Properties — row operations, multiplicativity: Worked examples

4.5.22 · D3 · Maths › Linear Algebra (Full) › Properties — row operations, multiplicativity

Yeh deep dive parent properties note par build karta hai aur cofactor/Leibniz definition, Gaussian Elimination & Row Echelon Form, Elementary Matrices, Invertibility & Singular Matrices, aur Volume, Orientation & the Determinant par lean karta hai.

Woh teen bookkeeping symbols jo hum use karenge

Neeche har worked example ek matrix ko triangular form mein reduce karta hai aur answer ek master formula se padh leta hai. Do chhote symbols saara bookkeeping karte hain — aao inhe ek baar plain words mein define kar lete hain, kahi bhi use karne se pehle.

Definition Swap-count

$s$ aur scale-product $\prod k$

$== s ==$ simply row swaps ki sankhya hai jo tum reduction ke dauran perform karte ho. Unhe apni ungliyon par gino: jab bhi tum do rows exchange karo, $s$ mein $1$ add karo. Kyunki ek single swap determinant ko $- 1$ se multiply karta hai (parent Rule 1), $s$ swaps karne se woh $(- 1)^{s}$ se multiply hota hai — jo even count ke liye $+ 1$ hota hai aur odd count ke liye $- 1$ .
$== \prod k ==$ (padho "sabhi $k$ 's ka product") har us scaling factor $k$ ka product hai jo tumne use kiya jab tumne $R_{i} \to k R_{i}$ form ka operation kiya. Agar tumne ek row ko $\frac{1}{3}$ se aur doosri ko $2$ se scale kiya, toh $\prod k = \frac{1}{3} \cdot 2 = \frac{2}{3}$ . Agar tumne kabhi koi row scale nahi ki, toh multiply karne ke liye koi factors nahi hain, isliye convention se $\prod k = 1$ . Kyunki ek row ko $k$ se scale karna determinant ko $k$ se multiply karta hai (parent Rule 2), kai rows ko scale karna use unke product $\prod k$ se multiply karta hai.

Dhyan do — kya count NAHI hota. Ek replacement $R_{i} \to R_{i} + k R_{j}$ (doosri row ka multiple add karna) ek scaling nahi hai: yeh determinant ko unchanged chodta hai, isliye iska $k$ kabhi $\prod k$ mein collect nahi hota aur $s$ ko kabhi touch nahi karta. Sirf standalone move $R_{i} \to k R_{i}$ hi $\prod k$ ko feed karta hai.

Un do symbols ke saath nail ho jaane ke baad, har example sirf careful counting hai.

Scenario matrix

Har determinant computation jo tum kabhi bhi miloge, in cells mein se kisi ek mein aayegi. Is page ka poora point yeh ensure karna hai ki ek bhi cell andheri na reh jaaye.

#	Case class	Kya cheez tricky banati hai	Covered by
C1	Pure replacement to triangular, koi swap/scale nahi	Bookkeeping: prove karo ki kuch lose nahi hua	Ex 1
C2	Ek swap aata hai → sign bookkeeping	$(- 1)^{s}$ factor mein off-by-one	Ex 2
C3	Tum pivot $1$ paane ke liye ek row scale karte ho → divide back karna padega	Scale undo karna bhool jana	Ex 3
C4	Degenerate / singular input → $det = 0$	Recognize karna ki ek zero pivot aata hai	Ex 4
C5	Sign / orientation case → negative determinant, flipped area	Minus sign ko "area got mirrored" ki tarah padhna	Ex 5 (figure)
C6	Mixed signs ke saath Multiplicativity $det A < 0 < det B$	Signs multiply hote hain, add nahi	Ex 6
C7	Poori matrix ka scalar multiple $det (k A) = k^{n} det A$ , negative $k$ bhi shamil	Exponent $n$ hai, aur $k$ negative ho sakta hai	Ex 7
C8	Inverse / power corollaries, ek limiting "almost singular" matrix bhi shamil	$det A^{- 1} = 1/ det A$ blow up karta hai jab $det A \to 0$	Ex 8
C9	Word problem — real-world area scaling	Geometry ko determinant mein translate karna	Ex 9 (figure)
C10	Exam twist — ek hi shot mein swap + scale + multiplicativity combine karo	Saari bookkeeping simultaneously karna	Ex 10

Table ko ek baar upar se neeche padho. Har row ek promise hai; das examples unhe sab nibhate hain.

Ex 1 — Cell C1: pure replacement, koi swap nahi, koi scale nahi

Worked example Statement

$det A$ compute karo $A = 123257 - 1 12$ sirf replacement operations ( $R_{i} \to R_{i} + k R_{j}$ ) use karke.

Forecast: Kyunki replacement kabhi determinant nahi badalta, answer sirf un pivots ka product hoga jo hum reach karte hain — padhne se pehle guess karo ki woh positive hoga ya negative.

Step 1. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 3 R_{1}$ . Yeh step kyun? Replacement $det$ ko bilkul unchanged chodta hai (yeh do equal rows wala ek term add karta hai, jo $0$ hai), isliye yeh ek "free" move hai. Hum ise pivot ke neeche pehle column ko zero out karne ke liye use karte hain. $\to 100211 - 1 35$ Step 2. $R_{3} \to R_{3} - R_{2}$ . Yeh step kyun? Same free move, ab upper-triangular form reach karne ke liye doosre pivot ke neeche ki entry clear karna. $\to 100210 - 1 32$ Step 3. Ek triangular matrix ke liye, $det$ = diagonal ka product. Humne sirf replacements use kiye, isliye swap-count $s = 0$ hai aur koi row-scalings nahi thi, jo $\prod k = 1$ deta hai. Master formula tab yeh padh'ta hai $det A = (- 1)^{0} \cdot \frac{1}{1} \cdot (1 \cdot 1 \cdot 2) = 2.$

Verify: Aao column 1 ke along cofactor-expand karein aur deliberately pehle trap dikhate hain. Ek tempting shortcut hai sirf top entry ka term likhna, $1 \cdot (5 \cdot 2 - 1 \cdot 7) = 1 \cdot 3 = 3$ — yeh galat hai, kyunki yeh chup-chaap column 1 mein $2$ aur $3$ se aane wale do doosre cofactor terms drop kar deta hai (woh original $A$ mein zero nahi hain; sirf reduction ke baad un entries ko zero kiya gaya tha). Honest full expansion teeno terms rakhta hai: $1 (5 \cdot 2 - 1 \cdot 7) - 2 (2 \cdot 2 - 1 \cdot 3) + (- 1) (2 \cdot 7 - 5 \cdot 3) = 1 (3) - 2 (1) - 1 (- 1) = 3 - 2 + 1 = 2.$ Toh correct value $2$ hai, jo reduction se match karta hai. ✓

Ex 2 — Cell C2: ek swap aata hai

Worked example Statement

$det B$ compute karo $B = 012213101$

Forecast: Top-left entry $0$ hai, isliye hume usable pivot paane ke liye swap karna zaroori hai. Ek swap matlab ek factor of $- 1$ . Apne final answer ka sign predict karo.

Step 1. $R_{1} \leftrightarrow R_{2}$ swap karo. Yeh step kyun? Tum $0$ par pivot nahi kar sakte. Ek swap ise fix karta hai, $det$ ko $- 1$ se multiply karne ki jaani-pehchaani cost par (Rule 1 se). Yeh hamara pehla (aur ek hi) swap hai, isliye swap-count ab $s = 1$ hai. $\to 102123011$ Step 2. $R_{3} \to R_{3} - 2 R_{1}$ (free replacement). $\to 100121011$ Step 3. $R_{3} \to R_{3} - \frac{1}{2} R_{2}$ (free replacement — dhyan do yeh row ka apne aap mein scaling nahi hai, yeh $R_{3}$ mein $R_{2}$ ka multiple add karna hai, isliye $det$ untouched hai aur yeh $s$ ya $\prod k$ ko feed nahi karta). $\to 100120 01 \frac{1}{2}$ Step 4. Pivots product $= 1 \cdot 2 \cdot \frac{1}{2} = 1$ . Koi scalings nahi toh $\prod k = 1$ ; swap factor $(- 1)^{s} = (- 1)^{1}$ apply karo: $det B = (- 1)^{1} \cdot \frac{1}{1} \cdot 1 = - 1.$

Verify: Leibniz directly: $0 (1 \cdot 1 - 0 \cdot 3) - 2 (1 \cdot 1 - 0 \cdot 2) + 1 (1 \cdot 3 - 1 \cdot 2) = 0 - 2 (1) + 1 (1) = - 1.$ ✓

Ex 3 — Cell C3: pivot $1$ banane ke liye row ko scale karna, phir divide back karna

Worked example Statement

$det C$ compute karo jahan $C = 310621954$ deliberately $R_{1}$ scale karte hue taaki uska pivot $1$ ho jaaye (taaki tum ise undo karne ki practice kar sako).

Forecast: Agar tum ek row ko $\frac{1}{3}$ se multiply karte ho toh poora determinant $3$ se divide ho jaata hai, isliye end mein tumhe $3$ se multiply back karna hoga. Guess karo: kya final $det$ $3$ ke multiple mein aata hai?

Step 1. $R_{1} \to \frac{1}{3} R_{1}$ . Yeh step kyun? Ek clean leading $1$ arithmetic simplify karta hai — lekin row ko $\frac{1}{3}$ se scale karna $det$ ko $\frac{1}{3}$ se multiply karta hai (Rule 2). Yeh ek genuine row-scaling hai, isliye yeh scale-product ko feed karta hai: $\prod k = \frac{1}{3}$ . Master formula ke anusaar hum end mein $det U$ ko is $\prod k$ se divide karenge, yaani $3$ se multiply karenge. $\to 110221354$ Step 2. $R_{2} \to R_{2} - R_{1}$ (free replacement). $\to 100201324$ Step 3. Doosra pivot ab $0$ hai. $R_{2} \leftrightarrow R_{3}$ swap karo; yeh hamara pehla swap hai, isliye swap-count $s = 1$ hai. $\to 100210342$ Step 4. Triangular: pivots product $det U = 1 \cdot 1 \cdot 2 = 2$ . Master formula $det C = (- 1)^{s} \cdot \frac{1}{\prod k} \cdot det U$ assemble karo $s = 1$ aur $\prod k = \frac{1}{3}$ ke saath: $det C = (- 1)^{1} \cdot \frac{1}{1/3} \cdot 2 = (- 1) \cdot 3 \cdot 2 = - 6.$

Verify: $C$ par direct Leibniz: $3 (2 \cdot 4 - 5 \cdot 1) - 6 (1 \cdot 4 - 5 \cdot 0) + 9 (1 \cdot 1 - 2 \cdot 0) = 3 (3) - 6 (4) + 9 (1) = 9 - 24 + 9 = - 6.$ ✓

Ex 4 — Cell C4: degenerate / singular input, $det = 0$

Worked example Statement

$det D$ compute karo $D = 125240361$

Forecast: Calculate karne se pehle rows $R_{1}$ aur $R_{2}$ ko dhyan se dekho. Dikh raha hai? Koi arithmetic ke bina answer predict karo.

Step 1. Notice karo $R_{2} = 2 R_{1}$ exactly. Yeh step kyun? Agar ek row doosri ka scalar multiple hai, toh matrix singular hai — uski rows linearly dependent hain, isliye unse bana parallelepiped flat squash ho jaata hai (zero volume). Hum ise mechanically confirm kar sakte hain: Step 2. $R_{2} \to R_{2} - 2 R_{1}$ (free replacement). $\to 105200301$ Ek zero row appear hoti hai. Multilinearity se, us zero row ko kisi bhi $k$ se scale karna use zero row hi chhod deta hai, phir bhi $det$ ko $k$ se multiply karna chahiye; ek hi number jo har $k$ se unchanged rehta hai woh $0$ hai. Isliye $det D = 0.$

Verify: Leibniz: $1 (4 \cdot 1 - 6 \cdot 0) - 2 (2 \cdot 1 - 6 \cdot 5) + 3 (2 \cdot 0 - 4 \cdot 5) = 1 (4) - 2 (- 28) + 3 (- 20) = 4 + 56 - 60 = 0.$ ✓

Ex 5 — Cell C5: negative determinant matlab orientation flip

Worked example Statement

Matrix $M = (0110)$ unit square ko ek naye shape mein map karta hai. $det M$ find karo aur bolo ki sign geometrically kya mean karti hai.

Forecast: $M$ sends $(1, 0) \mapsto (0, 1)$ aur $(0, 1) \mapsto (1, 0)$ — yeh axes swap karta hai. Padhne se pehle area factor aur uski sign guess karo.

Step 1. $M$ ek row swap se $I$ se dur hai: $R_{1} \leftrightarrow R_{2}$ swap karne se $I$ milta hai, jiska $det = 1$ hai. Yeh step kyun? $det I = 1$ normalization axiom hai, aur ek single swap ( $s = 1$ ) $(- 1)^{1}$ se multiply karta hai. $det M = (- 1)^{1} \cdot 1 = - 1.$ Step 2. Sign interpret karo. $∣ det M ∣ = 1$ matlab area preserved hai. Minus sign matlab orientation flip — image original ka ek mirror reflection hai, jaise square ko palat diya ho. Figure mein, dekho corner-labels $P, Q, R$ map ke baad clockwise jaate hain jabki pehle counter-clockwise the.

Figure — kya observe karna hai: BAAYI taraf, teal unit square ke corners $O \to P \to Q \to R$ counter-clockwise trace hote hain (purple arrow CCW curl karta hai). DAAHINI taraf, $M$ apply karne ke baad, exactly wahi corner labels ab clockwise trace karte hain (purple arrow CW curl karta hai), even though shape ka size same hai. Ghoomne ki direction ka yeh reversal exactly wahi hai jo ek negative determinant jaisa dikhta hai — plane ko pancake ki tarah palat diya gaya hai, sirf rotate nahi kiya.

Verify: Leibniz $2 \times 2$ : $0 \cdot 0 - 1 \cdot 1 = - 1.$ ✓

Ex 6 — Cell C6: mixed signs ke saath multiplicativity

Worked example Statement

$A = (1324)$ (toh $det A = - 2$ ) aur $B = (2013)$ (toh $det B = 6$ ). $det (A B)$ do tareekon se compute karo.

Forecast: Ek determinant negative hai, ek positive. Kya woh add hote hain ya multiply? $det (A B)$ ka sign predict karo.

Step 1. Matrices multiply karo. Yeh step kyun? Hum pehle $det (A B)$ "honest" tarike se chahte hain, shortcut check karne ke liye. $A B = (1 \cdot 2 + 2 \cdot 0 3 \cdot 2 + 4 \cdot 0 1 \cdot 1 + 2 \cdot 3 3 \cdot 1 + 4 \cdot 3) = (26715) .$ Step 2. $det (A B) = 2 \cdot 15 - 7 \cdot 6 = 30 - 42 = - 12.$ Step 3. Multiplicativity ke through shortcut: $det A \cdot det B = (- 2) (6) = - 12.$ Yeh kyun kaam karta hai: har map signed area ko scale karta hai, aur maps stack karna un factors ko multiply karta hai (ek flip times ek non-flip phir bhi ek flip hai → negative).

Verify: Dono routes $- 12$ dete hain, aur $(- 2) (6) = - 12$ . ✓

Ex 7 — Cell C7: $det (k A) = k^{n} det A$ , negative $k$ bhi shamil

Ex 8 — Cell C8: inverse, power, aur "almost singular" limit

Worked example Statement

Maano $A = (21 1 ε)$ . (a) $det A$ find karo. (b) $ε = 1$ ke liye $det (A^{- 1})$ aur $det (A^{3})$ find karo. (c) Describe karo kya hota hai $det (A^{- 1})$ ke saath jab $ε \to \frac{1}{2}$ .

Forecast: $det (A^{- 1}) = 1/ det A$ . $1/ det A$ ka kya hona chahiye jab $det A$ $0$ ki taraf jaata hai?

Step 1. $det A = 2 ε - 1 \cdot 1 = 2 ε - 1.$ Yeh step kyun? $2 \times 2$ matrix $(a c b d)$ ke liye Leibniz/cofactor definition $det = a d - b c$ deta hai — do "diagonal products" minus do "anti-diagonal products". Yahan $a = 2, b = 1, c = 1, d = ε$ , isliye $det A = 2 ε - 1$ . Step 2 (b), $ε = 1$ . $det A = 2 (1) - 1 = 1.$ Phir $det (A^{- 1}) = \frac{1}{d e t A} = \frac{1}{1} = 1, det (A^{3}) = (det A)^{3} = 1^{3} = 1.$ Kyun: $det (A^{- 1}) = 1/ det A$ aata hai $det A \cdot det A^{- 1} = det I = 1$ se; $det (A^{k}) = (det A)^{k}$ multiplicativity ko iterate karke milta hai. Step 3 (c), limit — sign matter karta hai. $det A = 2 ε - 1$ likho, isliye $det (A^{- 1}) = \frac{1}{2 ε - 1}$ . Jab $ε \to \frac{1}{2}$ denominator $\to 0$ hota hai, aur uski sign is baat par depend karti hai ki tum kis side se approach kar rahe ho:

Agar $ε ↓ \frac{1}{2}$ (upar se, $ε > \frac{1}{2}$ ), toh $2 ε - 1 > 0$ aur tiny hai, isliye $det (A^{- 1}) \to + \infty$ .
Agar $ε ↑ \frac{1}{2}$ (neeche se, $ε < \frac{1}{2}$ ), toh $2 ε - 1 < 0$ aur tiny hai, isliye $det (A^{- 1}) \to - \infty$ .

Toh blow-up $+ \infty$ hai right se aur $- \infty$ left se — two-sided limit exist nahi karta. Geometrically parallelogram opposite orientations se ek line ki taraf crush ho raha hai, isliye inverse map ka stretch factor bound ke baghair badh'ta hai, ek side par positive aur doosri par negative (orientation-flipped).

Verify: $ε = 1$ par: $det A = 1$ , $det A^{- 1} = 1$ , $det A^{3} = 1$ ; aur $det A = 2 ε - 1 = 0$ exactly $ε = \frac{1}{2}$ par. ✓

Ex 9 — Cell C9: word problem, real-world area scaling

Worked example Statement

Ek graphic designer transform $T = (3012)$ ek company logo par apply karta hai jiska original area $12 cm^{2}$ hai. Naya area kya hai, aur kya logo mirror ho gaya?

Forecast: Determinant area-scaling factor hai. Compute karne se pehle naya area predict karo.

Step 1. $det T = 3 \cdot 2 - 1 \cdot 0 = 6.$ Yeh step kyun? Determinant exactly linear map ka signed area-scaling factor hota hai, isliye kisi bhi shape ka area $∣ det T ∣$ se multiply hota hai. Step 2. Naya area $= ∣ det T ∣ \times 12 = 6 \times 12 = 72 cm^{2} .$ Step 3. $det T$ ki sign positive hai, isliye orientation preserved hai — logo stretch aur shear hua hai, mirror nahi hua. Figure unit square (area 1) ko area 6 ke parallelogram mein bante dikhata hai.

Figure — kya observe karna hai: Chhota teal square original unit cell hai (area $1$ ). $T$ apply karo aur woh orange parallelogram ban jaata hai: do basis arrows $T e_{1} = (3, 0)$ (purple, axis ke along) aur $T e_{2} = (1, 2)$ (ink, tilted) par land karte hain. Dhyan do arrows phir bhi original axes ki tarah same way turn karte hain — koi mirror flip nahi — isliye $det T = + 6$ positive hai. Orange region visibly teal square ki chhe copies rakhta hai, confirm karta hai ki area factor $det T = 6$ se badha. Apne real logo ke $12 cm^{2}$ ko usi $6$ se multiply karo.

Verify: $det T = 6 > 0$ aur $6 \times 12 = 72$ . Units: (dimensionless factor) $\times cm^{2} = cm^{2}$ . ✓

Ex 10 — Cell C10: exam twist, sab ek saath

Worked example Statement

Diya hai ki ek $3 \times 3$ matrix $P$ ke liye $det P = 4$ hai, aur $E_{1}$ ( $R_{1} \leftrightarrow R_{3}$ swap) aur $E_{2}$ ( $R_{2}$ ko $5$ se scale karo) elementary matrices hain, compute karo $det (2 E_{1} E_{2} P) .$

Forecast: Chaar effects stack karte hain: ek $3 \times 3$ par scalar $2$ (woh $2^{3}$ hai), ek swap ( $- 1$ ), ek scale-by- $5$ ( $5$ ), aur $det P = 4$ . Unhe multiply karo — add mat karo.

Step 1. Scalar peel karo: $det (2 X) = 2^{3} det X = 8 det X$ jahan $X = E_{1} E_{2} P$ . Yeh step kyun? $2 X$ saari $n = 3$ rows scale karta hai (Rule 2 / Cell C7), jo factor $2^{n} = 2^{3} = 8$ deta hai. Step 2. Multiplicativity use karo: $det (E_{1} E_{2} P) = det E_{1} \cdot det E_{2} \cdot det P$ . Yeh step kyun? $det (A B) = det A det B$ teeno factors par chain kiya. Step 3. Elementary determinants fill karo (parent ki table se): ek swap matrix ka $det E_{1} = - 1$ hota hai, ek scale-by- $5$ matrix ka $det E_{2} = 5$ hota hai, aur humein bataya gaya hai $det P = 4$ . Substitute karke, $det (E_{1} E_{2} P) = (- 1) \cdot 5 \cdot 4 = - 20.$ Step 4. Step 1 ke scalar factor ke saath combine karo: $det (2 E_{1} E_{2} P) = 8 \cdot (- 20) = - 160.$

Verify: $8 \times (- 1) \times 5 \times 4 = - 160$ . ✓

Active recall

Recall Kaunsa cell, kaunsa factor? (answers cover karo)

Tumne ek $det$ compute karte waqt 3 baar rows swap kiye — kya factor? ::: $(- 1)^{3} = - 1$ .
$det A = 2$ wale $4 \times 4$ ke liye $det (3 A)$ ? ::: $3^{4} \cdot 2 = 162$ .
$A$ ke do proportional rows hain — $det$ ? ::: $0$ (singular).
$det A = - 3$ , $det B = 2$ , toh $det (A B)$ ? ::: $- 6$ .
Jab $det A \to 0^{+}$ , $det (A^{- 1})$ kya karta hai? ::: $\to + \infty$ (aur $\to - \infty$ agar $det A \to 0^{-}$ ).
$det T = 6$ positive — mirrored hai ya nahi? ::: Mirror nahi (orientation preserved).