain, toh kya invert kar sakte hain?

Nahi — identical rows

det A = 0

force karti hain, toh machine space flatten kar deti hai aur koi inverse exist nahi karta; tumhe consistency analysis use karni hogi.

Ek single equation mein ek single unknown,

a x = b

: kya method ordinary algebra par reduce ho jaata hai?

Haan — yahan

A = [a]

det A = a

, aur

X = A^{- 1} B = \frac{b}{a}

, valid exactly jab

a \neq = 0

, jo

1 \times 1

case ke liye "zero se divide nahi kar sakte" mirror karta hai.

det A

bahut chota lekin non-zero hai (maano

1 0^{- 9}

): kya method phir bhi valid hai?

Mathematically haan — unique solution exist karta hai aur

X = A^{- 1} B

exact hai — lekin

\frac{1}{d e t A}

factor bahut bada hai, toh

B

mein tiny errors enormously amplify ho jaati hain (system "almost" un-invertible hai).

Agar

A^{- 1}

exist karta hai, toh kya

(A^{- 1})^{- 1} = A

Haan — reverse machine ko undo karna original machine restore karta hai, aur algebraically

A^{- 1} A = I

kehta hai ki

A

precisely

A^{- 1}

ka inverse hai.

Kya

det A = 0

kabhi ek unique solution ke saath coexist kar sakta hai?

Nahi —

det A = 0

ka matlab inverse fail hona hai, aur ek unique solution tumhe

A^{- 1}

reconstruct karne deta, jo ek contradiction hai. Zero determinant hamesha matlab "ya toh koi nahi ya infinitely many."

Connections

Determinants — $det A \neq = 0$ gate jis par har trap wapas aata hai.
Adjoint and Inverse of a Matrix — jahan transpose-the-cofactors ki slip rehti hai.
Consistency of Linear Systems — $det A = 0$ branch aur $(adj A) B$ test.
Cramer's Rule — same gate share karta hai, uniqueness par ek aur lens.
Cofactors and Minors — $adj (A)$ ke pieces.