Properties — row operations, multiplicativity

4.5.22 · Maths › Linear Algebra (Full)

WHAT are we studying?

WHY these rules hold — derivation from scratch

Hum EK foundational fact lete hain aur sab kuch derive karte hain. Determinant ek unique function $D (R_{1}, \dots, R_{n})$ hai rows ka, jo yeh satisfy karta hai:

(M) Multilinear: har row mein alag-alag linear hai,
(A) Alternating: $D = 0$ jab bhi do rows equal hon,
(N) Normalized: $D (I) = 1$ .

Neeche sab kuch isi se follow hota hai.

Rule 1: Do rows swap karna sign flip karta hai

Rows $u$ aur $v$ lo. Woh matrix consider karo jisme positions $i$ aur $j$ par $\dots, u + v, \dots, u + v, \dots$ rows hain. Kyunki do rows equal hain: $D (\dots, u + v, \dots, u + v, \dots) = 0.$ Dono positions mein multilinearity se expand karo: $D (u, u) + D (u, v) + D (v, u) + D (v, v) = 0.$ Lekin $D (u, u) = 0$ aur $D (v, v) = 0$ (equal rows). Toh: $D (u, v) + D (v, u) = 0 \Rightarrow == D (v, u) = - D (u, v) == .$

Yeh step kyun? Humne sirf "do equal rows zero deti hain" aur linearity use ki — bas itna hi kaafi hai. Sign flip koi extra axiom nahi hai; yeh ek consequence hai.

Rule 2: Ek row ko scale karna determinant ko scale karta hai

Multilinearity se (row $i$ mein linearity): $D (\dots, k R_{i}, \dots) === k D (\dots, R_{i}, \dots) == .$ Kyun? Us slot mein linear hone ka matlab hai $f (k x) = k f (x)$ directly.

Ek consequence: $det (k A) = k^{n} det A$ ek $n \times n$ matrix ke liye (saari $n$ rows scale hoti hain).

Rule 3: Replacement determinant ko unchanged chodta hai

$R_{i} \to R_{i} + k R_{j}$ karo (jahan $i \neq = j$ ). Slot $i$ mein multilinearity use karo: $D (\dots, R_{i} + k R_{j}, \dots) = D (\dots, R_{i}, \dots) + k D (\dots, R_{j}, \dots) .$ Doosre term mein $R_{j}$ slot $i$ aur slot $j$ dono mein hai → do equal rows → yeh $0$ hai. $\Rightarrow == D (\dots, R_{i} + k R_{j}, \dots) = D (\dots, R_{i}, \dots) == .$

Yeh step kyun? Yeh Gaussian elimination ka workhorse hai: tum entries clear kar sakte ho bina determinant kabhi change kiye.

Figure — Properties — row operations, multiplicativity

$det$ row reduction se compute karna

Multiplicativity kyun sach hai

Algebraic derivation

$A$ par har row operation = $A$ ko left se ek elementary matrix $E$ se multiply karna:

Swap $E$ : $det E = - 1$ .
Scale-by- $k$ $E$ : $det E = k$ .
Replacement $E$ : $det E = 1$ .

Rules 1–3 se, $E$ ko $A$ par apply karne se $det (E A) = det (E) det (A)$ milta hai — yeh har elementary $E$ ke liye sach hai.

Ab factor karo: agar $A$ invertible hai, $A = E_{1} E_{2} \dots E_{m}$ (elementaries ka product). Phir $det (A B) = det (E_{1} \dots E_{m} B) = det (E_{1}) \dots det (E_{m}) det (B) = det (A) det (B) .$ Agar $A$ singular hai, $det A = 0$ aur $A B$ bhi singular hai (rank $\leq$ rank $A < n$ ), toh $det (A B) = 0 = det A det B$ . $■$

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Self-test (answers dhako)

Row swap ka $det$ par effect? → $- 1$ se multiply karo.
$R_{i} \to R_{i} + k R_{j}$ ka effect? → kuch nahi.
$n \times n$ ke liye $det (k A)$ ? → $k^{n} det A$ .
$det (A B) =$ ? → $det A det B$ .
$det (A^{- 1}) =$ ? → $1/ det A$ .

Flashcards

Do rows swap karne ka det par effect

Determinant ko

- 1

se multiply karta hai.

R_{i} \to k R_{i}

ka det par effect

Determinant ko

k

se multiply karta hai.

R_{i} \to R_{i} + k R_{j}

(i≠j) ka det par effect

Koi change nahi — determinant invariant rehta hai.

Replacement det kyun nahi badalta?

Yeh ek term

k D (\dots, R_{j}, \dots, R_{j}, \dots)

add karta hai jisme do equal rows hain, jo 0 hota hai.

Swap sign kyun flip karta hai (derivation)?

D (u + v, u + v) = 0

expand karo;

D (u, v) + D (v, u) = 0

terms force karti hain

D (v, u) = - D (u, v)

n \times n

matrix ke liye

det (k A)

k^{n} det A

(har row scale hoti hai).

Multiplicativity batao

det (A B) = det A det B

det (A^{- 1})

det A

ke terms mein

1/ det A

, valid jab

A

invertible ho.

Elementary matrix ka

det

: swap / scale-by-k / replacement

- 1

k

1

det (A^{T})

det A

Equal — toh saare row rules columns par bhi kaam karte hain.

Triangular matrix ka determinant

Diagonal entries ka product.

Kya

det (A + B) = det A + det B

Nahi (counterexample

A = B = I_{2}

Recall Feynman: ek 12-saal ke bachche ko explain karo

Ek stretchy rubber square imagine karo. Ek matrix ek machine hai jo use kheeench ke aur teda karke ek parallelogram banati hai, aur determinant bas area kitne times bada hua hai (minus sign ke saath agar shape mirror ki tarah palat gayi).

Agar tum machine ki do instructions swap karo, shape palat jaati hai → minus sign.
Agar tum ek instruction double karo, woh direction do guna stretch hoti hai → area double hota hai.
Agar tum ek side ko doosri ke saath slide karo (shear/lean), parallelogram tedha hota hai lekin area same rehta hai → koi change nahi. Isi liye "ek row ka multiple doosri mein add karo" determinant par kuch nahi karta.
Agar tum machine $B$ phir machine $A$ chalao, total stretch bas dono stretches multiply hoti hain — yahi hai $det (A B) = det A \cdot det B$ .

Connections

Determinant — Definition (cofactor / Leibniz)
Elementary Matrices
Gaussian Elimination & Row Echelon Form
Invertibility & Singular Matrices
Volume, Orientation & the Determinant
Multilinear Alternating Forms
Eigenvalues — det as product of eigenvalues