Scalar multiplication: Question bank — Hinglish deep dive

2.6.4 · D5 · Maths › Matrices & Determinants — Introduction › Scalar multiplication

Ek rule yaad rakho jisko yahan sab kuch test karta hai: ek matrix $A = [a_{ij}]_{m \times n}$ aur ek real number $k$ ke liye, scalar multiple $k A$ har ek entry ko $k$ se multiply karta hai, aur shape $m \times n$ kabhi nahi badlti.

True or false — justify

Kya

k A

hamesha

A

ke same shape ka hota hai?

True. Scalar multiplication contents (numbers) ko touch karta hai lekin container (rows aur columns) ko kabhi nahi, isliye ek

m \times n

matrix

m \times n

hi rehta hai.

Agar

k A = O

(zero matrix) ho, toh kya zaruri hai ki

k = 0

ho?

False. Agar

A

pehle se hi zero matrix hai, toh

k A = O

kisi bhi

k

ke liye hoga. Toh

k A = O

ka matlab hai "

k = 0

A

zero matrix hai", sirf

k = 0

nahi.

Kya

(- 1) A

wahi object hai jo additive inverse

- A

hai jo

A + (- A) = O

satisfy karta hai?

True. Har entry ko

- 1

se multiply karne par har sign flip ho jaata hai, aur ek matrix ko uske sign-flipped form ke saath add karne par entry-by-entry sab zeros milte hain — yahi additive inverse ki definition hai.

Kya

k (A + B) = k A + k B

har real

k

aur har same-shaped

A, B

ke liye hold karta hai?

True. Yeh har cell ke andar ordinary distributive law

k (a_{ij} + b_{ij}) = k a_{ij} + k b_{ij}

par reduce ho jaata hai, jo sab reals ke liye hold karta hai.

Kya scalar multiplication is sense mein commutative hai ki

k A = A k

Effect mein True — dono ka matlab hai "har entry ko

k

se scale karo" — lekin

A k

unusual notation hai; ek scalar matrix nahi hota, toh yahan koi row–column product chhupa nahi hai.

Agar

2 A = 2 B

ho, toh kya hum

A = B

conclude kar sakte hain?

True. Har entry ko

2

se divide karo (

\frac{1}{2}

se multiply karo); kyunki

2

nonzero hai yeh har entry uniquely recover karta hai, toh matrices identical hone chahiye.

Agar

k A = k B

aur

k \neq = 0

ho, toh kya

A = B

hoga?

True. Ek nonzero scalar cancellable hota hai: dono sides ko entry-wise

\frac{1}{k}

se multiply karo aur

A

aur

B

exactly recover ho jaate hain.

Agar

k A = k B

sirf

k = 0

ke liye diya gaya ho, toh kya

A = B

hoga?

False.

0 \cdot A = O = 0 \cdot B

kisi bhi

A, B

ke liye hota hai, toh yeh equation iske baare mein koi information nahi deta ki

A

equals

B

hai ya nahi.

Kya

(k_{1} k_{2}) A = k_{1} (k_{2} A)

hota hai?

True. Har cell ke andar yeh sirf

(k_{1} k_{2}) a_{ij} = k_{1} (k_{2} a_{ij})

hai, jo real-number multiplication ki associativity hai.

Kya ek matrix ko

k

se scale karne par uska "size/magnitude" exactly

k

se scale hota hai?

Careful — sirf

k \geq 0

ke liye. Har entry

k

se multiply hoti hai, lekin agar

k < 0

ho toh entries ke signs bhi flip ho jaate hain, toh magnitude

∣ k ∣

se scale hota hai jabki direction reverse ho jaata hai.

Spot the error

3 [1324]

compute karne ke liye maine pehli entry multiply ki:

[3324]

Error: sirf ek entry scale hui. Scalar sabhi entries mein distribute hota hai, result hoga

[39612]

k \cdot A

mein

A \cdot B

jaisi row-by-column multiplication chahiye."

Error: yeh scalar multiplication ko matrix multiplication ke saath confuse karta hai. Ek scalar plain number hota hai, toh tum sirf har entry multiply karo — koi product ka sum nahi, koi dimension-matching rule nahi.

5 A

2 \times 3

matrix ko bada kar deta hai kyunki '5 times' ka matlab zyada numbers hai."

Error: scaling values badlti hai, count nahi. Ek

2 \times 3

matrix mein pehle bhi chhe entries hain aur baad mein bhi; har ek sirf

5

se multiply hoti hai.

\frac{1}{3} E

undefined hai kyunki tum matrix ko divide nahi kar sakte."

Error:

\frac{1}{3} E

number

\frac{1}{3}

(

3

ka reciprocal) se scalar multiplication hai, jo bilkul defined hai — har entry ko

3

se divide karo.

"Kyunki

k (A + B) = k A + k B

hai, toh yeh bhi follow karta hai ki

k (A B) = (k A) (k B)

Error: galat step scalar ko square karna hai. Sahi rule hai

k (A B) = (k A) B = A (k B)

— scalar ek baar appear karta hai, kyunki

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

hota hai.

- 2 [3 - 1] = [- 6 - 1]

Error: doosri entry ka sign handle nahi kiya gaya.

- 2 \cdot (- 1) = + 2

, toh answer hai

[- 6 2]

k A

jahan

k

khud ek

2 \times 2

matrix hai, yeh scalar multiplication hai."

Error: agar

k

matrix hai toh woh scalar nahi hai. Woh matrix multiplication hogi aur uske liye compatible dimensions chahiye, entry-wise scaling nahi.

Why questions

1 \cdot A = A

kyun sirf convention nahi balki zaruri hai?

Kyunki har entry

1 \cdot a_{ij} = a_{ij}

ban jaati hai; reals ka multiplicative identity har number ko unchanged chhod deta hai, toh poora matrix unchanged rehta hai.

Scalar multiplication matrix multiplication se "simpler" kyun hai?

Entries ke beech koi interaction nahi hota — har cell independently scale hoti hai. Matrix multiplication mein poori rows ko poori columns ke saath sums of products ke zariye combine kiya jaata hai.

Addition par distributivity ke liye

A

aur

B

ka same shape hona kyun zaruri hai?

Kyunki

k A + k B

matrix addition use karta hai, jo sirf tabhi defined hai jab dono matrices cell-for-cell dimensions match karti hon.

Ek negative scalar ko geometrically origin ke through reflection kyun read kiya jaata hai?

Ek matrix column ko vector maanke, negative number se multiply karne par har component ki direction reverse ho jaati hai, vector ko origin ke opposite side bhej deta hai — dekho linear transformations.

"Scalar division" ek alag operation kyun nahi hai?

k \neq = 0

se divide karna

\frac{1}{k}

(uske reciprocal) se multiply karne ke roop mein defined hai, toh yeh scalar multiplication mein already covered hai — koi naya rule nahi chahiye.

0 \cdot A

zero matrix kyun hota hai aur number

0

kyun nahi?

Kyunki operation ko same shape ka matrix return karna hota hai; har entry

0 \cdot a_{ij} = 0

ban jaati hai, producing ek

m \times n

grid of zeros, jise

O

likha jaata hai.

Sabhi chaar scalar properties ultimately ordinary numbers ke facts par kyun reduce hoti hain?

Kyunki har property ek ek cell par check ki jaati hai, aur ek single cell ke andar sirf real-number arithmetic hoti hai — matrix structure sirf in facts ko parallel mein carry karta hai.

Edge cases

k A

kya hota hai jab

A

1 \times 1

matrix ho?

Yeh

[k a_{11}]

hota hai — ek

1 \times 1

matrix ki scalar multiplication bilkul do ordinary numbers ko multiply karne jaisi behave karti hai, lekin result phir bhi matrix ke roop mein likha jaata hai.

Jab already zero entry wala matrix scale kiya jaaye toh kya hota hai?

Zero entry zero hi rehti hai (kyunki

k \cdot 0 = 0

) kisi bhi

k

ke liye, jabki nonzero entries normally scale hoti hain — koi entry skip nahi hoti, bas

0

par land karti hai.

Kya zero matrix aur kisi bhi scalar

k

ke liye

k O = O

hoga?

Haan. Har entry pehle se

0

hai, aur

k \cdot 0 = 0

, toh zero matrix ko kisi bhi cheez se scale karne par zero matrix hi milti hai.

Jab

k = 0

ho lekin

A

mein bahut badi entries hon toh

k A

kya deta hai?

Zero matrix

O

, chahe entries kitni bhi badi rahi hon — multiplier

0

har value ko annihilate kar deta hai.

Kya scalar

k

irrational ho sakta hai, jaise

2

π

Haan. Definition sirf

k \in R

require karti hai (dekho scalars), toh koi bhi real number, rational ho ya na ho, valid multiplier hai.

Agar do different scalars same result dein,

k_{1} A = k_{2} A

jahan

k_{1} \neq = k_{2}

, toh

A

kya hona chahiye?

Zero matrix.

(k_{1} - k_{2}) A = O

jahan

k_{1} - k_{2} \neq = 0

har entry ko

0

force karta hai, toh

A = O

hona chahiye.

Recall Lock in karne ke liye one-line summary

Scalar multiplication har entry ko $k$ se scale karta hai, shape kabhi nahi badlta, aur iske har "rule" ka matlab sirf real-number arithmetic hai jo ek saath har cell mein hoti hai.