Cramer's rule: Worked examples — Hinglish deep dive

4.5.24 · D3 · Maths › Linear Algebra (Full) › Cramer's rule

Yeh page Cramer's rule ka "sab kuch try karo" companion hai. Parent note ne formula $x_{i} = \frac{det A _{i}}{det A}$ banaya aur kyun kaam karta hai yeh explain kiya. Yahan hum isse har tarah ke input ke against stress-test karte hain jo tum exam mein dekh sakte ho: saaf numbers, negative aur zero coefficients, ek aisa system jahan $det A = 0$ (toh rule run hi nahi karega), ek limiting near-singular case, ek geometry problem, ek word problem, aur ek exam twist jisme letter parameter hai.

Scenario matrix

Har cell ko neeche kam se kam ek worked example cover karta hai. Agar koi case tumhe ulajha sakta hai, woh yahan maujood hai.

Cell	Kya special hai	Covered by
A. Clean 2×2, positive	saari entries positive, $det A > 0$	Example 1
B. Mixed signs 2×2	negative coefficients, negative $det A$	Example 2
C. Zero entries / basis-like	ek coefficient $0$ hai (ek column axis-aligned)	Example 3
D. Degenerate: $det A = 0$ (no solution)	rule run nahi ho sakta — inconsistent	Example 4
D'. Degenerate: $det A = 0$ (infinitely many)	rule run nahi ho sakta — $0/0$	Example 4b
E. Limiting / near-singular	$det A \to 0^{+}$ , dekho $x_{i}$ blast ho raha hai	Example 5
F. 3×3 full	teen unknowns, teen column swaps	Example 6
G. Geometry	do lines ka intersection = ek 2×2 system	Example 7
H. Word problem (units)	real quantities, units ka sanity-check	Example 8
I. Exam twist (parameter)	$A$ mein ek letter $k$ ; kab toot-ta hai?	Example 9

Prerequisites jo tum open rakhna chahoge: Determinants, Cofactor Expansion, aur degenerate cell ke liye Gaussian Elimination aur Invertible Matrix Theorem.

Example 1 — Cell A: clean positive 2×2

Worked example Cell A

Solve karo ${3 x + 2 y = 12 x + 4 y = 14$ .

Forecast: saare numbers positive aur chhote hain — compute karne se pehle guess karo ki $x$ aur $y$ saaf whole numbers mein aayenge ya nahi.

Step 1 — Box likho. $A = (3124)$ , $b = (1214)$ . Yeh step kyun? $A$ ke columns do "ingredients" hain; $b$ target hai. Figure mein box dekho.

Step 2 — Denominator $det A$ . $det A = (3) (4) - (2) (1) = 10$ . Yeh step kyun? Yeh original ingredient-box ka area hai. Yeh $\neq = 0$ hai, toh ek unique recipe exist karti hai — rule run karne ki permission hai.

Step 3 — Column 1 ko $b$ se swap karo. $A_{1} = (121424)$ , $det A_{1} = (12) (4) - (2) (14) = 48 - 28 = 20$ . Toh $x = \frac{20}{10} = 2$ . Yeh step kyun? $x$ -column ko $b$ se cover karne se measure hota hai ki "column 1 ka kitna hissa" $b$ banata hai.

Step 4 — Column 2 ko $b$ se swap karo. $A_{2} = (311214)$ , $det A_{2} = (3) (14) - (12) (1) = 42 - 12 = 30$ . Toh $y = \frac{30}{10} = 3$ . Yeh step kyun? Ab hum $y$ -column (column 2) ko $b$ se cover karte hain; us naye box ka area original box ke area ke upar exactly column 2 ki zaroorat ki matra hai, yaani $y$ .

Verify: $3 (2) + 2 (3) = 6 + 6 = 12$ ✓, $2 + 4 (3) = 2 + 12 = 14$ ✓.

Example 2 — Cell B: mixed signs

Worked example Cell B

Solve karo ${2 x - 3 y = 13 - x + 5 y = - 8$ .

Forecast: coefficients positive aur negative dono hain. Kya numerator mein negative signs saaf fractions denge?

Step 1 — Box. $A = (2 - 1 - 3 5)$ , $b = (13 - 8)$ . Yeh step kyun? Same setup; sirf signs ka fark hai, jo determinant automatically handle kar leta hai.

Step 2 — $det A$ . $det A = (2) (5) - (- 3) (- 1) = 10 - 3 = 7$ . Yeh step kyun? $det A = 7 \neq = 0$ — box ka genuine area hai, toh rule run karega. (Note: mixed-sign matrix ka determinant phir bhi positive aa sakta hai; yahan aisa hi hai.)

Step 3 — $x$ . $A_{1} = (13 - 8 - 3 5)$ , $det A_{1} = (13) (5) - (- 3) (- 8) = 65 - 24 = 41$ . Toh $x = \frac{41}{7}$ . Yeh step kyun? Column 1 ( $x$ -column) ko $b$ se cover karo; is box aur original box ka ratio $x$ hai. Integers zaroori nahi — fraction ek perfectly good answer hai.

Step 4 — $y$ . $A_{2} = (2 - 1 13 - 8)$ , $det A_{2} = (2) (- 8) - (13) (- 1) = - 16 + 13 = - 3$ . Toh $y = \frac{- 3}{7}$ . Yeh step kyun? Column 2 ( $y$ -column) ko $b$ se cover karo; us box ke area ka original box ke area se ratio $y$ deta hai. Negative result ka matlab sirf yeh hai ki box orientation flip ho gayi.

Verify: $2 (\frac{41}{7}) - 3 (- \frac{3}{7}) = \frac{82 + 9}{7} = \frac{91}{7} = 13$ ✓, aur $- (\frac{41}{7}) + 5 (- \frac{3}{7}) = \frac{- 41 - 15}{7} = \frac{- 56}{7} = - 8$ ✓.

Example 3 — Cell C: ek zero entry (axis-aligned column)

Worked example Cell C

Solve karo ${4 x + 0 \cdot y = 12 3 x + 2 y = 5$ .

Forecast: column 2 hai $(02)$ — yeh seedha $y$ -axis ki taraf point karta hai. Isse box ek clean rectangle banana chahiye. Pehli equation se akele $x$ guess karo.

Step 1 — Box. $A = (4302)$ , $b = (125)$ . Yeh step kyun? Zero ka matlab hai column 1 aur column 2 "kam ulajhe" hain; figure mein ek column flat aur doosra khada dikhta hai.

Step 2 — $det A$ . $det A = (4) (2) - (0) (3) = 8$ . Yeh step kyun? Zero ek product ko khatam kar deta hai; box area sirf $4 \times 2 = 8$ hai (rectangle-ish box ki base times height).

Step 3 — $x$ . $A_{1} = (12502)$ , $det A_{1} = (12) (2) - (0) (5) = 24$ . Toh $x = \frac{24}{8} = 3$ . Yeh step kyun? $x$ -column ko $b$ se cover karo; box ratio forecast confirm karta hai: $4 x = 12 \Rightarrow x = 3$ .

Step 4 — $y$ . $A_{2} = (43125)$ , $det A_{2} = (4) (5) - (12) (3) = 20 - 36 = - 16$ . Toh $y = \frac{- 16}{8} = - 2$ . Yeh step kyun? $y$ -column (column 2) ko $b$ se cover karo; is naye box aur original box ka ratio $y$ deta hai, aur negative sign flipped orientation record karta hai.

Verify: $4 (3) = 12$ ✓, $3 (3) + 2 (- 2) = 9 - 4 = 5$ ✓.

Example 4 — Cell D: degenerate, $det A = 0$ (no solution)

Worked example Cell D (rule refuse karta hai)

Solve karne ki koshish karo ${x + 2 y = 3 2 x + 4 y = 10$ .

Forecast: row 2 almost row 1 ki double hai. Kya yeh coincidence hai, ya warning sign?

Step 1 — $det A$ . $A = (1224)$ , $det A = (1) (4) - (2) (2) = 4 - 4 = 0$ . Yeh step kyun? Denominator hamesha pehle compute karo. Yahan yeh $0$ hai — Cramer's rule run nahi kar sakta.

Step 2 — Geometry padho. Do columns $(12)$ aur $(24)$ parallel hain (ek doosre ki double hai), toh box flat hai — zero area. Figure mein dono column-arrows ek hi line par dikhte hain. Yeh step kyun? Zero volume ka matlab hai columns har $b$ tak nahi pahunch sakte: ya toh koi solution nahi ya infinitely many.

Step 3 — Diagnose karo kaun sa. Equation 1 ko double karo toh $2 x + 4 y = 6$ milta hai, lekin equation 2 kehti hai $2 x + 4 y = 10$ . Kyunki $6 \neq = 10$ , do equations contradict karti hain — koi solution nahi hai. Yeh step kyun? Invertible Matrix Theorem ke hisaab se, $det A = 0$ ka matlab non-unique hai; hum elimination (Gaussian Elimination) use karte hain yeh dekhne ke liye ki yeh "no solution" branch hai.

Verify: $det A_{1} = det (31024) = 12 - 20 = - 8 \neq = 0$ . Nonzero numerator over zero denominator confirm karta hai inconsistent (blow-up, $0/0$ nahi). Cramer sahi taur par koi finite answer nahi deta.

Example 4b — Cell D': degenerate, $det A = 0$ (infinitely many)

Worked example Cell D' (doosri singular branch)

Solve karne ki koshish karo ${x + 2 y = 3 2 x + 4 y = 6$ .

Forecast: Example 4 jaisa hi left-hand side hai, lekin ab equation 2 ka right-hand side exactly equation 1 ka double hai. Guess karo ki yeh "no solution" hai ya "bahut zyada solutions".

Step 1 — $det A$ . $A = (1224)$ , $det A = (1) (4) - (2) (2) = 0$ . Yeh step kyun? Pehle jaisa hi flat box — denominator $0$ hai, toh Cramer phir bhi ek single number produce nahi kar sakta.

Step 2 — Numerators compute karo. $det A_{1} = det (3624) = (3) (4) - (2) (6) = 12 - 12 = 0$ , aur $det A_{2} = det (1236) = (1) (6) - (3) (2) = 6 - 6 = 0$ . Yeh step kyun? Jab dono numerators bhi vanish hote hain, toh Cramer ratios $0/0$ hain — undefined, lekin Example 4 ka ulta signal deta hai.

Step 3 — Diagnose karo. Equation 2 exactly $2 \times$ equation 1 hai, toh yeh koi nayi information nahi deti. Line $x + 2 y = 3$ par har point system solve karta hai — infinitely many solutions hain. Yeh step kyun? $0/0$ ( $nonzero /0$ nahi) infinitely-many branch ki fingerprint hai; free variable padhne ke liye Gaussian Elimination use karo.

Verify: $y = 0 \Rightarrow x = 3$ lo: check karo $2 (3) + 4 (0) = 6$ ✓. $y = 1 \Rightarrow x = 1$ lo: check karo $2 (1) + 4 (1) = 6$ ✓. Do alag valid solutions "infinitely many" confirm karte hain.

Example 5 — Cell E: near-singular limit

Worked example Cell E (dekho yeh blast kaise hota hai)

Solve karo ${x + y = 1 x + (1 + ε) y = 0$ chhote $ε > 0$ ke liye, phir $ε \to 0^{+}$ karo.

Forecast: jaise $ε \to 0$ hota hai do equations almost identical ho jaati hain. Guess karo ki $x, y$ bounded rehenge ya blast karenge.

Step 1 — $det A$ . $A = (11 1 1 + ε)$ , $det A = (1) (1 + ε) - (1) (1) = ε$ . Yeh step kyun? Yeh tiny denominator $ε$ hi poori kahani hai: kisi cheez ke paas divide karne se sab kuch magnify ho jaata hai.

Step 2 — $x$ . $A_{1} = (10 1 1 + ε)$ , $det A_{1} = (1) (1 + ε) - (1) (0) = 1 + ε$ . Toh $x = \frac{1 + ε}{ε}$ . Yeh step kyun? Column 1 ko $b = (1, 0)^{⊤}$ se cover karo; box ratio $x$ deta hai. Notice karo numerator $1$ ke paas rehta hai jabki denominator shrink hota hai — yahi blast force karta hai.

Step 3 — $y$ . $A_{2} = (1110)$ , $det A_{2} = (1) (0) - (1) (1) = - 1$ . Toh $y = \frac{- 1}{ε}$ . Yeh step kyun? Column 2 ko $b$ se cover karo; yahan numerator fixed $- 1$ hai, toh tiny $ε$ se divide karne se $y$ $- \infty$ ki taraf jaata hai.

Step 4 — Limit lo. Jaise $ε \to 0^{+}$ : $x = \frac{1 + ε}{ε} \to + \infty$ aur $y = \frac{- 1}{ε} \to - \infty$ . Yeh step kyun? Yeh parent note ki numerical-instability warning ko concrete bana deta hai: $ε$ mein ek chota sa change answer ko wildly swing kar deta hai.

Verify ( $ε = 0.1$ par): $x = 1.1/0.1 = 11$ , $y = - 1/0.1 = - 10$ . Check: $11 + (- 10) = 1$ ✓, $11 + 1.1 (- 10) = 11 - 11 = 0$ ✓. Huge numbers, tiny denominator — exactly wahi instability.

Example 6 — Cell F: full 3×3

Worked example Cell F

Solve karo $⎩ ⎨ ⎧ x + 2 y + z = 4 2 x + 5 y + z = 3 - x + y + 3 z = 5$ .

Forecast: teen unknowns ke liye char determinants chahiye (ek denominator, teen numerators). Mostly-positive entries se $det A$ ka rough sign guess karo.

Step 1 — $det A$ by Cofactor Expansion along row 1. $A = 12 - 1 251113$ . $det A = 1 14 (5 \cdot 3 - 1 \cdot 1) - 2 7 (2 \cdot 3 - 1 \cdot (- 1)) + 1 7 (2 \cdot 1 - 5 \cdot (- 1)) = 14 - 14 + 7 = 7.$ Yeh step kyun? Expansion signs alternate karte hain $+, -, +$ ; woh alternation wahi "alternating" property hai jo columns swap karne par sign flip karti hai.

Step 2 — $x$ : column 1 ko $b = (4, 3, 5)^{⊤}$ se cover karo. $A_{1} = 435251113$ . $det A_{1} = 4 (15 - 1) - 2 (9 - 5) + 1 (3 - 25) = 4 (14) - 2 (4) + (- 22) = 56 - 8 - 22 = 26$ . Toh $x = \frac{26}{7}$ . Yeh step kyun? $x$ -column ko $b$ se replace karne se woh box banta hai jiski volume, original volume $det A$ se divide karne par, exactly $x$ hai — "column ko cover karo" ka 3-D version.

Step 3 — $y$ : column 2 ko cover karo. $A_{2} = 12 - 1 435113$ . $det A_{2} = 1 (9 - 5) - 4 (6 + 1) + 1 (10 + 3) = 4 - 28 + 13 = - 11$ . Toh $y = \frac{- 11}{7}$ . Yeh step kyun? Ab $y$ -column $b$ carry karta hai; signed-volume ratio $y$ deta hai, aur negative sign record karta hai ki is swapped box ki orientation opposite hai.

Step 4 — $z$ : column 3 ko cover karo. $A_{3} = 12 - 1 251435$ . $det A_{3} = 1 (25 - 3) - 2 (10 + 3) + 4 (2 + 5) = 22 - 26 + 28 = 24$ . Toh $z = \frac{24}{7}$ . Yeh step kyun? Last column swap: $b$ ko $z$ -slot mein rakhna aur uski box volume ko $det A$ se divide karna $z$ deta hai, recipe complete karta hai.

Verify: $x + 2 y + z = \frac{26 - 22 + 24}{7} = \frac{28}{7} = 4$ ✓, $2 x + 5 y + z = \frac{52 - 55 + 24}{7} = \frac{21}{7} = 3$ ✓, $- x + y + 3 z = \frac{- 26 - 11 + 72}{7} = \frac{35}{7} = 5$ ✓.

Example 7 — Cell G: geometry (do lines ka milna)

Worked example Cell G

Find karo kahan lines $L_{1} : 2 x + y = 4$ aur $L_{2} : x - y = - 1$ cross karti hain.

Forecast: do non-parallel lines exactly ek point par milti hain. Compute karne se pehle unhe sketch karo aur crossing estimate karo.

Step 1 — Yeh ek 2×2 system hai. $A = (21 1 - 1)$ , $b = (4 - 1)$ . Yeh step kyun? "Intersection point" hi simultaneous solution hai — same object, alag words.

Step 2 — $det A$ . $det A = (2) (- 1) - (1) (1) = - 2 - 1 = - 3$ . Yeh step kyun? $\neq = 0$ matlab lines parallel nahi hain, toh exactly ek crossing exist karta hai.

Step 3 — $x$ . $A_{1} = (4 - 1 1 - 1)$ , $det A_{1} = (4) (- 1) - (1) (- 1) = - 4 + 1 = - 3$ . Toh $x = \frac{- 3}{- 3} = 1$ . Yeh step kyun? $x$ -column ko $b$ se cover karo; is box aur $det A$ ka ratio crossing point ka $x$ -coordinate deta hai.

Step 4 — $y$ . $A_{2} = (21 4 - 1)$ , $det A_{2} = (2) (- 1) - (4) (1) = - 2 - 4 = - 6$ . Toh $y = \frac{- 6}{- 3} = 2$ . Yeh step kyun? $y$ -column ko $b$ se cover karo; us box ko $det A$ se divide karna $y$ -coordinate deta hai — saath mein $(x, y)$ intersection hai.

Verify: point $(1, 2)$ : $2 (1) + 2 = 4$ ✓, $1 - 2 = - 1$ ✓ — yeh dono lines par lie karta hai, figure ke red dot se match karta hai.

Example 8 — Cell H: word problem with units

Worked example Cell H

Ek café do combos bechta hai. Combo A = 2 coffees + 1 muffin ki cost ₹230 hai. Combo B = 1 coffee + 3 muffins ki cost ₹290 hai. Ek coffee ( $c$ ) aur ek muffin ( $m$ ) ki price nikalo.

Forecast: coffees usually muffins se zyada cost karti hain, close to ₹100 each — solve karne se pehle guess karo.

Step 1 — Model banao. ${2 c + 1 m = 230 1 c + 3 m = 290$ , toh $A = (2113)$ , $b = (230290)$ . Yeh step kyun? Har column hai "us item ki kitni per combo"; $b$ rupees hai. $c, m$ ke units ₹ mein aate hain.

Step 2 — $det A$ . $(2) (3) - (1) (1) = 6 - 1 = 5$ . Yeh step kyun? Nonzero ⇒ do combos genuinely independent info dete hain, toh prices uniquely pin down hoti hain.

Step 3 — $c$ . $A_{1} = (23029013)$ , $det A_{1} = (230) (3) - (1) (290) = 690 - 290 = 400$ . Toh $c = \frac{400}{5} = 80$ (₹). Yeh step kyun? Coffee-column (column 1) ko rupee-vector $b$ se cover karo; box ratio coffee price isolate karta hai, kyunki $c$ hai "column 1 ka kitna hissa" $b$ banata hai.

Step 4 — $m$ . $A_{2} = (21230290)$ , $det A_{2} = (2) (290) - (230) (1) = 580 - 230 = 350$ . Toh $m = \frac{350}{5} = 70$ (₹). Yeh step kyun? Muffin-column (column 2) ko $b$ se cover karo; us box ko $det A$ se divide karna muffin price ₹ mein deta hai.

Verify (units + values): Combo A $= 2 (80) + 70 = 160 + 70 = 230$ ✓ ₹, Combo B $= 80 + 3 (70) = 80 + 210 = 290$ ✓ ₹. Dono answers positive aur sensibly priced hain.

Example 9 — Cell I: exam twist with a parameter

Worked example Cell I

Kin value(s) of $k$ par ${k x + 2 y = 4 3 x + k y = 6$ ka unique solution nahi hoga? Isko $k = 4$ par solve karo.

Forecast: unique solution tab exist karta hai jab $det A \neq = 0$ . "Fail" cases woh roots hain jahan $det A = 0$ hota hai. Guess karo kitne aisa $k$ hote hain.

Step 1 — $det A$ as a function of $k$ . $A = (k 3 2 k)$ , $det A = k \cdot k - (2) (3) = k^{2} - 6$ . Yeh step kyun? Denominator ab $k$ par depend karta hai; jahan bhi yeh vanish kare, Invertible Matrix Theorem kehta hai koi unique solution nahi.

Step 2 — Bad $k$ nikalo. $k^{2} - 6 = 0 \Rightarrow k = \pm 6$ . Yeh step kyun? Yeh exactly woh values hain jo column-box ko zero volume pe flatten kar deti hain — Cramer wahan ruk jaata hai.

Step 3 — $k = 4$ par solve karo (safe hai, kyunki $16 - 6 = 10 \neq = 0$ ). $det A = 10$ . $A_{1} = (4624)$ , $det A_{1} = 16 - 12 = 4$ , toh $x = \frac{4}{10} = \frac{2}{5}$ . $A_{2} = (4346)$ , $det A_{2} = 24 - 12 = 12$ , toh $y = \frac{12}{10} = \frac{6}{5}$ . Yeh step kyun? $k = 4$ ke saath box non-flat hai, toh hum rule run kar sakte hain: column 1 ko $b$ se cover karo aur $det A$ se divide karo $x$ ke liye, column 2 ko $b$ se cover karo aur divide karo $y$ ke liye — ratios actual solution dete hain.

Verify: $4 (\frac{2}{5}) + 2 (\frac{6}{5}) = \frac{8 + 12}{5} = \frac{20}{5} = 4$ ✓, $3 (\frac{2}{5}) + 4 (\frac{6}{5}) = \frac{6 + 24}{5} = \frac{30}{5} = 6$ ✓.

Recall Quick self-check

Do identical equations $x + y = 1, x + y = 1$ ka system kis cell mein aata hai, aur Cramer kya kehta hai? ::: Cell D' — $det A = 0$ aur har numerator bhi $0$ hai, toh yeh $0/0$ / infinitely-many-solutions branch hai (contrast karo Example 4 ke nonzero-over-zero contradiction se).

Recap of the scenario coverage

Connections

Cramer's rule — parent formula aur uska proof.
Determinants — har denominator ke peeche ka volume engine.
Cofactor Expansion — Example 6 mein 3×3 compute karne ke liye use hua.
Gaussian Elimination — Cells D aur D' mein reach karne ka tool.
Invertible Matrix Theorem — $det A \neq = 0 ⟺$ unique solution (Cells D, D', I).
Matrix Inverse — Cramer adjugate via inverse hai jo $b$ par apply hota hai.
Multilinear and Alternating Maps — kyun column swap karne par waise behave hota hai.