Matrix multiplication — definition, associativity, non-commutativity: Question bank

4.5.7 · D5 · Maths › Linear Algebra (Full) › Matrix multiplication — definition, associativity, non-commu

True ya false — justify karo

True/false bolo, phir reason do. Sirf verdict kaafi nahi.

A B

aur

B A

ka size hamesha same hota hai.

False. Agar

A

m \times n

ki hai aur

B

n \times p

ki hai to

A B

m \times p

ki hogi, lekin

B A

ke liye

p = m

hona bhi zaroori hai; jaise ek

2 \times 3

3 \times 2

se multiply karo to ek taraf

2 \times 2

milti hai, doosri taraf

3 \times 3

Agar dono

A B

aur

B A

exist karte hain aur same size ke hain, to

A B = B A

False. Same size zaroori hai par kaafi nahi;

(1011)

aur

(1101)

dono

2 \times 2

hain phir bhi unke products diagonal par alag hain.

Matrix multiplication associative hoti hai.

True. Yeh maps ko compose karne ko represent karta hai — "

C

phir

B

phir

A

" — ek clear pipeline; entrywise dono bracketings same double sum

\sum_{j, k} A_{ik} B_{k j} C_{j ℓ}

ke barabar hain.

Matrix multiplication addition par distribute hoti hai.

True.

A (B + C) = A B + A C

aur

(A + B) C = A C + B C

— lekin dhyan raho ki left/right side fix rakhni hoti hai kyunki order abhi bhi matter karta hai.

Agar

A B = A C

ho to

B = C

False. Cancellation ke liye

A

ka invertible hona zaroori hai.

A = (1000)

ke saath tum

B

ki doosri row ko freely change kar sakte ho bina

A B

ko chhue.

Sabhi square

A, B

ke liye

(A + B)^{2} = A^{2} + 2 A B + B^{2}

False. Expand karne par

A^{2} + A B + B A + B^{2}

milta hai;

A B + B A

2 A B

mein sirf tab merge kar sakte ho jab

A

aur

B

commute karein.

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

jab bhi inverses exist karein.

False. Order reverse hota hai:

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

, kyunki "

B

phir

A

" ko undo karne ke liye pehle

A

ko undo karna padta hai. Dekho Identity and Inverse Matrices.

(A B)^{⊤} = A^{⊤} B^{⊤}

False. Transpose bhi order reverse karta hai:

(A B)^{⊤} = B^{⊤} A^{⊤}

. Dekho Transpose.

A B = 0

A = 0

B = 0

forced hota hai.

False. Matrices mein zero divisors hote hain;

(0010)^{2} = 0

hai jabki matrix khud nonzero hai.

A^{2} = 0

A = 0

implied hota hai.

False. Same nilpotent example:

(0010)

square hokar zero deta hai par khud zero nahi hai.

Identity

I

matching size ki har square matrix ke saath commute karta hai.

True.

A I = I A = A

hamesha, kyunki "kuch nahi phir

A

" aur "

A

phir kuch nahi" — ek rare pair jo truly commute karta hai.

Koi bhi matrix apne powers ke saath commute karti hai:

A A^{2} = A^{2} A

True. Dono

A^{3}

ke barabar hain; ek matrix hamesha khud ke saath commute karti hai, aur associativity se hum brackets hata sakte hain. Dekho Matrix Powers and Diagonalization.

det (A B) = det (A) det (B)

bhi hold karta hai jabki

A B \neq = B A

True. Determinant multiplicative hota hai order ki parwah kiye bina, isliye

det (A B) = det (B A)

hamesha hota hai, chahe products khud alag kyun na hon. Dekho Determinants.

Why questions

$::$ underlying reason ke liye prompt karta hai, koi formula nahi.

Row-times-column rule choose ki nahi gayi balki forced hai — kyun?

Yahi ek rule hai jo

(A B) x = A (B x)

ko har

x

ke liye satisfy karti hai; "

A B

= dono maps ko compose karo" ki demand

x_{j}

ke coefficient ko exactly

\sum_{k} A_{ik} B_{k j}

pin kar deti hai.

Inner dimensions ka match karna zaroori kyun hai?

Har entry

A

ki length-

n

row aur

B

ki length-

n

column ka dot product hai; agar woh lengths alag hoon to sum ke liye terms ko pair karna possible hi nahi hai. Dekho Dot Product.

Associativity "free" kyun hai jabki commutativity nahi?

Functions ko ek fixed order mein compose karna hamesha associative hota hai — "

C

phir

B

phir

A

" ka matlab ek hi hai chahe tum kaise bhi bracket karo — lekin order khud pipeline ko badal deta hai, isliye swap generally allowed nahi hai.

Product ko invert karne par order kyun reverse hota hai?

"Pehle moje phir joote pehno" ko undo karne ke liye pehle joote utaro:

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

, isliye last-applied map ka undo pehle aata hai.

Ek nonzero matrix square hokar zero kaise de sakti hai jabki nonzero number kabhi nahi de sakta?

Ek matrix sab kuch ek subspace mein map kar sakti hai jise doosri application poori tarah collapse kar deti hai (jaise shear phir re-shear onto zero); numbers ke paas koi aisi "hidden direction" nahi hoti.

Non-commutativity associativity ya distributivity ko kyun nahi todti?

Un laws mein kabhi

A

aur

B

ko swap nahi kiya jaata; woh sirf regroup ya split karte hain sums ko, jo application ke fixed order ko respect karta hai.

det (A B) = det (B A)

kyun hold karta hai jabki

A B \neq = B A

Determinant kisi map ki total volume-scaling measure karta hai, aur scaling factors order ki parwah kiye bina multiply hote hain, isliye dono products volume ko

det A \cdot det B

se scale karte hain. Dekho Determinants.

Edge cases

Degenerate aur boundary inputs jo topic quietly assume karta hai ki tum handle kar sakte ho.

A B

kya hoga jab

A

1 \times n

ki ho aur

B

n \times 1

ki ho?

1 \times 1

matrix — ek single number, exactly row aur column ka dot product; yahi atomic case hai jisme ek general product ki har entry reduce hoti hai.

A B

kya hoga jab

A

n \times 1

ki ho aur

B

1 \times n

ki ho?

n \times n

matrix (ek outer product) jiska rank at most

1

hoga;

B A

mein swap karne par ek

1 \times 1

scalar milta hai — ek stark shape-based non-commutation.

Kya zero matrix sab kuch ke saath commute karti hai?

Haan:

0 \cdot A = A \cdot 0 = 0

hamesha jab shapes allow karein, kyunki us map ke saath compose karna jo sab vectors ko

0

bhejta hai, dono ways mein

0

deta hai.

Agar

A

m \times n

ki ho jahan

m \neq = n

ho, to kya

A^{2}

exist kar sakta hai?

Nahi:

A^{2} = AA

ke liye inner dimensions

n

aur

m

ka match karna zaroori hai, isliye sirf square matrices ke powers hote hain. Dekho Matrix Powers and Diagonalization.

Kya commutator

[A, B] = A B - B A

kabhi bhi guaranteed zero hota hai bina

A, B

ke special hue?

Nahi — yeh zero hota hai precisely tab jab woh commute karein; generic square matrices ke liye yeh ek nonzero matrix hai jo exactly measure karta hai ki order kitna matter karta hai.

Agar

A

B

ke saath commute kare aur

B

C

ke saath commute kare, to kya

A

necessarily

C

ke saath commute karega?

Nahi — commuting transitive nahi hai; identity sab kuch ke saath commute karti hai phir bhi jo matrices usse linked hain woh ek doosre ke saath commute nahi karna zaroori.

Kisi bhi square matrix

A

I

se dono sides par multiply karne se

I

ke role ke baare mein kya pata chalta hai?

A I = I A = A

dikhata hai ki

I

"do-nothing" map hai, multiplicative identity hai, aur un kuch cheezon mein se ek hai jo

A

ke saath commute karna guaranteed hai. Dekho Identity and Inverse Matrices.

Connections

Matrix Multiplication — definition, associativity, non-commutativity — woh parent jise yeh bank stress-test karta hai.
Linear Transformations — composition view jo har trap resolve karta hai.
Identity and Inverse Matrices — inverse-of-product order reversal.
Transpose — transpose-of-product order reversal.
Dot Product — har entry row·column sum ke roop mein.
Determinants — multiplicative despite non-commutativity.
Matrix Powers and Diagonalization — jab matrices sach mein commute karein.

Question bank — Matrix multiplication — definition, associativity, non-commutativity

True ya false — justify karo

Error dhundo

Why questions

Edge cases

Connections