Worked examples — Matrix multiplication — conditions, process, non-commutativity

2.6.5 · D3 · Maths › Matrices & Determinants — Introduction › Matrix multiplication — conditions, process, non-commutativi

Yeh page ek drill hai har tarah ke matrix-multiplication situation ke liye jo tum encounter kar sakte ho. Hum pehle ek table pe saare case classes map karte hain, phir har cell ke liye ek example work karte hain taaki koi bhi scenario tumhare liye naya na rahe. Zyaadatar examples ke saath ek figure hai — picture usually "why" ko arithmetic se zyaada fast explain karti hai.

Shuru karne se pehle, parent note se ek reminder — product ki har entry usi tarah banti hai:

Is poori page mein zero matrix ko $0$ (boldface) likha gaya hai, taaki use kabhi $O$ naam ke variable se confuse na karo.

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The scenario matrix

Neeche har problem in cells mein se exactly ek mein aati hai. Har "Example" cell us section ka naam hai jahan tum jump kar sakte ho.

#	Case class	Kya tricky banata hai	Worked in
A	Rectangular × rectangular (valid)	shapes differ karti hain, dimensions track karne padte hain	Example 1
B	Matrix × column vector	yahi hai "apply a transformation" ka matlab	Example 2
C	Non-commutativity ($AB\ne BA$)	order se answer badal jaata hai	Example 3
D	Zero / degenerate inputs	$0$ se multiply karna; $AB=0$ jab $A,B\ne 0$	Example 4
E	Identity & sign patterns	negatives, charon sign combinations	Example 5
F	Undefined product	inner dimensions clash → STOP	Example 6
G	Word problem (real units)	numbers se meaning padhna	Example 7
H	Exam twist ($AB=BA$ does hold)	jab commuting actually sach ho, aur kyun	Example 8

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[!example] Example 1 — Case A: rectangular × rectangular

$$A={\left[\begin{array}{ccc}2& -1& 0\\ 3& 4& 5\end{array}\right]}_{2\times 3},B={\left[\begin{array}{cc}1& 0\\ 2& -3\\ 4& 1\end{array}\right]}_{3\times 2}$$

$AB$ compute karo. Neeche ki figure shapes ko lock hote dikhati hai — do red inner numbers touch karte hain, aur green outer numbers answer ki size ban jaate hain.

Forecast: Aage padhne se pehle — answer ki shape kya hai, aur kya yeh allow bhi hai? (Agli line cover karo.)

Steps:

1. Shape check. $A$ is $2\times 3$, $B$ is $3\times 2$. Inner numbers match karte hain ($3=3$) ✓. Answer is $2\times 2$. Yeh step kyun? Agar reds clash karte toh hum ruk jaate; compute karne ke liye koi product hai hi nahi — exactly woh dominoes-don't-meet situation jo figure mein draw hai.

2. $C_{11}$ = row 1 of $A$ · column 1 of $B$: $C_{11}=(2)(1)+(-1)(2)+(0)(4)=2-2+0=0$ Yeh step kyun? Hum row $[2,-1,0]$ ko column $[1,2,4]$ ke against slide karte hain, matched slots multiply karte hain, add karte hain. Woh akela number position (1,1) mein rehta hai.

3. $C_{12}$ = row 1 · column 2: $C_{12}=(2)(0)+(-1)(-3)+(0)(1)=0+3+0=3$ Yeh step kyun? $A$ ka wahi row, lekin ab $B$ ka doosra column — yahi cheez ise answer ka column 2 banati hai.

4. $C_{21}$ = row 2 · column 1: $C_{21}=(3)(1)+(4)(2)+(5)(4)=3+8+20=31$ Yeh step kyun? Answer ki doosri row fill karne ke liye hum $A$ ke doosre row par switch karte hain; $B$ ka column unchanged rehta hai, toh hum column 1 mein hi hain.

5. $C_{22}$ = row 2 · column 2: $C_{22}=(3)(0)+(4)(-3)+(5)(1)=0-12+5=-7$ Yeh step kyun? Last empty slot: $A$ ka row 2 aur $B$ ka column 2 milke $2\times 2$ grid complete karte hain.

$AB=\left[\begin{array}{cc}0& 3\\ 31& -7\end{array}\right]$

Verify: Shape is $2\times 2$ jaisa predict kiya tha ✓. $C_{21}$ ka quick sanity check: sabse bade positive terms ($8+20=28$) plus $3$ give $31$ — us row/column mein koi negatives nahi, toh ek bada positive expected hai ✓.

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[!example] Example 2 — Case B: matrix times a column vector

Ek matrix ka vector par act karna hi poori wajah hai ki multiplication is tarah define ki gayi hai: yeh ek transformation apply karti hai.

$$A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right],\vec{x}=\left[\begin{array}{c}3\\ 2\end{array}\right]$$

$A\vec{x}$ compute karo aur geometrically describe karo ki usne kya kiya.

Figure mein kya dhundhna hai: lavender arrow input $\vec{x}=(3,2)$ hai; coral arrow output $A\vec{x}=(-2,3)$ hai; mint arc unke beech $90^{\circ }$ counter-clockwise turn mark karta hai. Dhyan do ki dono arrows ki length same hai — transformation sirf ghoomati hai, stretch nahi karti.

Forecast: Yeh particular $A$ ek famous transformation hai. Compute karne se pehle guess karo ki yeh arrow $(3,2)$ ke saath kya karega.

Steps:

1. Shape check. $A$ is $2\times 2$, $\vec{x}$ is $2\times 1$. Match ✓. Result is $2\times 1$ (phir se ek column vector). Yeh step kyun? Ek transformation ko vector return karna chahiye, aur $2\times 1$ confirm karta hai wahi.

2. First entry = row 1 · the column: $(0)(3)+(-1)(2)=-2$ Yeh step kyun? $A$ ka row 1, arrow ke naye $x$-component ko decide karta hai.

3. Second entry = row 2 · the column: $(1)(3)+(0)(2)=3$ $A\vec{x}=\left[\begin{array}{c}-2\\ 3\end{array}\right]$ Yeh step kyun? $A$ ka row 2, naye $y$-component ko decide karta hai; har output slot ek row of $A$ dotted with $\vec{x}$ hai — row batati hai ki input ke components ko kaise mix karo.

4. Interpret. Arrow $(3,2)$ ban gaya $(-2,3)$ — figure mein mint arc dekho: yeh ek quarter turn counter-clockwise (90°) ghoom gaya. Yeh $A$ hai rotation-by-90° matrix.

Verify: 90° rotation length preserve karni chahiye. $\sqrt{3^2+2^2}=\sqrt{13}$ aur $\sqrt{(-2{)}^2+3^2}=\sqrt{13}$ ✓. Dono arrows perpendicular bhi rehne chahiye: $(3)(-2)+(2)(3)=0$ ✓ (dot product zero = right angle). Yeh Linear transformations se jodta hai.

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[!example] Example 3 — Case C: order matters ($AB\ne BA$)

$$A=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right](\text{a shear}),B=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right](\text{a stretch})$$

$AB$ aur $BA$ dono compute karo.

Figure mein kya dhundhna hai: hum corner point $(1,1)$ (unit square ka top) track karte hain. Left panel mein stretch then shear ($AB$) hota hai aur point $(4,1)$ pe land karta hai; right panel mein shear then stretch ($BA$) hota hai aur same start $(6,1)$ pe land karta hai. Do alag final places = order matters.

Forecast: Kya tum expect karte ho ki yeh equal honge? (Almost never — guess karo kaunsa entry disagree karega.)

Steps:

1. $AB$ (shear-of-a-stretch): $AB=\left[\begin{array}{cc}(1)(2)+(2)(0)& (1)(0)+(2)(1)\\ (0)(2)+(1)(0)& (0)(0)+(1)(1)\end{array}\right]=\left[\begin{array}{cc}2& 2\\ 0& 1\end{array}\right]$ Yeh step kyun? Right-to-left padhne par, $AB$ matlab "pehle $B$ (stretch) karo, phir $A$ (shear)" — figure ka left panel.

2. $BA$ (stretch-of-a-shear): $BA=\left[\begin{array}{cc}(2)(1)+(0)(0)& (2)(2)+(0)(1)\\ (0)(1)+(1)(0)& (0)(2)+(1)(1)\end{array}\right]=\left[\begin{array}{cc}2& 4\\ 0& 1\end{array}\right]$ Dono kyun compute karein? Kyunki "matrices numbers hain, numbers commute karte hain" ek trap hai. Top-right entry alag hai ($2$ vs $4$) — right panel mein corner aur door tak jaata hai.

3. Conclusion: $AB\ne BA$. $(1,2)$ entry ek taraf $2$ hai aur doosri taraf $4$.

Verify: Teeno entries mein agree karte hain aur ek mein differ karte hain, toh $AB-BA=\left[\begin{array}{cc}0& -2\\ 0& 0\end{array}\right]\ne 0$ ✓. Yeh exactly woh non-commutativity hai jiske baare mein parent warn karta hai: cheezein alag order mein karne se alag result milta hai.

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[!example] Example 4 — Case D: zero & degenerate inputs

Do sub-cases jo logon ko surprise karte hain.

(4a) Zero matrix $0=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ se multiply karo:

$$A=\left[\begin{array}{cc}5& -3\\ 7& 2\end{array}\right],A0=?$$

(4b) Do non-zero matrices jinka product zero hai:

$$P=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right],Q=\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right],PQ=?$$

Figure mein kya dhundhna hai: panel geometrically dikhata hai ki $PQ=0$ kyun hai — $P$ poore plane ko ek single line (mint) par squash karta hai, aur $Q$ ke output arrows (coral) exactly us direction pe land karte hain jise $P$ kill karta hai, toh $P$ unhe origin pe flatten kar deta hai.

Forecast: (4b) ke liye, kya do matrices jo clearly not zero hain, multiply hokar zero matrix de sakti hain? (Ordinary numbers ke saath yeh impossible hai.)

Steps:

1. (4a) $A0$ ki har entry, $A$ ki row dotted with zeros ka column hai, toh har term $0$ hai: $A0=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$ Yeh step kyun? Parent se "zero property" $A\cdot 0=0$ confirm karta hai — all-zero list ke saath dot product hamesha $0$ hota hai.

2. (4b) $PQ$ entry by entry compute karo: $PQ=\left[\begin{array}{cc}(1)(1)+(1)(-1)& (1)(1)+(1)(-1)\\ (1)(1)+(1)(-1)& (1)(1)+(1)(-1)\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$ Yeh kyun matter karta hai: Haan! $PQ=0$ even though $P\ne 0$ aur $Q\ne 0$. Toh tum matrices ko numbers ki tarah cancel nahi kar sakte.

Verify: (4b) ke liye har entry $1-1=0$ hai ✓. Note karo $\det P=1\cdot 1-1\cdot 1=0$ aur $\det Q=(1)(-1)-(1)(-1)=0$: dono degenerate hain (zero determinant). Sirf aise "flattening" matrices zero ko multiply hokar zero de sakti hain bina khud zero hue — ek hint ki na toh $P$ aur na $Q$ ka koi inverse hai.

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[!example] Example 5 — Case E: identity aur sare sign combinations

$$A=\left[\begin{array}{cc}-2& 3\\ 4& -5\end{array}\right],I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],S=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$$

$AI$ aur $AS$ compute karo. (Yeh hume charon sign patterns se force karta hai: $-++-$.)

Figure mein kya dhundhna hai: ek test arrow (lavender) $I$ se unmoved rehta hai, aur $S=-I$ se apni exact opposite (coral, origin ke through pointing) pe jaata hai — ek $180^{\circ }$ flip.

Forecast: $I$ se multiply karne par kya hota hai? $S=-I$ ke baare mein kya?

Steps:

1. $AI$: $AI=\left[\begin{array}{cc}(-2)(1)+(3)(0)& (-2)(0)+(3)(1)\\ (4)(1)+(-5)(0)& (4)(0)+(-5)(1)\end{array}\right]=\left[\begin{array}{cc}-2& 3\\ 4& -5\end{array}\right]=A$ Yeh step kyun? Identity matrix matrices ka "multiply-by-1" hai; $A$ ki har row khud ko pick out karti hai. Kuch nahi badlata.

2. $AS$ (yahan $S=-I$): $AS=\left[\begin{array}{cc}(-2)(-1)+(3)(0)& (-2)(0)+(3)(-1)\\ (4)(-1)+(-5)(0)& (4)(0)+(-5)(-1)\end{array}\right]=\left[\begin{array}{cc}2& -3\\ -4& 5\end{array}\right]=-A$ Yeh step kyun? $-I$ se multiply karna har sign flip karta hai — dekho kaise charon entries mein se har ek reverse hoti hai, $+\to -$ aur $-\to +$ cover karte hue.

Verify: $AI=A$ exactly ✓. Aur $AS+A=0$ kyunki $AS=-A$: add karne par sab zeros aata hai ✓ (uses Matrix addition and scalar multiplication).

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[!example] Example 6 — Case F: undefined product (STOP)

$$A={\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]}_{2\times 3},B={\left[\begin{array}{cc}7& 8\\ 9& 10\end{array}\right]}_{2\times 2}$$

$AB$ compute karo.

Figure mein kya dhundhna hai: shapes side by side rakhe hain. $AB$ ke liye do inner numbers ($3$ aur $2$) red gap ke saath draw kiye gaye hain — yeh meet nahi karte, toh product form nahi ho sakta. $BA$ ke liye inner numbers ($2$ aur $2$) touch karte hain (green), toh woh order legal hai.

Forecast: Sirf shapes dekho — kya yeh legal hai?

Steps:

1. Shape check. $A$ is $2\times 3$, $B$ is $2\times 2$. Inner numbers hain $3$ aur $2$ — yeh clash karte hain ($3\ne 2$) ✗. Yeh step kyun? $A$ ki row $i$ mein $3$ entries hain, lekin $B$ ke har column mein sirf $2$ entries hain. Alag length ki lists ke beech dot product form nahi ho sakta.

2. Conclusion: $AB$ undefined hai. Kuch compute mat karo.

3. Flip it: kya $BA$ legal hai? $B$ is $2\times 2$, $A$ is $2\times 3$: inner $2=2$ ✓, toh $BA$ defined hai aur $2\times 3$ hai. Ek order kaam karta hai, doosra nahi. Flip kyun dikhayein? Yeh hammer home karne ke liye ki "undefined" order aur shape ke baare mein hai, matrices ke "bad" hone ke baare mein nahi. Wahi do matrices doosri taraf theek multiply ho jaati hain — isliye hamesha specific order ke shapes check karne chahiye jo tumse pucha gaya ho.

Verify: Check karne ke liye koi number nahi — check structural hai: $3\ne 2$ toh "undefined" correct aur complete answer hai. Isliye parent dimensions pehle likhne par insist karta hai.

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[!example] Example 7 — Case G: word problem real units ke saath

Ek shop do smoothies banati hai. Recipe matrix $R$ (rows = smoothie, columns = fruit: banana, apple) aur price matrix $P$ (rows = fruit, columns = store 1, store 2):

$$R=\left[\begin{array}{cc}2& 1\\ 1& 3\end{array}\right]\begin{array}{c}\text{smoothie A}\\ \text{smoothie B}\end{array},P=\left[\begin{array}{cc}20& 25\\ 30& 28\end{array}\right]\begin{array}{c}\text{banana}\\ \text{apple}\end{array}(\text{rupees})$$

Har smoothie ki cost har store mein nikalo.

Figure mein kya dhundhna hai: chaar resulting costs ek labelled grid (smoothie × store) ke roop mein dikhaye gaye hain, taaki tum har entry ka meaning dekh sako sirf ek number ke bajaye.

Forecast: Kaun sa product — $RP$ ya $PR$ — yahan meaning rakhta hai? Pehle units match karo.

Steps:

1. Units match karo. $R$ ke columns = fruits; $P$ ke rows = fruits. Yeh line up hote hain, toh $RP$ meaningful product hai ($2\times 2$ times $2\times 2\to 2\times 2$: smoothie × store). Yeh step kyun? Inner index "same cheez" (fruit) hona chahiye warna sum nonsense hai — tum "bananas × ₹/apple" add nahi kar sakte.

2. Smoothie A ki cost ($R$ ka row 1):

   • Store 1: $(2)(20)+(1)(30)=40+30=70$
   • Store 2: $(2)(25)+(1)(28)=50+28=78$ Yeh step kyun? $R$ ka row 1, smoothie A ke fruit counts hain; ek store ke price column ke saath dot karne par us store mein us smoothie ka bill milta hai.

3. Smoothie B ki cost ($R$ ka row 2):

   • Store 1: $(1)(20)+(3)(30)=20+90=110$
   • Store 2: $(1)(25)+(3)(28)=25+84=109$ Yeh step kyun? Hum $R$ ke row 2 (smoothie B ki recipe) par switch karte hain; har store ka price column phir se ek total deta hai, answer ki doosri row fill karta hai. $RP=\left[\begin{array}{cc}70& 78\\ 110& 109\end{array}\right]\text{(rupees)}$

Verify: Units: (fruit count)×(₹ per fruit) = ₹ ✓. Smoothie B zyaada apples use karti hai (store 1 par mahenga fruit), toh uski store-1 cost ($110$) smoothie A ki ($70$) se zyaada honi chahiye — hai bhi ✓.

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[!example] Example 8 — Case H: exam twist, jab $AB=BA$ does hold

$$A=\left[\begin{array}{cc}3& 0\\ 0& 5\end{array}\right],B=\left[\begin{array}{cc}2& 0\\ 0& 7\end{array}\right](\text{both diagonal})$$

Dikhao $AB=BA$, aur explain karo ki kyun order yahan matter karna band kar deta hai.

Figure mein kya dhundhna hai: dono matrices sirf $x$- aur $y$-axes ke along stretch karti hain (koi tilting nahi). Kyunki har ek apni axis par independently act karta hai, unhe kisi bhi order mein karne par har axis do factors ke product se scale hota hai — dono taraf same.

Forecast: Humne kaha ki matrices usually commute nahi karti. Yeh do exception kyun ho sakti hain?

Steps:

1. $AB$: $AB=\left[\begin{array}{cc}(3)(2)& 0\\ 0& (5)(7)\end{array}\right]=\left[\begin{array}{cc}6& 0\\ 0& 35\end{array}\right]$ Yeh step kyun? Diagonal matrices ke liye sirf matching diagonal slots interact karte hain, toh product ka diagonal bas $3\cdot 2$ aur $5\cdot 7$ hai.

2. $BA$: $BA=\left[\begin{array}{cc}(2)(3)& 0\\ 0& (7)(5)\end{array}\right]=\left[\begin{array}{cc}6& 0\\ 0& 35\end{array}\right]$ Kyun match karte hain: same diagonal numbers multiply hote hain, aur ordinary number multiplication commutative hai ($3\cdot 2=2\cdot 3$). Koi off-diagonal mixing nahi hoti.

3. Conclusion: $AB=BA$. Commuting ek special case hai, rule nahi — yahan kaam karta hai kyunki har matrix sirf axes ke along stretch karti hai, aur stretch-then-stretch mein koi order dependence nahi hoti.

Verify: Dono products equal hain $\left[\begin{array}{cc}6& 0\\ 0& 35\end{array}\right]$ toh $AB-BA=0$ ✓. Example 3 se contrast karo, jahan ek shear ne axes mix kiye aur commutativity fail ho gayi.

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[!recall]- Scenario check — answers cover karo

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Connections

• Linear transformations — Example 2 ka rotation, $A\vec{x}$ ka geometric meaning hai.
• Identity matrix — Example 5 dikhata hai $AI=A$.
• Determinants — Example 4b ke zero products dono ka zero determinant hai.
• Inverse of a matrix — Example 4 ke degenerate matrices ka koi inverse nahi hai.
• Matrix addition and scalar multiplication — Example 5 ke sign check mein use kiya gaya.
• System of linear equations — matrix-vector products (Example 2), $A\vec{x}=\vec{b}$ ko power karte hain.