Question bank — Solving 2×2 systems using Cramer's rule

Galat. Ye sirf tab apply hota hai jab $D\ne 0$; agar $D=0$ toh dono lines parallel ya identical hain, isliye koi unique point return karne wala nahi hai aur division $D_x/D$ undefined hai.

Sahi — lekin sirf tab jab $D\ne 0$ ho. Tab $x=D_x/D=0/D=0$. Agar $D=0$ bhi ho, toh $x$ zero nahi hai, is rule se undefined hai, isliye ye statement tumhe trap kar sakta hai.

Galat. $D=0$ ka sirf matlab hai "koi unique solution nahi." Ye do cases mein split hota hai: koi solution nahi (parallel distinct lines) ya infinitely many solutions (same line), ye decide hota hai is baat se ki $D_x,D_y$ bhi zero hain ya nahi.

Galat. Rows ko swap karne se $D$, $D_x$, aur $D_y$ teeno ke signs ek saath flip ho jaate hain, isliye har ratio $D_x/D$ aur $D_y/D$ unchanged rehta hai — same lines, same crossing point.

Galat. Ye pehla column replace karta hai, kyunki $x$ ke coefficients column 1 mein neeche tak rehte hain. "Variable = column," kabhi row nahi.

Sahi. $D\ne 0$ ke saath dono ratios $0/D=0$ hain, isliye $(0,0)$ unique solution hai — ye exactly ek homogeneous system hai jiska sirf ek hi answer origin hai, dekho Consistency of linear systems.

Galat. Equation 1 ko $5$ se scale karne par har determinant ki top row $5$ se multiply ho jaati hai, isliye $D,D_x,D_y$ teeno $5$ se scale hote hain; ratios aur "$D=0$ hai ya nahi" ka verdict unchanged rehta hai.

Galat. $D\ne 0$ guarantee karta hai ki dono lines exactly ek point par cross karti hain, isliye solution hamesha exist karta hai aur unique hota hai. Failure sirf tab hoti hai jab $D=0$ ho.

Error zero se divide karna hai — $D_x/0$ undefined hai, "bada" nahi. Jab $D=0$ ho toh rukna chahiye aur system ko classify karne ke liye $D_x,D_y$ inspect karo, na ki zero se divide karo.

Galat column. $y$ ka column 2 hai, isliye $D_y$ mein column 2 ko constant column se swap karte hain; wahan column 1 swap karne par tumhe galti se $D_x$ mil jaata hai.

Sign flip hai — rule hai main-diagonal product minus anti-diagonal product, usi order mein. Is student ne $D$ negate kar diya; agar woh $D_x,D_y$ bhi negate nahi karte toh answers galat signs ke saath aate hain.

Zero denominator ko numerator mein cancel nahi kar sakte. $-8/0$ undefined hai; $D=0$ aur $D_x\ne 0$ koi solution nahi hone ki signature hai, koi value nahi.

Identical lines har jagah overlap karti hain, isliye infinitely many solutions milte hain, ek nahi. Yahan $D=D_x=D_y=0$ hai, aur solution ko ek poori line ke roop mein describe karte hain, jaise $x=c_1-b_{1}y$ jahan $y$ free ho.

$D$ purely coefficient matrix $A$ ($a$'s aur $b$'s) se banta hai. Constants andar laane par $D$, $D_x$ ya $D_y$ ban jaata hai aur poora ratio kharab ho jaata hai.

Ek row ka $0=0$ tak collapse hona ek sahi redundant statement hai — iska matlab hai dono equations same line hain, jo infinitely many solutions deta hai ($D=D_x=D_y=0$), yani inconsistent ka bilkul ulta.

Kyunki agar $D=0$ hai toh Cramer's rule koi value de nahi sakta; pehle $D$ compute karne se tum jaldi ruk sakte ho aur degenerate case classify karne lag jaate ho, zero se divide karne ki jagah.

$y$ ka elimination karne par $(a_1{b}_2-a_2{b}_1)x=c_1{b}_2-c_2{b}_1$ bachta hai; right side $c_1{b}_2-c_2{b}_1$ wahi determinant hai jisme $A$ ka column 1 constants se swap hua ho — algebra khud tumhare liye column swap bana deta hai.

$D$ dono coefficient rows se bane parallelogram ka signed area hai; zero area ka matlab hai rows same direction mein point kar rahi hain, yaani dono lines ka slope equal hai aur woh parallel ya identical hain.

$D=0$ dono degenerate cases mein common hai. Sirf $D_x$ aur $D_y$ faisla karte hain: sab zero matlab same line (infinite solutions), koi bhi nonzero matlab parallel distinct lines (koi solution nahi).

Column swap determinant ki antisymmetry hai — ye parallelogram ko reflect karta hai, uski orientation reverse kar deta hai aur isliye signed area ka sign change ho jaata hai. Isliye $D_x,D_y$ mein column order matter karta hai.

Inverse Inverse of a 2×2 matrix mein $1/\det A=1/D$ ka factor hota hai, aur use $c$ se multiply karne par term by term $D_x/D$ aur $D_y/D$ ratios reproduce hote hain — same $1/D$, same numerators.

Row 1 ko $k$ se scale karne par $D,D_x,D_y$ teeno usi $k$ se multiply hote hain, isliye ratios $D_x/D$ aur $D_y/D$ mein $k$ divide out ho jaata hai — geometrically tumne same line ko dobara likha, move nahi kiya.

Tab $D_x=D_y=0$ automatically hota hai. Agar $D\ne 0$ toh ek hi solution $x=y=0$ hai (origin); agar $D=0$ toh origin se hoke infinitely many solutions hain.

$D=D_x=D_y=0$, isliye infinitely many solutions; ek equation rakhte hain aur ek variable free chodte hain, jaise $y$ free aur $x$ usse express karte hain.

$A$ ki top row sab zeros hai, isliye $D=0$ forced ho jaata hai; "equation" ya toh impossible hai ($c_1\ne 0$, koi solution nahi) ya vacuous hai ($c_1=0$, redundant) — ye unique intersection ke liye kabhi usable line contribute nahi karta.

Nahi. $D\ne 0$ hai kyunki lines parallel nahi hain; intersection sirf ittefaq se origin par hai, jo $D_x=D_y=0$ ke roop mein dikhta hai jabki $D$ nonzero rehta hai.

Haan, $D$ ek signed area hai, isliye negative ho sakta hai; sign sirf coefficient rows ki orientation record karta hai. Kuch galat nahi hai — ratios $D_x/D,D_y/D$ phir bhi correct unique solution dete hain.

Haan. Tiny lekin nonzero $D$ ka matlab hai lines ek baar cross karti hain, isliye system ka genuine unique solution hai; ise $0$ treat karne par galti se system degenerate declare ho jaata hai.