Question bank — Properties of determinants

False. Linearity (P5) ek row at a time kaam karti hai jab baaki rows frozen hain; poori matrices add karne se har row ek saath badal jaati hai, isliye yeh factor nahi ho sakta. Counterexample $A=B=I_2$: $\det (2I_2)=4\ne 1+1$.

False. $A$ ko $k$ se scale karna sabhi $n$ rows ko scale karta hai, aur har row ek factor of $k$ contribute karta hai (P4), isliye $\det (kA)=k^n\det A$.

False. Zero row $\det =0$ force karta hai (P7), lekin converse fail karta hai: $\det =0$ ka matlab sirf yeh hai ki rows linearly dependent hain, jaise $\left|\begin{array}{cc}1& 2\\ 2& 4\end{array}\right|=0$ bina kisi zero row ke.

True. P1 se har row rule ek column rule hai, aur rows swap karna sign flip karta hai (P2), isliye columns swap karna bhi karta hai.

Sirf square matrices ke determinants hote hain, isliye premise theek hai sirf square $A$ ke liye; un ke liye P1 deta hai $\det (A^{\top })=\det A$ kyunki permutation sum ko ${\sigma }^{-1}$ se reindex karna har signed term ko preserve karta hai.

True. Product rule se $\det (AB)=\det A\det B=\det B\det A=\det (BA)$, kyunki do numbers commute karte hain chahe matrices na karein.

False. Permutation sum off-diagonal entries bhi use karta hai. $\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|=-1\ne 0$.

False. "Slide Saves" (P6) ke liye alag row ka multiple add karna zaroori hai. $R_i\to R_i+k{R}_i=(1+k)R_i$ row ko scale karta hai, isliye P4 se determinant $(1+k)$ se multiply ho jaata hai.

Saamaanyatah: False. $-A=(-1)A$ isliye $\det (-A)=(-1{)}^n\det A$: even $n$ ke liye unchanged, odd $n$ ke liye negate.

True. $A{A}^{-1}=I$ aur product rule se, $\det A\cdot \det (A^{-1})=\det I=1$, isliye $\det (A^{-1})=1/\det A$ (jo yeh bhi dikhata hai ki inverse exist karne ke liye $\det A\ne 0$ zaroori hai). Dekho Inverse of a Matrix via Adjoint.

Swap ignore kar diya gaya. Slide (P6) safe hai, lekin swap (P2) running value ko $-1$ se multiply karta hai; yeh bhoolna final answer ka sign flip kar deta hai.

Agar $3$ sirf ek row se aaya tha, toh $3$ se divide karna sahi hai (P4). Lekin agar aapne har row se $3$ factor kiya toh aapne $3^3=27$ nikala, isliye $27$ se divide karna hoga, $3$ se nahi.

Proportional hona bhi isse khatam kar deta hai: $R_2$ se $5$ bahar nikalo (P4) aur do equal rows mil jaati hain, jo $\det =0$ force karti hain (P3). Trigger literal equality nahi, dependence hai.

Powers products hain, scalings nahi. P9 se, $\det (A^2)=\det A\cdot \det A=(\det A{)}^2$, na ki $2\det A$.

Har swap sign flip karta hai (P2). Agar computation ke dauran swap karo toh har swap ke liye $-1$ ka factor track karna hoga; warna "aasaan" wala route galat sign dega.

Inverting ka sign flips se koi lena-dena nahi hai. Sahi relation hai $\det (A^{-1})=1/\det A$ product rule se.

Permutation sum ka har term us row se exactly ek entry pick karta hai; har term mein zero factor available hone se, saare terms zero ho jaate hain (yeh exactly P7 hai, yaani P4 with $k=0$).

Jo bhi permutation diagonal se neeche jaata hai use khaali triangle se ek zero entry leni padti hai, jo us term ko kill kar deta hai; sirf identity permutation bachta hai, jo $a_{11}{a}_{22}\cdots a_{nn}$ deta hai (P8).

Additivity (P5) se yeh original plus $k$ times ek aise determinant mein split ho jaata hai jisme do equal rows hain; doosra $0$ hai (P3), isliye value "saved" ho jaati hai — exactly isi tarah P6 prove hota hai. Neeche sliding-card picture dekho.

Kyunki $\det (A^{\top })=\det A$ (P1): transpose karna rows ko columns mein badal deta hai bina value change kiye, isliye rows ke liye jo bhi statement prove ho woh verbatim columns par bhi transfer ho jaati hai.

Swap kuch nahi badalta (rows identical thi) phir bhi value negate karni padti hai (P2), isliye $\det =-\det$, aur sirf $0$ hi apne khud ke negative ke barabar hota hai.

Magnitude batata hai ki area/volume kitna stretch hota hai, aur sign record karta hai ki orientation preserve hoti hai $(+)$ ya mirror-flip hoti hai $(-)$. Neeche wali picture ek positive-area square aur uski flipped, negative-area mirror image dikhati hai — dekho Linear Transformations & Scaling of Area.

Convention se yeh $1$ hota hai — permutation sum mein exactly ek term hota hai, empty product, jo $1$ equals karta hai. Yeh rules jaise $\det (AB)=\det A\det B$ aur cofactor expansion ko tab bhi kaam karta rehne deta hai jab ek submatrix mein rows khatam ho jaayein.

Sirf $a$ khud — permutation sum mein ek hi identity term $a_{11}$ hai, isliye koi cross-multiplication nahi hoti.

$1$, kyunki yeh triangular hai aur saare diagonal entries $1$ hain (P8); yeh "no stretching, no flip" baseline hai.

Do rows dependent hain, isliye unit square ek line par squash ho jaata hai: zero area, koi inverse nahi. Yeh exactly Area of a Triangle using Determinants ka flat ho jaana hai.

Yeh zero row ban jaata hai, aur $\det =0$ (P4 with $k=0$, jo P7 hai); map ek direction completely collapse kar deta hai.

Haan — entries ki positivity dependence ke baare mein kuch nahi kehti. $\left|\begin{array}{cc}1& 1\\ 1& 1\end{array}\right|=0$ ke equal rows hain chahe har entry positive ho (P3).

Sirf $n=1$ ke liye, jahan $\det (kA)=ka=k\det A$; sabhi $n\ge 2$ ke liye sahi factor $k^n$ hai, isliye naive rule ek genuine trap hai.