Matrix multiplication — conditions, process, non-commutativity: Exercises

2.6.5 · D4 · Maths › Matrices & Determinants — Introduction › Matrix multiplication — conditions, process, non-commutativi

Yeh page ek ladder hai. Har rung pichle se thoda zyada maangta hai. Problem ko pehle khud try karo — solutions deliberately collapse karke rakhe hain taaki page tumhe actually test kar sake.

Yahan har cheez parent note ke ek hi rule par tikhi hai:

Figure — Matrix multiplication — conditions, process, non-commutativity

Yahan laal path $A$ ki row $i$ hai jo across slide kar rahi hai, aur $B$ ki laal column $j$ se mil rahi hai jo neeche slide kar rahi hai. Jahan woh cross karte hain, answer ka ek number janam leta hai.

Level 1 — Recognition

(Kya tum yeh bhi bata sakte ho ki koi product legal hai ya nahi, aur uski shape kya hogi?)

Exercise 1.1

Har pair ke liye batao ki $A B$ defined hai ya nahi, aur agar hai toh $A B$ ki size kya hogi.

(a) $A$ hai $2 \times 3$ , $B$ hai $3 \times 5$ .
(b) $A$ hai $4 \times 2$ , $B$ hai $4 \times 2$ .
(c) $A$ hai $1 \times 3$ (ek row), $B$ hai $3 \times 1$ (ek column).
(d) $A$ hai $3 \times 1$ , $B$ hai $1 \times 3$ .

Recall Solution 1.1

Rule: $A_{m \times n}$ times $B_{n^{'} \times p}$ ke liye, humein $n = n^{'}$ chahiye; answer $m \times p$ hoga.

(a) inner numbers $3$ aur $3$ match karte hain → defined, size $2 \times 5$ . ✓
(b) inner numbers $2$ aur $4$ match nahi karte → undefined. ✗
(c) inner numbers $3$ aur $3$ match karte hain → defined, size $1 \times 1$ (ek single number — yeh exactly ek dot product hai).
(d) inner numbers $1$ aur $1$ match karte hain → defined, size $3 \times 3$ (ek poora grid — column times row se!).

Dhyan do ki (c) aur (d) mein wahi do matrices hain lekin ulti order mein, aur bilkul alag shapes milti hain. Order toh shape ke liye bhi matter karta hai.

Level 2 — Application

(Rule ko haath se chalao, saari entries, koi shortcut nahi.)

Exercise 2.1

$A B$ compute karo jahan $A = [20 - 1 3], B = [15 4 - 2] .$

Recall Solution 2.1

Dono $2 \times 2$ hain, inner numbers match karte hain, answer $2 \times 2$ hoga. Har entry = ( $A$ ki row)·( $B$ ki column).

$C_{11} = (2) (1) + (- 1) (5) = 2 - 5 = - 3$ $C_{12} = (2) (4) + (- 1) (- 2) = 8 + 2 = 10$ $C_{21} = (0) (1) + (3) (5) = 0 + 15 = 15$ $C_{22} = (0) (4) + (3) (- 2) = 0 - 6 = - 6$ $A B = [- 3 15 10 - 6]$

Exercise 2.2

Row aur column ka product compute karo: $[25 - 1] 3 - 2 4 .$

Recall Solution 2.2

$1 \times 3$ times $3 \times 1$ → answer $1 \times 1$ hoga. Bas ek dot product: $(2) (3) + (5) (- 2) + (- 1) (4) = 6 - 10 - 4 = - 8.$ Answer: $[- 8]$ . Yeh sabse chhota possible matrix product hai — woh atom jis se saare bade products bane hain.

Exercise 2.3

$A$ hai $2 \times 3$ , $B$ hai $3 \times 2$ : $A = [1 - 1 0321], B = 401 1 - 2 1 .$ $A B$ nikalo.

Recall Solution 2.3

Answer $2 \times 2$ hoga. $C_{11} = (1) (4) + (0) (0) + (2) (1) = 4 + 0 + 2 = 6$ $C_{12} = (1) (1) + (0) (- 2) + (2) (1) = 1 + 0 + 2 = 3$ $C_{21} = (- 1) (4) + (3) (0) + (1) (1) = - 4 + 0 + 1 = - 3$ $C_{22} = (- 1) (1) + (3) (- 2) + (1) (1) = - 1 - 6 + 1 = - 6$ $A B = [6 - 3 3 - 6]$

Level 3 — Analysis

(Ab tumhe sirf calculate nahi karna, yeh samajhna hai ki kyun aisa hota hai.)

Exercise 3.1

$A = [1324]$ aur $B = [2102]$ lekar, dono $A B$ aur $B A$ compute karo aur confirm karo ki woh alag hain. Phir ek sentence mein batao yeh kaunsa geometric fact demonstrate karta hai (Linear transformations).

Recall Solution 3.1

$A B = [(1) (2) + (2) (1) (3) (2) + (4) (1) (1) (0) + (2) (2) (3) (0) + (4) (2)] = [41048]$ $B A = [(2) (1) + (0) (3) (1) (1) + (2) (3) (2) (2) + (0) (4) (1) (2) + (2) (4)] = [27410]$ $A B \neq = B A$ . Matlab: Matrices transformations hain, aur "pehle $B$ karo phir $A$ " aam taur par ek alag combined transformation hai "pehle $A$ karo phir $B$ " se — composition ki order matter karti hai.

Exercise 3.2

Maano $D = [0010]$ . $D^{2} = D D$ compute karo. Answer kya batata hai "zero divisors" ke baare mein — kya ek non-zero matrix square hokar zero matrix ban sakti hai?

Recall Solution 3.2

$D^{2} = [(0) (0) + (1) (0) (0) (0) + (0) (0) (0) (1) + (1) (0) (0) (1) + (0) (0)] = [0000] .$ Toh ek non-zero matrix zero matrix mein square ho sakti hai. Ordinary numbers mein, $x^{2} = 0 \Rightarrow x = 0$ ; woh guarantee matrices ke liye nahi chalti. Aisi $D$ ko nilpotent kehte hain, aur iska hona exactly wahi reason hai kyun tum matrices ko freely "cancel" nahi kar sakte (unlike invertible waalon ke).

Exercise 3.3

Ek aisi $2 \times 2$ matrix $A$ dhoondo (zero matrix nahi, identity nahi) jis ke liye $A^{2} = A$ ho. Verify karo.

Recall Solution 3.3

Jis matrix ke liye $A^{2} = A$ ho use projection kehte hain. Try karo: $A = [1000] .$ $A^{2} = [(1) (1) + (0) (0) (0) (1) + (0) (0) (1) (0) + (0) (0) (0) (0) + (0) (0)] = [1000] = A . ✓$ Geometrically yeh har vector ko $x$ -axis par flatten kar deta hai; jo cheez already flat hai use flatten karne se kuch nahi badalta, isliye ise do baar apply karna ek baar apply karne ke barabar hai.

Level 4 — Synthesis

(Multiplication ko doosre tools ke saath combine karo.)

Exercise 4.1

Verify karo ki $(A B)^{T} = B^{T} A^{T}$ (dekho Matrix transpose properties) in matrices ke liye: $A = [1021], B = [3410] .$ Yahan $M^{T}$ (transpose) ka matlab hai "rows aur columns flip karo": row $i$ , col $j$ waali entry row $j$ , col $i$ par chali jaati hai.

Recall Solution 4.1

Pehle $A B$ : $A B = [(1) (3) + (2) (4) (0) (3) + (1) (4) (1) (1) + (2) (0) (0) (1) + (1) (0)] = [11410] .$ Transpose (diagonal ke across flip karo): $(A B)^{T} = [11140] .$ Ab doosri side. $B^{T} = [3140]$ , $A^{T} = [1201]$ . $B^{T} A^{T} = [(3) (1) + (4) (2) (1) (1) + (0) (2) (3) (0) + (4) (1) (1) (0) + (0) (1)] = [11140] .$ Dono match karte hain. Note karo ki order ulta ho jaata hai: yeh $B^{T} A^{T}$ hai, $A^{T} B^{T}$ nahi.

Exercise 4.2

Check karo ki $det (A B) = det (A) det (B)$ (Determinants) in matrices ke liye: $A = [2113], B = [0 - 1 12],$ jahan $2 \times 2$ matrix ke liye $det [a c b d] = a d - b c$ .

Recall Solution 4.2

$det A = (2) (3) - (1) (1) = 5$ . $det B = (0) (2) - (1) (- 1) = 1$ . Toh determinants ka product hai $5 \cdot 1 = 5$ . Ab $A B$ : $A B = [(2) (0) + (1) (- 1) (1) (0) + (3) (- 1) (2) (1) + (1) (2) (1) (1) + (3) (2)] = [- 1 - 3 47] .$ $det (A B) = (- 1) (7) - (4) (- 3) = - 7 + 12 = 5.$ ✓ Barabar, jaisa promise tha. (Determinant multiplicative hai — yeh ek rare jagah hai jahan order nahi kaattta, kyunki scalars commute karte hain.)

Exercise 4.3

Matrix-vector multiplication (System of linear equations) use karke system solve karo. Diya gaya hai: $A = [21 1 - 1], x = [x y], b = [51],$ $A x = b$ se represent hone wale do scalar equations likho, phir $x, y$ solve karo.

Recall Solution 4.3

$A x$ ka matlab hai: $A$ ki row 1 ko $x$ se dot karo, phir $A$ ki row 2 ko $x$ se dot karo. $[2 x + y x - y] = [51] \Rightarrow {2 x + y = 5 x - y = 1$ Dono equations jodo: $3 x = 6 \Rightarrow x = 2$ . Back-substitute karo: $2 - y = 1 \Rightarrow y = 1$ . Check: $2 (2) + 1 = 5$ ✓ aur $2 - 1 = 1$ ✓. Toh $x = [21]$ .

Level 5 — Mastery

(Ek aisa problem jo upar ki sab cheezein use karta hai.)

Exercise 5.1

Maano: $R = [01 - 1 0] (ek 9 0^{\circ} rotation), S = [2001] (ek horizontal stretch) .$ Unit vector $e = [10]$ positive $x$ -axis ki taraf point karta hai.

(a) $R S$ aur $S R$ compute karo. Confirm karo ki $R S \neq = S R$ .
(b) Transformation "pehle stretch phir rotate" ko $e$ par apply karo. Kaun sa matrix product yeh represent karta hai — $R S$ ya $S R$ ? Resulting vector compute karo.
(c) " $e$ par pehle rotate phir stretch" apply karo aur result compute karo.
(d) Ek sentence mein explain karo ki geometrically kya fark dikh raha hai.

Recall Solution 5.1

(a) $R S = [(0) (2) + (- 1) (0) (1) (2) + (0) (0) (0) (0) + (- 1) (1) (1) (0) + (0) (1)] = [02 - 1 0]$ $S R = [(2) (0) + (0) (1) (0) (0) + (1) (1) (2) (- 1) + (0) (0) (0) (- 1) + (1) (0)] = [01 - 2 0]$ $R S \neq = S R$ . ✓

(b) "Pehle $S$ , phir $R$ " ka matlab hai $S$ ko vector ke sabse paas apply karo: $R (S e) = (R S) e$ . Toh $RS$ use karo. Step 1 — stretch: $S e = [2001] [10] = [20]$ (arrow $x$ ke along double ho jaata hai). Step 2 — usse rotate karo: $R [20] = [02]$ . Result: $[02]$ (upar point karta hai, length 2). $(R S) e$ se bhi same: $[02 - 1 0] [10] = [02]$ . ✓

(c) "Pehle $R$ , phir $S$ " = $(S R) e$ . Step 1 — rotate: $R e = [01]$ (arrow ab upar point karta hai, length 1 hi rahti hai). Step 2 — horizontally stretch: $S [01] = [01]$ (horizontal stretch ek vertical arrow par kuch nahi karta!). Result: $[01]$ . $(S R) e$ se bhi same: $[01 - 2 0] [10] = [01]$ . ✓

(d) Pehle stretch karne par arrow tab lambaa hota hai jab woh abhi $x$ ke along hai, isliye stretch kaam aata hai (final length 2); pehle rotate karne par arrow horizontal stretch se pehle $x$ -axis se hat jaata hai, isliye stretch ke paas pakadne ko kuch nahi aur length 1 rahti hai. Wahi do operations, ulti order, alag vector — yahi hai non-commutativity, seedhi aankhon se dikhti hui.

[!recall]- Jaane se pehle ek-line self-check

A B C x

product mein

x

par pehle kaun act karta hai?

C

(sabse daayein, vector ke sabse paas); phir

B

; phir

A

A_{4 \times 3}

aur

B_{3 \times 6}

ke liye,

A B

ki size?

4 \times 6

Kya

(A B)^{T} = A^{T} B^{T}

hota hai?

Nahi — order ulta ho jaata hai:

(A B)^{T} = B^{T} A^{T}

Connections

Matrix multiplication — conditions, process, non-commutativity — parent rules jin par yeh drills practice karate hain
Determinants — Exercise 4.2 mein use kiya
Matrix transpose properties — Exercise 4.1 mein use kiya
System of linear equations — Exercise 4.3 mein use kiya
Linear transformations — Level 5 ke peeche ki geometry
Inverse of a matrix — transpose jaisa hi order-reversal idea
Identity matrix — woh "kuch nahi karo" transformation jo $A^{2} = A$ ke peeche chhupa hai