Question bank — Determinant of 2×2 matrix

False — har side ko $2$ se scale karne par columns $2$ se scale hote hain, isliye area $2\times 2=4$ se scale hota hai; $\det (2A)=2^2\det (A)=4$ ek $2\times 2$ matrix ke liye.

False — $\det =1$ sirf yeh kehta hai ki area aur orientation preserved hai; ek rotation ya shear jaise $\left[\begin{array}{cc}1& 5\\ 0& 1\end{array}\right]$ ka bhi $\det =1$ hota hai lekin woh identity nahi hai.

False — $\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$ mein koi zero nahi hai phir bhi $\det =4-4=0$; jo matter karta hai woh yeh hai ki ek column doosre ka multiple hai ya nahi, na ki entries zero hain ya nahi.

False — ek row swap determinant ko $-1$ se multiply kar deta hai kyunki yeh do column vectors ki orientation reverse kar deta hai ($ad-bc\to cb-da$).

True — transpose karne se $b\leftrightarrow c$ swap hote hain, aur $ad-cb$ wahi number hai jo $ad-bc$ hai; area unchanged rehta hai.

False — magnitude $|-3|=3$ area factor deta hai (areas triple hote hain); minus sign sirf orientation flip report karta hai, shrinking nahi.

False — determinant additive nahi hota. E.g. $A=B=I$ lene par $\det (A+B)=\det (2I)=4$ lekin $\det A+\det B=1+1=2$.

True — parallel columns ek line span karte hain, parallelogram nahi; enclosed area zero hota hai (dekho Linear independence).

True — kyunki $\det (A)\det (A^{-1})=\det (I)=1$, isliye $\det (A^{-1})=1/\det (A)$ milta hai, yahi reason hai ki jab $\det (A)=0$ ho to $A^{-1}$ exist nahi kar sakta.

True — dono $\det (A)\det (B)$ ke barabar hain, aur numbers ki multiplication commute karti hai, chahe matrix products $AB$ aur $BA$ alag hon.

Diagonals ko galat order mein subtract kiya gaya; sahi hai main-diagonal-first, $ad-bc$. Unka answer exact negative hai — ek determinant jo reversed orientation ke saath compute kiya gaya.

Unhone trace (diagonal ka sum) compute kiya, determinant nahi. Determinant hai $2\cdot 3-0\cdot 0=6$; trace aur determinant alag quantities hain.

$\det =0$ ka matlab singular hai (columns linearly dependent hain), empty nahi. $\left[\begin{array}{cc}0& 1\\ 0& 2\end{array}\right]$ ka $\det =0$ hai phir bhi yeh zero matrix se bahut alag hai.

Diagonal entries ko swap karna tha ($a\leftrightarrow d$), unhe waise hi rakhna nahi tha; sahi adjugate hai $\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ (dekho Inverse of a 2×2 matrix).

$A$ ko $5$ se multiply karne par dono columns $5$ se scale hote hain, isliye area $5^2$ se scale hota hai: $\det (5A)=25\det (A)$. Scalar matrix size ke barabar ki power ke saath bahar aata hai.

$\det (AB)=\det (A)\det (B)=4\cdot 0=0$; singular matrix se involve koi bhi product singular hota hai. Ek flattening map ke saath compose karne par flatten ho hi jaata hai.

Zero determinant plane ko ek line par collapse kar deta hai, isliye map reversible nahi hai; alag inputs ek hi output par land karte hain, jisse ya to koi solution nahi hoga ya infinitely many honge (yeh Cramer's rule ka boundary case hai).

Rotation plane ko rigidly ghoomata hai bina stretch ya flip kiye, isliye area preserved ($+$) hota hai factor $1$ ke saath — na scaling na orientation reversal.

Area magnitude $|ad-bc|$ se aati hai; sign record karta hai ki column-1-then-column-2 counterclockwise sweep karta hai ($+$) ya clockwise ($-$), yaani kya plane pancake ki tarah flip hua.

Determinant sirf area-scaling-plus-flip summary capture karta hai, shape aur rotation discard kar deta hai; ek shear aur ek rotation dono area preserve kar sakte hain, equal determinants dete hain chahe alag act karte hon.

Dependent columns parallel hote hain (ek doosre ka scalar multiple hai), isliye jo parallelogram ye span karte hain uski zero width hoti hai aur hence zero area — precisely $\det =0$ (dekho Linear independence).

Transposing parallelogram ko reflect karta hai, uska area aur orientation relationship preserve karta hai; ek row swap re-label karta hai ki kaun sa vector "pehla" hai, sweep direction reverse ho jaati hai aur hence sign change hota hai.

$\det (A)=0$: zero column ek zero-length side contribute karta hai, isliye parallelogram degenerate ho jaata hai ek segment (ya point) mein jiska koi area nahi.

Zero — identical columns sabse extreme parallel case hain; $3\cdot 5-3\cdot 5=0$, isliye "parallelogram" sirf ek line hai.

Yeh $ad$ ban jaata hai (diagonal entries ka product) kyunki axes independently $a$ aur $d$ se stretch hote hain, area $a\times d$ se scale hota hai bina kisi shearing ke.

Nahi — invertibility equivalent hai $\det \ne 0$ se. Agar $B$ singular hai to $\det (B)=0$, jo $\det (A)=0$ force karta hai, jisse $A$ bhi singular ho jaata.

Yeh $k$ se ek baar scale hota hai: $\det$ ban jaata hai $k\det (A)$, kyunki parallelogram ki sirf ek side $k$ se stretch hoti hai, area us single factor se change hota hai.

Haan — invertibility ko sirf $\det \ne 0$ chahiye. Negative determinant (maano $-13$) ka matlab orientation flip hai, lekin map perfectly reversible hai.

Ek line segment ya single point mein — orientation-preserving ($\det >0$) aur orientation-reversing ($\det <0$) transformations ke beech ka collapse boundary (dekho Transformations and scaling).