Scalar multiplication: Exercises — Hinglish deep dive

2.6.4 · D4 · Maths › Matrices & Determinants — Introduction › Scalar multiplication

Yeh Scalar Multiplication of Matrices ki practice jagah hai. Neeche har problem ka complete, revealed solution ek collapsible callout mein chhupa hua hai — pehle pen se khud try karo, phir kholna. Exercises paanch levels mein upar chadhti hain: Recognition → Application → Analysis → Synthesis → Mastery. Levels skip mat karo; har rungs yeh assume karta hai ki tum uske neeche wala already jaante ho.

Reminders jinhe tum use karoge:

Addition/subtraction entry-by-entry hoti hai aur sirf same shape ki matrices ke beech — dekho matrix addition.
Properties $k (A + B) = k A + k B$ , $(k_{1} + k_{2}) A = k_{1} A + k_{2} A$ , $(k_{1} k_{2}) A = k_{1} (k_{2} A)$ properties of matrix operations se aati hain.

Level 1 — Recognition

Kya tum rule spot karke use ek baar apply kar sakte ho?

L1.1 Maano $A = [20 - 1 4]$ . $3 A$ compute karo.

L1.2 In mein se kaun sa $(- 2) [3 - 5]$ ke barabar hai? (a) $[- 6 10]$ (b) $[- 6 - 10]$ (c) $[610]$

L1.3 $7 B$ ki dimensions batao agar $B$ ek $3 \times 5$ matrix hai.

Recall Solution L1.1

Har entry ko $3$ se multiply karo: $3 A = [3 \cdot 2 3 \cdot 0 3 \cdot (- 1) 3 \cdot 4] = [60 - 3 12]$ Note karo ki $3 \cdot 0 = 0$ — zeros zero hi rehte hain, lekin tum unhe phir bhi "touch" karte ho.

Recall Solution L1.2

Har entry ko $- 2$ se multiply karo: $(- 2) (3) = - 6$ aur $(- 2) (- 5) = + 10$ (negative times negative is positive). Answer: (a) $[- 6 10]$ .

Recall Solution L1.3

Scalar multiplication kabhi bhi shape nahi badalta. Isliye $7 B$ abhi bhi $3 \times 5$ hai.

Level 2 — Application

Rule ko addition aur fractions ke saath chain karo.

L2.1 Maano $A = [1324]$ , $B = [02 - 1 5]$ . $2 A + 3 B$ compute karo.

L2.2 Maano $E = [84 - 12 20]$ . $\frac{1}{4} E$ compute karo.

L2.3 L2.1 ke $A, B$ use karke $5 A - 5 B$ compute karo, phir check karo ki yeh $5 (A - B)$ ke barabar hai.

Recall Solution L2.1

Pehle har matrix ko scale karo, phir corresponding entries add karo (dono $2 \times 2$ hain, isliye addition legal hai). $2 A = [2648], 3 B = [06 - 3 15]$ $2 A + 3 B = [2 + 0 6 + 6 4 + (- 3) 8 + 15] = [212123]$

Recall Solution L2.2

$\frac{1}{4}$ se multiply karna har entry ko $4$ se divide karne jaisa hai: $\frac{1}{4} E = [8/4 4/4 - 12/4 20/4] = [21 - 3 5]$

Recall Solution L2.3

Direct route: $5 A = [5151020], 5 B = [010 - 5 25]$ $5 A - 5 B = [5 - 0 15 - 10 10 - (- 5) 20 - 25] = [55 15 - 5]$ Distributivity se check karo: $A - B = [11 3 - 1]$ , aur $5 (A - B) = [55 15 - 5]$ . ✓ Match karte hain — yahi hai $k (A - B) = k A - k B$ action mein.

Level 3 — Analysis

Ab tumhe backwards reason karna hoga ya ek property test karni hogi.

L3.1 Scalar $k$ dhundho jisse ki $k [39 - 6 12] = [13 - 2 4]$ .

L3.2 Verify karo ki $(k_{1} k_{2}) A = k_{1} (k_{2} A)$ , $k_{1} = 2$ , $k_{2} = - 3$ , $A = [1 - 2 05]$ ke liye, dono sides compute karke.

L3.3 $k$ ki kaunsi value(s) ke liye $k A$ zero matrix ke barabar hoga, jahaan $A = [4 - 1 72]$ ? Justify karo.

Recall Solution L3.1

Right side ki har entry left entry times $k$ hai. Ek pair compare karo: $k \cdot 3 = 1 \Rightarrow k = \frac{1}{3}$ . Check karo ki yeh har jagah kaam karta hai: $\frac{1}{3} \cdot (- 6) = - 2$ ✓, $\frac{1}{3} \cdot 9 = 3$ ✓, $\frac{1}{3} \cdot 12 = 4$ ✓. $k = \frac{1}{3}$ (Agar ek bhi entry disagree karta, toh koi ek scalar exist hi nahi karta.)

Recall Solution L3.2

Left side: $k_{1} k_{2} = 2 \cdot (- 3) = - 6$ , isliye $(- 6) A = [- 6 12 0 - 30]$ Right side: pehle $k_{2} A = - 3 A = [- 3 6 0 - 15]$ , phir $2 \cdot (- 3 A) = [- 6 12 0 - 30]$ Dono sides identical hain ✓ — yeh sirf ordinary number multiplication ki associativity hai jo har cell ke andar ho rahi hai.

Recall Solution L3.3

$k A = O$ ka matlab hai har entry $k \cdot a_{ij} = 0$ . Kyunki $A$ mein non-zero entries hain (jaise $a_{11} = 4$ ), humein chahiye $k \cdot 4 = 0$ , jo force karta hai $k = 0$ . $k = 0 only$ Kyunki $A \neq = O$ , koi non-zero scalar ise annihilate nahi kar sakta. (Contrast: agar $A$ already $O$ hota, toh koi bhi $k$ kaam karta.)

Level 4 — Synthesis

Scalar multiplication ko geometry aur structure ke saath combine karo.

Neeche diya unit square apne corners ek matrix ke columns mein collect karke rakha gaya hai. Dekho scaling iske saath kya karta hai.

L4.1 Points $P = (1, 0)$ aur $Q = (0, 1)$ $M = [1001]$ ke columns mein store hain. $2 M$ compute karo aur geometrically describe karo ki dono points ke saath kya hua.

L4.2 Usi $M$ ka use karke $(- 1) M$ compute karo aur points par geometric effect describe karo.

L4.3 Matrix $X$ ke liye solve karo: $2 X + [4862] = [102012]$ .

Recall Solution L4.1

$2 M = [2002]$ Columns ab $(2, 0)$ aur $(0, 2)$ hain. Dono points apni apni directions mein origin se do guna door chale gaye — poori picture size mein double ho gayi (factor $2$ se expansion). Figure mein outer teal square dekho. Stretching se yeh connection exactly wahi hai jo linear transformations precise karta hai.

Recall Solution L4.2

$(- 1) M = [- 1 0 0 - 1]$ Columns $(- 1, 0)$ aur $(0, - 1)$ ban jaate hain: har point origin ki opposite side par phank diya jaata hai — yeh ek reflection through the origin hai ( $180°$ ka turn). Figure mein plum points dekho.

Recall Solution L4.3

$X$ ko ordinary equation ki tarah isolate karo. Dono sides se known matrix subtract karo: $2 X = [102012] - [4862] = [6 - 6 - 6 10]$ Ab har entry ko $2$ se divide karo ( $\frac{1}{2}$ se multiply karo): $X = \frac{1}{2} [6 - 6 - 6 10] = [3 - 3 - 3 5]$

Level 5 — Mastery

General reasoning, edge cases, aur proof-flavoured problems.

L5.1 Maano $A$ koi bhi $2 \times 2$ matrix hai. Prove karo ki $3 A - A = 2 A$ , entry-wise definition use karke.

L5.2 Ek matrix $A$ satisfy karti hai $k A = A$ ek scalar $k \neq = 1$ ke liye. $A$ ke baare mein kya sach hona chahiye? Prove karo.

L5.3 Saare scalars $k$ dhundho jisse $k^{2} A = 9 A$ , jahaan $A = [13 2 - 1]$ (note karo $A \neq = O$ ).

Recall Solution L5.1

Likho $A = [a_{ij}]$ . Left side ki ek single entry dekho: $(3 A - A)_{ij} = 3 a_{ij} - a_{ij} = (3 - 1) a_{ij} = 2 a_{ij} = (2 A)_{ij} .$ Isme $(k_{1} - k_{2}) A = k_{1} A - k_{2} A$ (scalar subtraction par distributivity) use hui. Kyunki har entry match karti hai, $3 A - A = 2 A$ . ∎

Recall Solution L5.2

Entry-wise, $k A = A$ ka matlab hai $k a_{ij} = a_{ij}$ har $i, j$ ke liye, yaani $(k - 1) a_{ij} = 0$ . Kyunki $k \neq = 1$ , factor $(k - 1) \neq = 0$ . Ek non-zero factor ke saath product zero hone ke liye, doosra factor zero hona chahiye: $a_{ij} = 0$ sabhi $i, j$ ke liye. Isliye $A = O$ (zero matrix). ∎

Recall Solution L5.3

$k^{2} A = 9 A$ ka matlab hai $(k^{2} - 9) A = O$ . Kyunki $A \neq = O$ (usmein non-zero entries hain), scalar vanish karna chahiye: $k^{2} - 9 = 0 \Rightarrow k^{2} = 9 \Rightarrow k = 3 or k = - 3.$ Dono kaam karte hain: $3^{2} = 9$ aur $(- 3)^{2} = 9$ . $k = 3 or k = - 3$

Wrap-up

Recall Har level ki one-line summary

L1 rule apply karo ::: har entry ko $k$ se multiply karo L2 add/fractions ke saath combine karo ::: har matrix ko poora scale karo, phir entry-wise add karo L3 backwards reason karo ::: ek single $k$ ko har entry equation satisfy karni hogi L4 geometry & equations ::: scaling = stretch/reflect; isolate karo phir reciprocal-scale karo L5 proofs & edge cases ::: $(scalar) A = O$ ko "scalar $= 0$ ya $A = O$ " mein todho

Scaling aur adding master karne ke baad agla stop: matrices ek doosre ko kaise multiply karti hain matrix multiplication mein — yeh genuinely different operation hai, aur "L1 trap" ka grown-up version ek classic source hai.