Exercises — Matrix multiplication — definition, associativity, non-commutativity

Pehle, in do shabdon ka matlab. Har matrix ko (rows)$\times$(columns) likho: $A$ hai $\underset{\text{rows}}{\underbrace{2}}\times \underset{\text{cols}}{\underbrace{3}}$. Jab tum ek product $XY$ ko $({\text{rows}}_X\times {\text{cols}}_X)({\text{rows}}_Y\times {\text{cols}}_Y)$ ki tarah likhte ho, toh inner dimensions woh do numbers hain jo beech mein ek doosre ke paas hote hain — ${\text{cols}}_X$ aur ${\text{rows}}_Y$ — aur outer dimensions woh do hain jo bahar hote hain — ${\text{rows}}_X$ aur ${\text{cols}}_Y$. Rule yeh hai: $XY$ banane ke liye inner dimensions match karni chahiye (taaki $X$ ki har row mein utni hi entries hon jitni $Y$ ke har column mein, jisse dot product line up ho sake), aur result apni (rows)$\times$(columns) ke liye outer dimensions leta hai.

• $AB$: $(2\times 3)(3\times 4)$ → inner $3=3$ ✓ → size (outer) $2\times 4$. Defined.
• $BA$: $(3\times 4)(2\times 3)$ → inner $4\ne 2$ ✗. Defined nahi.
• $BC$: $(3\times 4)(4\times 2)$ → inner $4=4$ ✓ → size $3\times 2$. Defined.
• $CA$: $(4\times 2)(2\times 3)$ → inner $2=2$ ✓ → size $4\times 3$. Defined.
• $AC$: $(2\times 3)(4\times 2)$ → inner $3\ne 4$ ✗. Defined nahi. Humne kya kiya: bas do inner numbers match kiye aur do outer numbers copy kiye. Kuch compute nahi kiya — yeh sirf bookkeeping hai.

$(AB{)}_{21}$ = $A$ ka row $2$, $B$ ke column $1$ ke saath dot kiya. $A$ ka row 2 hai $(0,3)$; $B$ ka column 1 hai $(4,6)$: $(AB{)}_{21}=0\cdot 4+3\cdot 6=18.$ $(AB{)}_{12}$ = $A$ ka row $1$, $B$ ke column $2$ ke saath dot kiya. $A$ ka row 1 hai $(2,-1)$; $B$ ka column 2 hai $(5,7)$: $(AB{)}_{12}=2\cdot 5+(-1)\cdot 7=10-7=3.$ Index $ij$ = (row $i$)(column $j$) kyun padha jaata hai: pehla index batata hai konsa output row, doosra batata hai konsa output column. Output row decide hota hai $A$ ke kis row ko slide karoge; output column decide hota hai $B$ ke kis column par drop karoge.

Char dot products, $A$ ki har row ko $B$ ke har column ke against.

• $(AB{)}_{11}=(2,-1)\cdot (4,6)=8-6=2$
• $(AB{)}_{12}=(2,-1)\cdot (5,7)=10-7=3$ (L1.2 se)
• $(AB{)}_{21}=(0,3)\cdot (4,6)=0+18=18$ (L1.2 se)
• $(AB{)}_{22}=(0,3)\cdot (5,7)=0+21=21$ $AB=\left(\begin{array}{cc}2& 3\\ 18& 21\end{array}\right).$

Inner dims $3=3$ ✓, toh $AB$ hai $2\times 2$. Har entry shared index $k=1,2,3$ par ek teen-term dot product hai.

• $(AB{)}_{11}=(1,2,-1)\cdot (2,0,4)=2+0-4=-2$
• $(AB{)}_{12}=(1,2,-1)\cdot (1,-1,2)=1-2-2=-3$
• $(AB{)}_{21}=(0,1,3)\cdot (2,0,4)=0+0+12=12$
• $(AB{)}_{22}=(0,1,3)\cdot (1,-1,2)=0-1+6=5$ $AB=\left(\begin{array}{cc}-2& -3\\ 12& 5\end{array}\right).$

(a) Row·column. Har output entry $M$ ki ek row ka $\mathbf{x}=(2,-1)$ ke saath dot product hai: $M\mathbf{x}=\left(\begin{array}{c}1\cdot 2+2\cdot (-1)\\ 3\cdot 2+4\cdot (-1)\\ 5\cdot 2+6\cdot (-1)\end{array}\right)=\left(\begin{array}{c}0\\ 2\\ 4\end{array}\right).$ (b) Column combination. $M\mathbf{x}$ hai column 1 ki $2$ copies minus column 2 ki $1$ copy: $2\left(\begin{array}{c}1\\ 3\\ 5\end{array}\right)-1\left(\begin{array}{c}2\\ 4\\ 6\end{array}\right)=\left(\begin{array}{c}2-2\\ 6-4\\ 10-6\end{array}\right)=\left(\begin{array}{c}0\\ 2\\ 4\end{array}\right).$ Dono kyun kaam karte hain: row·column rule aur "columns ko vector ki entries se weight karo" — dono same arithmetic ko alag tarike se group karte hain. View (b) woh hai jo matrices ko linear transformations ke roop mein reveal karta hai — output hamesha columns ka ek weighted mix hota hai.

Pehle $AB=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& 2\\ 2& 5\end{array}\right)$. Phir $(AB)C=\left(\begin{array}{cc}1& 2\\ 2& 5\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 5& 2\end{array}\right)$. Ab doosra bracketing: $BC=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 0\end{array}\right)$. Phir $A(BC)=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 5& 2\end{array}\right)$. Equal ✓. Kyun guaranteed hai: dono ka matlab hai "pehle $C$ karo, phir $B$, phir $A$" — same pipeline; sirf kis order mein multiply karne ke liye pause kiya woh badla, maps act karne ka order nahi.

Parent note ka rule: "$B$ phir $A$" ko undo karne ke liye pehle $A$ undo karo, toh $(AB{)}^{-1}=B^{-1}{A}^{-1}.$ Compute karo: $B^{-1}{A}^{-1}=\left(\begin{array}{cc}1& 0\\ -3& 1\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ -5& -2\end{array}\right).$ Ordering ka proof (kyun $A^{-1}{B}^{-1}$ nahi): identity aur associativity use karke, $(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=AI{A}^{-1}=A{A}^{-1}=I.✓$ Inner $B{B}^{-1}$ cancel hota hai kyunki woh adjacent hai; reversed order hi cancellation ko line up karta hai. Galat order $A^{-1}{B}^{-1}$ beech mein ek mismatched $B{A}^{-1}$ chhod dega jo collapse nahi hoga.

$AB=\left(\begin{array}{cc}1& 2\\ 0& 3\end{array}\right)\left(\begin{array}{cc}4& 0\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}6& 4\\ 3& 6\end{array}\right)$, toh $(AB{)}^{\top }=\left(\begin{array}{cc}6& 3\\ 4& 6\end{array}\right)$. Ab transpose side. $B^{\top }=\left(\begin{array}{cc}4& 1\\ 0& 2\end{array}\right)$, $A^{\top }=\left(\begin{array}{cc}1& 0\\ 2& 3\end{array}\right)$: $B^{\top }{A}^{\top }=\left(\begin{array}{cc}4& 1\\ 0& 2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 2& 3\end{array}\right)=\left(\begin{array}{cc}6& 3\\ 4& 6\end{array}\right).$ Match ✓. Order phir se reverse kyun hota hai: transpose rows↔columns swap karta hai, aur kyunki $(AB{)}_{ij}$ $A$ ke row $i$ ko $B$ ke column $j$ ke saath pair karta hai, transpose karne se woh $A$ ka column $i$ ban jaata hai (= $A^{\top }$ ka row $i$)... iska ek hi consistent reassembly hai $B^{\top }{A}^{\top }$.

$A+B=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$. $(A+B{)}^2=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}5& 4\\ 4& 5\end{array}\right).$ Sahi expansion order rakhti hai: $(A+B{)}^2=A^2+AB+BA+B^2$. L3.2 se, $AB=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ aur $BA=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$. Saath hi $A^2=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$, $B^2=\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)$. Sum: $A^2+AB+BA+B^2=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)+\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)=\left(\begin{array}{cc}5& 4\\ 4& 5\end{array}\right)✓$ Galat $A^2+2AB+B^2$ mein $AB+BA=\left(\begin{array}{cc}3& 2\\ 2& 3\end{array}\right)$ ki jagah $2AB=\left(\begin{array}{cc}4& 2\\ 2& 2\end{array}\right)$ use hota, jo $\left(\begin{array}{cc}6& 4\\ 2& 4\end{array}\right)$ deta — galat, kyunki yahan $AB\ne BA$ hai.

Lo $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\ne 0$. $N^2=N\cdot N$ har entry ke liye row·column dot product se compute karo — koi shortcuts nahi.

• $(N^2{)}_{11}=$ row 1 $(0,1)\cdot$ col 1 $(0,0)=0\cdot 0+1\cdot 0=0$.
• $(N^2{)}_{12}=$ row 1 $(0,1)\cdot$ col 2 $(1,0)=0\cdot 1+1\cdot 0=0$.
• $(N^2{)}_{21}=$ row 2 $(0,0)\cdot$ col 1 $(0,0)=0\cdot 0+0\cdot 0=0$.
• $(N^2{)}_{22}=$ row 2 $(0,0)\cdot$ col 2 $(1,0)=0\cdot 1+0\cdot 0=0$. Dhyan do ki $N$ ka bottom row sab zeros hai, toh koi bhi dot product jo use karta hai automatically $0$ hai — yeh $(N^2{)}_{21}$ aur $(N^2{)}_{22}$ ko kill kar deta hai. Top row bachta hai, lekin iska jo bhi nonzero piece hai (position 2 mein $1$) hamesha $N$ ke columns se $0$ se milta hai, toh woh bhi khatam ho jaata hai. Is liye $N^2=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).$ Numbers ke liye kyun nahi: reals ke liye, $x^2=0\Rightarrow x=0$ (koi zero divisors nahi). Matrices mein zero divisors hote hain, toh "$XY=0\Rightarrow X=0$ ya $Y=0$" galat hai. Geometrically $N$ plane ko $x$-axis par collapse karta hai, phir jo bacha woh line ko origin par push kar deta hai — do squashes sab kuch maar dete hain.

$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ likho. Hypothesis kehti hai ki $AM=MA$ har $M$ ke liye; khaas taur par yeh specially chosen test matrices ke liye bhi hold karna chahiye, aur har choice $a,b,c,d$ par ek constraint nikaalti hai. Test 1: $E=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$. $AE=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}a& 0\\ c& 0\end{array}\right),EA=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}a& b\\ 0& 0\end{array}\right).$ $AE=EA$ entrywise set karo: $(2,1)$ entries dete hain $c=0$; $(1,2)$ entries dete hain $b=0$. Toh $A$ diagonal hona chahiye, $A=\left(\begin{array}{cc}a& 0\\ 0& d\end{array}\right)$. Test 2: $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ (swap matrix). Ab diagonal $A$ use karke: $AS=\left(\begin{array}{cc}a& 0\\ 0& d\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}0& a\\ d& 0\end{array}\right),SA=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}a& 0\\ 0& d\end{array}\right)=\left(\begin{array}{cc}0& d\\ a& 0\end{array}\right).$ $AS=SA$ set karo: $(1,2)$ entries force karte hain $a=d$. Is liye $A=\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right)=aI$, yani $A=cI$ jahan $c=a$. Proof kyun complete hota hai. Sab $M$ ke saath commute karna ek bahut strong demand hai — phir bhi humein sirf do cleverly chosen $M$ chahiye the saari char entries pin down karne ke liye: Test 1 ne off-diagonal entries ko khatam kiya, Test 2 ne do diagonal entries ko ek saath bandh kiya. Conversely, $cI$ obviously har cheez ke saath commute karta hai, kyunki $(cI)M=cM=M(cI)$ kisi bhi $M$ ke liye. Toh "sab ke saath commute karne wale" matrices ka set exactly $\{cI\}$ hai — na zyada, na kam. $■$

$2\times 2$ ke liye, $\det \left(\begin{array}{cc}p& q\\ r& s\end{array}\right)=ps-qr$ (dekho Determinants). $\det A=1\cdot 4-2\cdot 3=-2$. $\det B=0\cdot 2-1\cdot (-1)=1$. Product $=-2$. Ab $AB=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}-2& 5\\ -4& 11\end{array}\right)$, toh $\det (AB)=(-2)(11)-(5)(-4)=-22+20=-2$. ✓ Kyun remarkable hai: determinants area-scaling measure karte hain. Do maps karna scaling ko multiply karta hai — aur scalars multiply karna order care nahi karta, toh $\det (AB)=\det (BA)=\det A\det B$ hamesha hota hai, chahe matrices khud commute na karein. Order-sensitivity matrix mein hai, uske determinant mein nahi.