Determinants — cofactor expansion along any row - column: Question bank

4.5.21 · D5 · Maths › Linear Algebra (Full) › Determinants — cofactor expansion along any row - column

Shuru karne se pehle, ek word jis par hum baar baar rely karte hain: ek minor $M_{ij}$ woh determinant hai jo row $i$ aur column $j$ delete karne ke baad bachta hai; ek cofactor $C_{ij} = (- 1)^{i + j} M_{ij}$ woh minor hai jisme ek checkerboard sign chipka diya gaya hai. Agar dono thoda bhi shaky lag raha hai, toh pehle parent ka §1 dobara padho — neeche sab kuch assume karta hai ki yeh dono tumhare dimaag mein pictures ki tarah zinda hain (ek row aur ek column ko haath se cover karo; bacha hua grid hi minor hai).

True ya False — justify karo

Sirf verdict se kuch nahi milta — reason hi asli cheez hai.

T/F. Tum ek determinant ko kisi bhi row ya kisi bhi column ke along expand kar sakte ho aur hamesha same number milega.

True. Determinant ek single Leibniz permutation sum hai (har term row aur column dono mein se ek entry pick karta hai), aur us sum ko "row-

i

ki entry kis column mein gayi" ke hisaab se group karna row-

i

expansion deta hai — grouping choice free hai, isliye sabhi lines agree karti hain.

T/F. Saari zeros wali row ke along expand karne par determinant

0

milta hai.

True. Us expansion ke har term mein (ek zero entry)

\times

(uska cofactor)

= 0

hota hai, isliye sum

0

hota hai — jo is fact se match karta hai ki zero row matrix ko singular bana deti hai.

T/F. Minor

M_{ij}

aur cofactor

C_{ij}

hamesha alag numbers hote hain.

False. Yeh sirf sign

(- 1)^{i + j}

se differ karte hain, jo

+ 1

hota hai jab bhi

i + j

even ho (jaise positions

(1, 1), (2, 2), (1, 3)

), toh wahan minor aur cofactor identical hote hain.

T/F. Expand karne se pehle do rows swap karne par determinant unchanged rehta hai, isliye rows kis order mein likhte ho koi fark nahi padta.

False. Ek row swap determinant ko

- 1

se multiply karta hai (yeh Leibniz sum mein har permutation ki parity flip karta hai). Sirf ek row ka multiple doosri row mein add karna

det

ko untouched chhoda hai.

T/F. Agar tum expand karne ke liye "wrong" row choose karo, toh zyada nonzeros hone ki wajah se tumhe ek alag (bada) number mil sakta hai.

False. Zyada nonzeros ka matlab zyada arithmetic hai, alag answer nahi. Final number invariant hai; sirf tumhari mehnat badlti hai — yahi toh poora 80/20 point hai.

T/F. Ek

3 \times 3

matrix ke liye, cofactor expansion teen

2 \times 2

determinants produce karta hai chahe koi bhi line choose karo.

Count mein True, cost mein nahi. Hamesha teen cofactor terms milte hain, lekin koi bhi term jiska entry

0

ho woh compute karne mein kuch contribute nahi karta — isliye do zeros wali line ko sirf ek real

2 \times 2

evaluation chahiye.

T/F. Sign

(- 1)^{i + j}

is baat par depend karta hai ki tum sweep ko "row expansion" kaho ya "column expansion."

False. Sign sirf original matrix mein entry ki position

(i, j)

par depend karta hai. Grid se

i

aur

j

padho; tum kis direction mein chalo yeh irrelevant hai.

T/F.

det A = det A^{T}

, isliye

A

ka column expansion,

A^{T}

ke row expansion ke barabar hai.

True. Transpose karna rows aur columns swap karta hai lekin Leibniz sum invariant rehta hai (har permutation apne inverse se map hoti hai, same sign ke saath), isliye

A

ka column-

j

expansion literally

A^{T}

ka row-

j

expansion hai.

T/F. Ek

3 \times 3

matrix ki ek row ko

5

se multiply karne par uska determinant

5

se multiply ho jaata hai.

True. Multilinearity ke zariye, determinant har row mein alag se linear hai, isliye ek row ko

k

se scale karne par

det

exactly

k

se scale hota hai (

k^{3}

se nahi).

T/F. Agar kisi matrix mein do identical rows hain, toh koi clever expansion line choose karke nonzero determinant milega.

False. Do equal rows har line ke liye

det = 0

force karti hain — unhe swap karne se kuch nahi badlta phir bhi sign flip hona chahiye, isliye

det = - det \Rightarrow det = 0

. Koi bhi expansion isse escape nahi kar sakta.

Error dhundho

Har statement mein ek specific mistake hai. Use naam do.

Ek student likhta hai

C_{12} = + M_{12}

kyunki "checkerboard symmetric hai."

Error sign mein hai: position

(1, 2)

par

i + j = 3

(odd) hai, isliye

C_{12} = (- 1)^{3} M_{12} = - M_{12}

. Checkerboard alternate karta hai, yeh sab

+

nahi hota.

"Column 2 ke along expand karne ke liye, main har entry ko column-2 signs

-, +, -

upar se neeche multiply karta hoon."

Column 2 ke neeche signs

-, +, -

yahan sahi hain, lekin diya gaya reason galat hai: yeh

(- 1)^{i + 2}

se aate hain jab

i

1, 2, 3

run karta hai, yaani entry ka actual row index — kisi memorised "column pattern" se nahi.

(i, j)

track karo, direction nahi.

3 \times 3

det

ke liye

3 \times 3 = 9

cofactors chahiye."

Ek single expansion

3

cofactors use karta hai (chosen line ki har entry ke liye ek), jisme se har ek khud ek

2 \times 2

determinant hai. Koi

9

nahi hai; recursion har level par

n

terms hai,

n^{2}

nahi.

Ek student row 2 ke along expand karta hai lekin yeh bhool jaata hai ki row

(040)

hai aur dutifully teeno cofactors compute karta hai.

Yeh sign error nahi balki wasted-effort error hai: do entries

0

hain, isliye do terms vanish ho jaate hain. Sirf

4 \cdot C_{22}

bachta hai — pen uthane se pehle "zeros chase karo."

R_{2} \to R_{2} - 2 R_{1}

karne ke baad mujhe determinant ko

\frac{1}{2}

se multiply karna hoga undo karne ke liye."

Kuch undo nahi karna hai. Ek row ka multiple doosri row mein add karna

det

ko bilkul change nahi karta, isliye koi correction factor nahi chahiye. (Poori row ko scale karna woh operation hai jiske liye factor ki zaroorat padti.)

1 \times 1

block

[a]

ka cofactor

(- 1)^{1 + 1} \cdot det (empty) = 0

hai."

Base case

det [a] = a

hai,

0

nahi. Akeli entry ka minor empty matrix ka determinant hai, jo

1

define hai, isliye recursion sahi se

a

par bottom out karta hai, kabhi

0

par nahi.

Why questions

Parent note yeh assert karta hai; yahan tumhe inhe explain karna hai.

Ek sentence mein: sign

(- 1)^{i + j}

aata hi kyun hai?

Yeh us permutation ki parity (sign) hai jo entry

(i, j)

ko top-left pivot corner mein shuffle karne ke liye chahiye — Leibniz sum mein woh sign pehle se hai, aur

(- 1)^{i + j}

use sirf package karta hai.

Leibniz sum ko

σ (i)

ke hisaab se group karne par koi term kabhi kyon nahi khoye ya duplicate nahi hoti?

Kyunki har permutation ke liye,

σ (i)

ek single well-defined column hai, isliye

n!

terms mein se har ek exactly ek bucket

j = σ (i)

mein jaati hai — ek clean partition.

Zyada zeros wali line ke along expand karna legitimate kyun hai aur "shortcut cheat" nahi?

Theorem guarantee karta hai ki har line same value deti hai, isliye zero-rich line choose karna ek true equality use karna hai jo sasta hota hai — answer identical hai, sirf kaam kam hota hai.

det A = det A^{T}

se hum rows aur columns ko equal footing par kyun treat kar sakte hain?

Transpose karna determinant invariant rakhta hai, isliye row expansion ke liye prove koi bhi statement,

A^{T}

par apply karke columns mein verbatim transfer ho jaati hai — koi alag proof nahi chahiye.

Cofactors directly matrix ke inverse mein kyun jaate hain?

Adjugate cofactor matrix ka transpose hai, aur

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

— isliye har cofactor literally

A^{- 1}

ke ek entry mein numerator hai.

Jis matrix ka

det = 0

ho, uska inverse kyun nahi hota?

Formula

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

mein

det A

se divide karna hota hai; jab

det A = 0

toh yeh undefined hai, jo geometry mirror karta hai — map ek dimension collapse kar deta hai aur reverse nahi ho sakta.

Edge cases

Woh degenerate inputs jinke liye recipe ko phir bhi survive karna padta hai.

1 \times 1

matrix

[a]

ka determinant kya hai, aur kya cofactor expansion apply hota hai?

Yeh sirf

a

hai — yeh recursion ka base case hai, jiske neeche aur koi minor nahi lena.

2 \times 2

matrix ke liye cofactor expansion kya deta hai, aur yeh true base workhorse kyun hai?

Row 1 ke along expand karo:

a_{11} a_{22} - a_{12} a_{21}

, yaani

a d - b c

. Har bada determinant inhi mein bottom out karta hai, isliye ise memorise karna non-negotiable hai.

Agar poori row zero ho, toh kya matrix invertible hai, aur har expansion kya return karta hai?

Invertible nahi: us zero row ke along expand karne par turant

0

milta hai, aur

det = 0

matlab singular.

n \times n

zero matrix ka determinant kya hai?

Kisi bhi

n \geq 1

ke liye

0

, kyunki kisi bhi (sab-zero) line ke along expand karne par zeros add hote hain; geometrically yeh map sab kuch ek point mein crush kar deta hai.

Kya lower-triangular matrix ke liye column 1 ke along expand karne par cofactor expansion sahi sign deta hai?

Haan — sirf top-left entry "free choice" hai jiska cofactor

+ M_{11}

hai, aur recurse karne par diagonal peel hoti jaati hai, isliye

det

= diagonal entries ka product, sign ke saath.

Kya

0 \times 0

(empty) matrix ka determinant defined hai, aur woh kya hai?

Convention se yeh

1

hai (empty product), jo exactly

1 \times 1

minor recursion ko consistent banata hai.

Recall Ek-line self test

"Kya main saari zeros wali column ke along expand kar sakta hoon aur $0$ par trust kar sakta hoon?" ::: Haan — har term zero hai, answer genuinely $0$ hai, aur matrix singular hai.

Connections

Determinants — cofactor expansion along any row - column — woh parent recipe jinhe yeh traps stress-test karti hain.
Leibniz formula for determinants — woh definition jis par har "why" answer trace back karta hai.
Properties of determinants (multilinearity, alternation) — swap/scale/add rules ka source.
Adjugate matrix and inverse — jahan cofactors rehne jaate hain.
Invertibility and singular matrices — $det = 0$ ka asal matlab.
Cramer's rule — cofactors jo linear systems solve karte hain.