Visual walkthrough — Determinant of 3×3 matrix — cofactor expansion

2.6.8 · D2 · Maths › Matrices & Determinants — Introduction › Determinant of 3×3 matrix — cofactor expansion

Hum sirf yeh assume karte hain ki tum numbers ka ek table padhna jaante ho. Baaki sab — yahan tak ki "determinant" ka matlab bhi — hum yahan rebuild karte hain.

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Step 1 — Matrix kya hai, aur uska determinant kya measure karta hai?

WHAT. Ek matrix bas numbers ka ek grid hota hai jo rows (across jaane wali) aur columns (neeche jaane wali) mein arrange hota hai. Ek 3×3 matrix mein 3 rows aur 3 columns hote hain — kul nau numbers. Hum har number ko $a_{ij}$ kehte hain: letter $i$ woh row hai jisme woh hai, aur $j$ woh column hai.

$A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

Yahan $a_{11}$ (padho "a-one-one") row 1, column 1 mein rehta hai — top-left. $a_{23}$ row 2, column 3 mein rehta hai.

WHY. Hum teen rows ko teen arrows ki tarah padhte hain jo 3D space mein point karte hain. Pehli row ek arrow hai, doosri row doosra, teesri row teesra. Determinant ek single number hai jo slanted box (ek parallelepiped — ek squished shoe-box) ki volume measure karta hai jo yeh teen arrows span karte hain. Bada box → bada number. Flat box → zero.

PICTURE. Teen arrows origin se nikl rahe hain; woh jo box cage karte hain uski volume hum chahte hain.

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Step 2 — Woh tool jo humre paas pehle se hai: 2×2 area

WHAT. 3D volume se pehle, 2D area yaad karo. Ek flat plane mein do arrows ek parallelogram (ek leaning rectangle) cage karte hain. Ek 2×2 matrix ke liye uski area ek clean formula hai, 2×2 determinant se:

$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$

Term by term: $ad$ main-diagonal product hai (top-left times bottom-right). $bc$ anti-diagonal product hai (top-right times bottom-left). Hum subtract karte hain kyunki $bc$ "wrong-way" overlap measure karta hai jo cancel hona chahiye.

WHY yeh tool. 3×3 ki poori strategy yeh hai ki kuch naya invent na karo. Agar hum 3D box ko un pieces mein chop kar sakein jinki sizes sirf 2×2 areas hain, toh kaam ho gaya — hum already areas jaante hain.

PICTURE. Parallelogram, uske diagonal-product rectangles shaded, ek red mein.

Recall

$\left|\begin{array}{cc}5& 6\\ -1& 2\end{array}\right|$ ki value kya hai? ::: $5\cdot 2-6\cdot (-1)=10+6=16$

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Step 3 — Box ko top row ke saath slice karo

WHAT. Pehli row dekho, $a_{11},a_{12},a_{13}$. Hum claim karte hain ki poori volume in teen numbers mein se har ek ko, ek ek karke, lene se aur use ek chhote 2×2 slice ki area se multiply karne se banti hai.

WHY. Volume of a box = (ek direction mein kitna door jaata hai) × (cross-section area jo woh saath khichta hai). Top-row entries "kitna door" ke pieces hain; 2×2 areas "cross-sections" hain. Toh hum top row ke across chalte hain aur teen volume-slabs collect karte hain, phir unhe add karte hain.

$a_{1j}$ se belonging 2×2 slice ko uska minor $M_{1j}$ kehte hain: row 1 (kyunki hum already use kar chuke hain) aur column $j$ (kyunki $a_{1j}$ wahan rehta hai) ko cover karo, aur baaki bache chaar numbers ka determinant lo.

PICTURE. Row 1 highlighted; $a_{11}$ ke liye hum uski row aur column grey out karte hain aur surviving 2×2 block red glow karta hai.

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Step 4 — Teen slices, draw out karke

WHAT. Har top-row entry ke liye covering karo baari baari:

$M_{11}=\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|,M_{12}=\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|,M_{13}=\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|$

• $M_{11}$: row 1 & column 1 cover karo → columns 2,3 survive karte hain.
• $M_{12}$: row 1 & column 2 cover karo → columns 1,3 survive karte hain (notice karo gap — column 2 gone hai).
• $M_{13}$: row 1 & column 3 cover karo → columns 1,2 survive karte hain.

WHY. Har minor woh cross-sectional area hai jo bachti hai jab hum "spend" karte hain ek top-row number ko ek direction mein. Teen numbers, teen cross-sections.

PICTURE. 3×3 teen baar side by side dikhaya gaya hai, har ek mein alag column red mein strike through hai.

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Step 5 — Minus sign kahan se paida hota hai

WHAT. Agar hum naively $a_{11}{M}_{11}+a_{12}{M}_{12}+a_{13}{M}_{13}$ add karein toh hume galat answer milega. Asli rule sign alternate karta hai:

$\det (A)=+a_{11}{M}_{11}-a_{12}{M}_{12}+a_{13}{M}_{13}$

Middle term subtract hota hai.

WHY. Surviving block ke columns dekho. $M_{11}$ ke liye humne columns $(2,3)$ order mein rakhe — natural order. $M_{12}$ ke liye humne $(1,3)$ rakhe: inhe "natural" left-to-right box ke saath line up karne ke liye humne column 1 ko column 2 ke past swap kiya. Ek swap box ko uske mirror image mein flip karta hai, jo volume ka sign flip karta hai. Woh single swap hi minus sign hai. $M_{13}$ ke liye, columns $(1,2)$ ko even number of swaps chahiye, toh plus hai phir se.

Hum is bookkeeping ko $(-1{)}^{i+j}$ mein package karte hain. Top row ke liye $i=1$:

$(-1{)}^{1+1}=+1,(-1{)}^{1+2}=-1,(-1{)}^{1+3}=+1$

$a_{ij}$ times uska sign times uska minor cofactor $C_{ij}=(-1{)}^{i+j}{M}_{ij}$ hai.

PICTURE. Checkerboard sign grid, top-row cells $+-+$ red mein picked out, aur ek chhota "one swap = flip" cartoon.

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Step 6 — Poora formula assemble karo aur test karo

WHAT. Steps 3–5 ko saath rakhne se parent ka headline result milta hai:

$\boxed{\det (A)=a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|}$

WHY test. Ek picture ka koi matlab nahi agar arithmetic toot jaaye. Lo $A=\left[\begin{array}{ccc}2& -3& 1\\ 5& 0& 6\\ -1& 4& 2\end{array}\right].$

• $M_{11}=\left|\begin{array}{cc}0& 6\\ 4& 2\end{array}\right|=0\cdot 2-6\cdot 4=-24$, sign $+$.
• $M_{12}=\left|\begin{array}{cc}5& 6\\ -1& 2\end{array}\right|=10+6=16$, sign $-$.
• $M_{13}=\left|\begin{array}{cc}5& 0\\ -1& 4\end{array}\right|=20-0=20$, sign $+$.

$\det (A)=2(-24)-(-3)(16)+1(20)=-48+48+20=20$

PICTURE. Teen signed slabs $-48,+48,+20$ ka bar chart jo red total $20$ mein stack ho raha hai.

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Step 7 — Koi bhi row kaam karti hai (aur zeros kyun chunne chahiye)

WHAT. Kuch bhi nahi tha jo humein top row slice karne ke liye force kare. Wohi argument kisi bhi row ya column ko slice karta hai. General rule:

$\det (A)={\sum }_{j=1}^3{a}_{ij}{C}_{ij}(\text{koi bhi fixed row }i\text{ chuno})$

Yahan $\sum$ ("sigma") ka matlab hai "inhe $j=1,2,3$ ke liye add karo." Chhon choices (teen rows, teen columns) sab identical number dete hain — properties dekho.

WHY. Kyunki yeh sab usi box ki volume measure karte hain. Toh woh row/column chuno jisme sabse zyada zeros hain — ek zero apna poora slab minor compute karne se pehle hi khatam kar deta hai. Ke liye $B=\left[\begin{array}{ccc}3& 0& 2\\ 1& 5& 0\\ 4& 0& 7\end{array}\right],$ column 2 mein do zeros hain, toh sirf middle survive karta hai: $\det (B)=5\cdot (-1{)}^{2+2}\left|\begin{array}{cc}3& 2\\ 4& 7\end{array}\right|=5(21-8)=65$

PICTURE. Column 2 highlighted; uske do zero-slabs faded ho gaye hain, sirf red middle slab bachti hai.

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Step 8 — Degenerate case: ek flat box zero deta hai

WHAT. Kya ho agar teen arrows 3D ke poore space mein nahi pahunchte — agar woh ek single flat sheet mein lie karte hain? Tab box ki koi thickness nahi, volume $=0$, toh $\det =0$. Example: $C=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right],\text{row 3}=2\cdot \text{row 2}-\text{row 1}.$

Kyunki row 3 baaki dono se bani hai, uska arrow pehle dono ke plane mein lie karta hai. Cofactor expansion confirm karta hai: $\det (C)=3(-3)+6(6)+9(-3)=-9+36-27=0$

WHY it matters. Zero determinant signal hai ki transformation 3D ko ek flat plane mein collapse kar raha hai — matrix ka koi inverse nahi hai, aur Cramer's rule jaise systems exactly yahan break karte hain.

PICTURE. Teen arrows ek tilted plane mein flat lie kar rahe hain; "box" ek red pancake hai zero height ke saath.

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Ek-picture summary

Upar sab kuch ek single frame mein: top row upar across, har entry ek signed 2×2 slab neeche fire kar rahi hai, teen slabs box ki volume mein sum ho rahe hain.

Recall Feynman retelling — seedha bolo

Numbers ki teen rows teen arrows hain; determinant woh volume hai us slanted box ka jo yeh cage karte hain. Ise measure karne ke liye, main top row ke across chalta hoon. Har top number batata hai box kitni door ek direction mein jaati hai; woh chhota 2×2 area jo us number ki row aur column chhupane ke baad bachta hai, woh cross-section batata hai jo woh saath kheechta hai. Distance times area = volume ka ek slab. Main teen slabs collect karta hoon — lekin middle wala mirror-flipped nikalta hai (ek column ko doosre ke past hop karna pada), toh main use subtract karta hoon: plus, minus, plus. Slabs add karo aur mujhe volume mil jaati hai. Main kisi bhi row ya column se start kar sakta hoon aur same answer milega, toh main wahan start karta hoon jahan zeros hain, kaam bachane ke liye. Aur agar teen arrows sab ek sheet mein flat lie karte hain, toh box ek pancake hai jisme koi height nahi — volume zero, determinant zero.