Properties — row operations, multiplicativity: Exercises

4.5.22 · D4 · Maths › Linear Algebra (Full) › Properties — row operations, multiplicativity

Intuition Is page ko kaise use karein

Ye problems ek ladder ki tarah chadhte hain. Har rung pichle se zyada maangta hai:

L1 Recognition — kya aap pehchaan sakte ho ki kaun-sa rule apply hota hai?
L2 Application — kya aap rule ko ek concrete matrix par run kar sakte ho?
L3 Analysis — kya aap rules ko combine karke bookkeeping (signs, scalings) track kar sakte ho?
L4 Synthesis — kya aap ek saath kai tools mein ideas ko chain kar sakte ho?
L5 Mastery — kya aap sirf numbers ki jagah general structure ke baare mein reason kar sakte ho?

Har answer ek collapsible Solution callout ke andar hai, toh unhe cover karo aur self-test karo. Ye page teen row rules aur multiplicativity assume karta hai jo Properties — row operations, multiplicativity se aati hain. Agar koi symbol unfamiliar lage, toh wo parent note usse zero se build karta hai.

Do quick reminders jinhe aap baar baar use karoge (har symbol parent mein earn kiya gaya hai):

Level 1 — Recognition

L1.1

$det A$ par har operation ka effect batao, aur corresponding elementary matrix $E$ ka determinant do: (a) $R_{1} \leftrightarrow R_{3}$ , (b) $R_{2} \to 5 R_{2}$ , (c) $R_{1} \to R_{1} - 4 R_{3}$ .

Recall Solution

Ek elementary matrix woh hoti hai jo aapko identity matrix $I$ (wo matrix jisme diagonal par $1$ s hain, baaki jagah $0$ s hain) par EK row operation karne se milti hai. Kyunki operation ko $A$ par karna matlab hai $A$ ko left se $E$ se multiply karna, factor $det E$ hi effect hai.

(a) Swap $\Rightarrow$ $det$ ko $- 1$ se multiply hota hai; $det E = - 1$ .
(b) Scale by $5$ $\Rightarrow$ $det$ ko $5$ se multiply hota hai; $det E = 5$ .
(c) Replacement $\Rightarrow$ $det$ unchanged rehta hai; $det E = 1$ .

L1.2

Kuch compute kiye bina, sign relationship decide karo: agar $A^{'}$ , $A$ hai jisme rows $1$ aur $2$ swap kiye gaye hain, aur $det A = 7$ hai, toh $det A^{'}$ kya hai?

Recall Solution

Ek swap sign ek baar flip karta hai. Toh $det A^{'} = - 7$ .

Level 2 — Application

L2.1

$det A$ compute karo row-reducing karke triangular form mein, har operation track karte hue: $A = 121253035 .$

Recall Solution

Step (kyun): column 1 ko pivot ke neeche clear karo replacements use karke — ye kabhi $det$ change nahi karte. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - R_{1}$ : $100211035 .$ Step: column 2 ki last entry clear karo: $R_{3} \to R_{3} - R_{2}$ (replacement, koi change nahi): $100210032 .$ Step: ab triangular hai. Koi swap nahi, koi scaling nahi, toh $det A = 1 \cdot 1 \cdot 2 = 2$ .

L2.2

Given $A$ ek $4 \times 4$ matrix hai jisme $det A = 3$ hai, $det (2 A)$ find karo.

Recall Solution

Poore matrix ko $2$ se multiply karna matlab hai har $4$ rows ko $2$ se scale karna. Har scaling ek factor of $2$ bahar nikalti hai: $det (2 A) = 2^{4} det A = 16 \cdot 3 = 48.$

L2.3

$A = (1324)$ , $B = (0110)$ . Verify karo ki $det (A B) = det A det B$ .

Recall Solution

$det A = 1 \cdot 4 - 2 \cdot 3 = - 2$ . $det B = 0 \cdot 0 - 1 \cdot 1 = - 1$ . $A B = (2413)$ , toh $det (A B) = 2 \cdot 3 - 1 \cdot 4 = 2$ . Check: $det A det B = (- 2) (- 1) = 2$ . ✓

Level 3 — Analysis

L3.1

$det B$ compute karo reduction se, aur aapko zaroori ek swap use karna hoga. Sign track karo. $B = 012210131 .$

Recall Solution

Top-left entry $0$ hai, isliye hum isse directly pivot ke roop mein use nahi kar sakte. Step (kyun): $R_{1} \leftrightarrow R_{2}$ swap karo taaki ek nonzero pivot upar aaye. Ye $1$ swap hai, toh $s = 1$ . $102120311 .$ Step: column 1 clear karo: $R_{3} \to R_{3} - 2 R_{1}$ (replacement, koi change nahi): $100 12 - 2 31 - 5 .$ Step: column 2 clear karo: $R_{3} \to R_{3} + R_{2}$ (replacement): $100120 31 - 4 .$ Step: pivots ka product $= 1 \cdot 2 \cdot (- 4) = - 8$ . Swap sign apply karo: $det B = (- 1)^{s} \cdot (- 8) = (- 1)^{1} \cdot (- 8) = 8.$

L3.2

Aap ek $3 \times 3$ matrix $C$ ko identity tak reduce karte ho. Is dauraan aapne kiya: 1 swap, ek row ko $\frac{1}{2}$ se scale kiya, ek aur row ko $3$ se scale kiya, aur kaafi saare replacements. $det C$ kya tha?

Recall Solution

$I$ tak reduce karne ka matlab hai final triangular product $det I = 1$ hai. Bookkeeping undo karo. Har operation ne current determinant ko apne factor se multiply kiya. $det C$ se start karke $1$ par khatam hote hue: $1 = (- 1)^{1} \cdot \frac{1}{2} \cdot 3 \cdot det C (replacements contribute \times 1) .$ Toh $(- 1) \cdot \frac{3}{2} \cdot det C = 1 \Rightarrow det C = \frac{1}{- \frac{3}{2}} = - \frac{2}{3} .$

L3.3

$A$ invertible hai jisme $det A = 4$ hai. $det (A^{- 1})$ , $det (A^{3})$ , aur $det (2 A^{- 1})$ find karo $A$ ke size $3 \times 3$ ke liye.

Recall Solution

$det (A^{- 1}) = \frac{1}{det A} = \frac{1}{4}$ , kyunki $det A \cdot det A^{- 1} = det I = 1$ (dekho Invertibility & Singular Matrices).
$det (A^{3}) = (det A)^{3} = 4^{3} = 64$ (multiplicativity do baar apply ki).
$det (2 A^{- 1}) = 2^{3} det (A^{- 1}) = 8 \cdot \frac{1}{4} = 2$ (teen rows $2$ se scale ki).

Level 4 — Synthesis

L4.1 (geometric)

Do $2 \times 2$ matrices unit square par act karti hain (wo square jiske corners $(0, 0), (1, 0), (1, 1), (0, 1)$ hain): $S = (2001), H = (1011) .$ (a) Har ek square ke saath geometrically kya karta hai? (b) $det S$ aur $det H$ compute karo. (c) $H S$ ke area factor ko predict karne ke liye multiplicativity use karo, phir $det (H S)$ compute karke confirm karo.

Figure — Properties — row operations, multiplicativity

Recall Solution

(a) Figure dekho. $S$ horizontal direction ko $2$ se stretch karta hai (ek $2 \times 1$ rectangle) — ek genuine area change. $H$ ek shear hai (matrix form mein ek replacement operation): ye top edge ko sideways slide karta hai lekin base aur height wahi rakhta hai, isliye parallelogram ka area square ke barabar hai. Ye "replacement $det$ change nahi karta" ka geometric face hai. (b) $det S = 2 \cdot 1 - 0 = 2$ . $det H = 1 \cdot 1 - 1 \cdot 0 = 1$ . (c) Multiplicativity predict karta hai ki pehle $S$ phir $H$ karne (map $H S$ ) ka area factor $det H \cdot det S = 1 \cdot 2 = 2$ hai. Confirm: $H S = (1011) (2001) = (2011)$ , aur $det (H S) = 2 \cdot 1 - 1 \cdot 0 = 2$ . ✓ Shear ne koi area nahi badhaya.

L4.2

Prove karo ki agar $A$ mein do identical rows hain, toh $det A = 0$ hai, sirf ek replacement use karke. Phir illustrate karo $A = 335119442 .$

Recall Solution

Proof: maano rows $i$ aur $j$ equal hain. Replacement $R_{i} \to R_{i} - R_{j}$ karo. Replacement $det$ change nahi karta. Lekin ab row $i$ saari zeros hai, aur ek matrix jisme zero row ho uska $det = 0$ hota hai (ek zero row matlab ek direction kuch nahi raha — zero volume). Toh $det A = 0$ . Illustration: rows $1$ aur $2$ equal hain. $R_{2} \to R_{2} - R_{1}$ se ek zero row $2$ milti hai; isliye $det A = 0$ .

L4.3

$P$ ek $3 \times 3$ permutation matrix hai jo $I$ se rows ki cyclic reordering $(R_{1}, R_{2}, R_{3}) \to (R_{2}, R_{3}, R_{1})$ se mili hai. Swaps count karke $det P$ find karo.

Recall Solution

$3$ rows ki ek cyclic shift $2$ swaps se banayi ja sakti hai: $I$ rows $(1, 2, 3)$ se start karo. Swap $R_{1} \leftrightarrow R_{2} \to (2, 1, 3)$ ; swap $R_{2} \leftrightarrow R_{3} \to (2, 3, 1)$ — target ordering. Wo $2$ swaps hain, toh $det P = (- 1)^{2} = 1$ .

Level 5 — Mastery

L5.1

Maano $A$ ek $n \times n$ matrix hai jisme $det A = d$ hai. $det ((c A)^{- 1})$ ke liye ek single formula do $c$ , $d$ , $n$ mein (maano $A$ invertible hai, $c \neq = 0$ ). Phir $n = 3, c = 2, d = 5$ ke liye evaluate karo.

Recall Solution

Tools se isse alag karo: $det (c A) = c^{n} det A = c^{n} d, det ((c A)^{- 1}) = \frac{1}{d e t ( c A )} = \frac{1}{c ^{n} d} .$ $n = 3, c = 2, d = 5$ ke liye: $\frac{1}{2 ^{3} \cdot 5} = \frac{1}{40}$ .

L5.2

Maano $A$ ek $n \times n$ matrix hai aur $A^{2} = A$ (aisi matrix ko idempotent kehte hain). $det A$ ki possible values kya hain? Multiplicativity se justify karo, aur eigenvalues se connect karo.

Recall Solution

$A^{2} = A$ ke dono sides par $det$ apply karo. Multiplicativity se $det (A^{2}) = (det A)^{2}$ , aur right side $det A$ hai: $(det A)^{2} = det A \Rightarrow (det A)^{2} - det A = 0 \Rightarrow det A (det A - 1) = 0.$ Toh $det A \in {0, 1}$ . Connection: ek idempotent ke eigenvalues saare $0$ ya $1$ hote hain, aur $det A$ eigenvalues ka product hai. Agar koi eigenvalue $0$ hai toh product $0$ hai (singular projection); agar saare $1$ hain toh product $1$ hai (tab $A = I$ ). Dono cases match karte hain.

L5.3

Prove karo ki agar $Q$ ek $n \times n$ matrix hai jo elementary matrices ke product ke roop mein bana hai $Q = E_{1} E_{2} \dots E_{m}$ jahan exactly $s$ of them swaps hain, $t$ of them scale karte hain (respectively) $k_{1}, \dots, k_{t}$ se, aur baaki replacements hain, toh $det Q = (- 1)^{s} \prod_{r = 1}^{t} k_{r} .$

Recall Solution

Ye kyun true hai: multiplicativity kehta hai $det (E_{1} E_{2} \dots E_{m}) = det E_{1} \cdot det E_{2} \dots det E_{m}$ . Har factor ek known number hai:

har swap contribute karta hai $det E = - 1$ — $s$ hain, jo deta hai $(- 1)^{s}$ ;
har scale-by- $k_{r}$ contribute karta hai $det E = k_{r}$ — jo deta hai $\prod_{r} k_{r}$ ;
har replacement contribute karta hai $det E = 1$ — product untouched rehta hai. Saare factors multiply karke: $det Q = (- 1)^{s} (\prod_{r} k_{r}) \cdot 1 = (- 1)^{s} \prod_{r} k_{r}$ . $■$ Ye "REF bookkeeping" formula ke peeche ka abstract engine hai — dekho Gaussian Elimination & Row Echelon Form.

Recall Ek-line ladder recap (cover karo aur recite karo)

L1: effect ka naam batao $\to$ swap $- 1$ , scale $k$ , replacement $1$ .
L2: reduce karo & pivots multiply karo; $det (2 A)$ ke liye $2^{n}$ dhyaan rakho.
L3: $(- 1)^{s}$ aur har scaling factor carry karo; divisions pay back karo.
L4: shear = area-preserving; identical rows $\Rightarrow det = 0$ .
L5: $det (c A)^{- 1} = \frac{1}{c ^{n} d}$ ; idempotent $\Rightarrow det \in {0, 1}$ ; $det Q = (- 1)^{s} \prod k_{r}$ .