Solving 2×2 systems using Cramer's rule: Worked examples

2.6.11 · D3 · Maths › Matrices & Determinants — Introduction › Solving 2×2 systems using Cramer's rule

Agar neeche koi symbol unfamiliar lage, toh machine khud parent note mein spell out hai; yahan hum sirf use use karte hain, baar baar.

Scenario matrix

Har 2×2 system exactly inhi "cells" mein se ek mein aata hai. Ise ek checklist ki tarah padho — end tak, har row mein ek fully worked example tagged hoga. (Yaad raho $a, b, c$ upar define kiye gaye $a x + b y = c$ ke pieces hain.)

Cell	Kya khaas hai	Example
A. Clean unique	$D \neq = 0$ , answers whole numbers hain	Ex 1
B. Negative & fraction	$D \neq = 0$ lekin $x, y$ negative / fractional aate hain	Ex 2
C. Zero in a coefficient	koi $a, b, c = 0$ — board par ek slot blank hai	Ex 3
D. Zero unknown	$x, y$ mein se ek $0$ hai (toh $D_{x}$ ya $D_{y} = 0$ , lekin $D \neq = 0$ )	Ex 4
E. Degenerate — no solution	$D = 0$ , koi $D_{x}, D_{y} \neq = 0$ (parallel distinct lines)	Ex 5
F. Degenerate — infinitely many	$D = D_{x} = D_{y} = 0$ (same line)	Ex 6
G. Word problem	real quantities with units	Ex 7
H. Exam twist — literal / symbolic	numbers ki jagah letters	Ex 8

Related classification Consistency of linear systems mein hai; "same/parallel/crossing lines" ki geometry wahi picture hai jo wahan use hoti hai — aur woh exactly wahi picture hai jo neeche figure mein hai.

Figure s01 teen fates ko side by side draw karta hai. Left mein, do lines alag slope ki (blue aur pink) single yellow dot par cross karti hain — unique-solution case ( $D \neq = 0$ ). Middle mein, do blue/pink lines parallel chalti hain aur kabhi nahi milti — no solution ( $D = 0$ , koi numerator $\neq = 0$ ). Right mein, blue line aur dashed pink line exactly ek doosre ke upar hain — infinitely many solutions ( $D = D_{x} = D_{y} = 0$ ). Yeh picture dimag mein rakho: neeche har example basically yahi pooch raha hai "mein inme se kaunsa hoon?"

Figure — Solving 2×2 systems using Cramer's rule

Cell A — Clean unique solution

Worked example Ex 1 · whole-number answer

$5 x + 2 y = 16, 3 x - y = 5$ (Yeh eq.(1) left par, eq.(2) right par hai.)

Forecast: Har line ko $y = slope \cdot x + \dots$ mein rewrite karo: eq.(1) deta hai $y = \frac{16 - 5 x}{2}$ , slope $- \frac{5}{2}$ ; eq.(2) deta hai $y = 3 x - 5$ , slope $3$ . Alag slopes ⇒ ek baar cross karenge ⇒ ek tidy answer ki expect karo. Guess: kya $D$ positive hoga ya negative?

$D$ compute karo. $D = 53 2 - 1 = (5) (- 1) - (2) (3) = - 5 - 6 = - 11.$ Yeh step kyun? Hamesha pehle $D$ : agar yeh $0$ hota, toh Cramer apply nahi hota aur hum ruk jaate.
$D_{x}$ compute karo — column 1 ( $x$ -column) ko constants $[165]$ se replace karo. $D_{x} = 165 2 - 1 = (16) (- 1) - (2) (5) = - 16 - 10 = - 26.$ Yeh step kyun? $x$ column 1 ka owner hai, toh hum $x$ ko isolate karne ke liye column 1 swap karte hain.
$D_{y}$ compute karo — column 2 replace karo. $D_{y} = 53165 = (5) (5) - (16) (3) = 25 - 48 = - 23.$ Yeh step kyun? $y$ column 2 ka owner hai, toh $y$ ko isolate karne ke liye hum us column ko (column 1 nahi) constants ke liye swap karte hain — step 2 ka mirror image.
Divide karo. $x = \frac{- 26}{- 11} = \frac{26}{11}, y = \frac{- 23}{- 11} = \frac{23}{11} .$ Division last kyun? Steps 1–3 sirf "spread" numbers measure karte the; actual coordinate do spreads ka ratio hai. Swapped-column determinant ko $D$ se divide karna use real answer mein rescale karta hai — woh final division hi woh jagah hai jahan geometry wapas numbers ban jaati hai.

Verify: eq.(2) mein plug in karo: $3 \cdot \frac{26}{11} - \frac{23}{11} = \frac{78 - 23}{11} = \frac{55}{11} = 5.$ ✓ (Forecast $D < 0$ sahi nikla.)

Cell B — Negative aur fractional answers

Worked example Ex 2 · har sign dhyan se dekho

$- 2 x + 4 y = 2, 3 x + y = - 5$ (eq.(1) upar, eq.(2) neeche.)

Forecast: Constants neeche/left taraf kheench rahe hain — $x$ negative expect karo. Leading minus se mat daro; yeh sirf $a_{1} = - 2$ hai (eq.(1) ka $a$ ).

$D$ . $- 2 3 41 = (- 2) (1) - (4) (3) = - 2 - 12 = - 14.$ Kyun: main-diagonal $(- 2) (1)$ minus anti-diagonal $(4) (3)$ — order matter karta hai, har baar usi order mein.
$D_{x}$ (col 1 → $[2 - 5]$ ): $2 - 5 41 = (2) (1) - (4) (- 5) = 2 + 20 = 22.$ Kyun: $(4) (- 5) = - 20$ , aur $- 20$ subtract karna add $20$ karna hai — classic sign trap.
$D_{y}$ (col 2 → constants): $- 2 3 2 - 5 = (- 2) (- 5) - (2) (3) = 10 - 6 = 4.$ Yeh step kyun? $y$ column 2 mein rehta hai, toh hum column 2 swap karte hain (column 1 ko original $x$ -coefficients ke saath chodke) $y$ ka numerator banane ke liye.
Divide karo. $x = \frac{22}{- 14} = - \frac{11}{7}, y = \frac{4}{- 14} = - \frac{2}{7} .$ Division last kyun? Negative $D = - 14$ dono positive numerators ka sign flip karta hai, aur yahi wajah hai ki answers negative aate hain — $x, y$ ka sign tab tak pata nahi chalta jab tak yeh last ratio na liya jaaye.

Verify: eq.(1): $- 2 (- \frac{11}{7}) + 4 (- \frac{2}{7}) = \frac{22 - 8}{7} = \frac{14}{7} = 2.$ ✓

Cell C — Zero coefficients (board par blank slots)

Is cell mein blank slot ke teen flavours hain: $a = 0$ (ek variable missing), $b = 0$ (doosra variable missing), aur $c = 0$ (zero constant). Hum teeno dikhate hain taaki zeros ki koi bhi arrangement kabhi nai na lage.

Worked example Ex 3a ·

$a = 0$ : ek equation mein variable missing $0 \cdot x + 3 y = 9, 2 x - y = 1$ (eq.(1) hai $3 y = 9$ ; eq.(2) hai $2 x - y = 1$ .) Zero coefficient bilkul allowed hai — woh sirf determinant mein ek $0$ dalta hai.

Forecast: Sirf eq.(1) se, $y = 3$ . Toh expect karo $y = 3$ aur Cramer ko agree karna chahiye.

$D$ . $02 3 - 1 = (0) (- 1) - (3) (2) = 0 - 6 = - 6.$ Kyun: $0$ main-diagonal term ko zap kar deta hai, sirf anti-diagonal bachta hai. Phir bhi $D \neq = 0$ , toh unique solution exist karta hai.
$D_{x}$ (col 1 → $[91]$ ): $91 3 - 1 = (9) (- 1) - (3) (1) = - 9 - 3 = - 12.$ Yeh step kyun? $x$ column 1 ka owner hai, toh hum $x$ ka numerator banane ke liye woh column constants ke liye swap karte hain.
$D_{y}$ : $0291 = (0) (1) - (9) (2) = - 18.$ Yeh step kyun? Hum column 1 (original $x$ -column, yahan $[02]$ ) rakhte hain aur sirf column 2 ko constants ke liye swap karte hain — yahi cheez is determinant ko $x$ ka nahi, $y$ ka numerator banati hai.
Divide karo. $x = \frac{- 12}{- 6} = 2, y = \frac{- 18}{- 6} = 3.$ Division last kyun? Sirf $D = - 6$ se divide karne ke baad hi $D_{y} = - 18$ collapse hokar $y = 3$ banta hai — woh forecast jo humne seedha eq.(1) se kiya tha use match karta hai. Determinants akele answer nahi the; ratio hai.

Verify: $y = 3$ forecast se match karta hai; eq.(2): $2 (2) - 3 = 1.$ ✓

Worked example Ex 3b ·

$b = 0$ aur $c = 0$ saath mein $2 x + 0 \cdot y = 6, x + 4 y = 0$ (eq.(1) hai $2 x = 6$ , toh yahan $b_{1} = 0$ ; eq.(2) mein constant $c_{2} = 0$ hai.) Ek saath do tarah ke zeros.

Forecast: Sirf eq.(1) se, $x = 3$ . Phir eq.(2) deta hai $3 + 4 y = 0 \Rightarrow y = - \frac{3}{4}$ . Expect karo $x = 3, y = - \frac{3}{4}$ .

$D$ . $2104 = (2) (4) - (0) (1) = 8 - 0 = 8.$ Kyun: is baar $b_{1} = 0$ anti-diagonal term ko kill karta hai; $D = 8 \neq = 0$ , unique solution.
$D_{x}$ (col 1 → $[60]$ ): $6004 = (6) (4) - (0) (0) = 24.$ Yeh step kyun? $x$ column 1 ka owner hai, toh hum woh column constants ke liye swap karte hain (zero constant $c_{2} = 0$ simply bottom slot mein baithta hai).
$D_{y}$ (col 2 → constants): $2160 = (2) (0) - (6) (1) = - 6.$ Yeh step kyun? $y$ column 2 ka owner hai, toh hum column 1 ko original $x$ -coefficients $[21]$ ke saath rakhte hain aur sirf column 2 swap karte hain — zero constant ab upar wale slot mein aata hai.
Divide karo. $x = \frac{24}{8} = 3, y = \frac{- 6}{8} = - \frac{3}{4} .$ Division last kyun? Bare determinants $24$ aur $- 6$ answer nahi hain; sirf $D = 8$ se divide karne par woh $x = 3, y = - \frac{3}{4}$ mein rescale hote hain, exactly woh forecast jo humne kiya tha.

Verify: eq.(1): $2 (3) = 6$ ✓; eq.(2): $3 + 4 (- \frac{3}{4}) = 3 - 3 = 0$ ✓.

Cell D — Ek zero unknown (answer khud zero hai)

Worked example Ex 4 ·

$x = 0$ , lekin $D \neq = 0$ $4 x + 5 y = 15, 2 x - 3 y = - 9$ (eq.(1) upar, eq.(2) neeche.)

Forecast: Yeh constants ek unknown ko vanish kara sakte hain. Parent note ka trap note karo: " $D_{x} = 0$ " ka matlab $D = 0$ nahi hai — yeh genuinely matlab ho sakta hai $x = 0$ .

$D$ . $42 5 - 3 = (4) (- 3) - (5) (2) = - 12 - 10 = - 22.$ Kyun: $D \neq = 0$ , toh ek unique answer hai — jo bhi aaye woh trustworthy hai.
$D_{x}$ (col 1 → constants): $15 - 9 5 - 3 = (15) (- 3) - (5) (- 9) = - 45 + 45 = 0.$ Yeh kyun theek hai: $D_{x} = 0$ jabki $D = - 22 \neq = 0$ deta hai $x = 0/ (- 22) = 0$ . Ek bilkul valid zero answer, koi breakdown nahi.
$D_{y}$ : $42 15 - 9 = (4) (- 9) - (15) (2) = - 36 - 30 = - 66.$ Yeh step kyun? Column 2 ( $y$ -column) ko constants ke liye swap kiya jaata hai jabki column 1 original $x$ -coefficients rakhta hai — wahi placement exactly ise $y$ ka numerator banati hai.
Divide karo. $x = \frac{0}{- 22} = 0, y = \frac{- 66}{- 22} = 3.$ Division last kyun? Yahan " $D_{x} = 0$ " aur " $x = 0$ " ka difference visible hota hai: sirf $0$ ko nonzero $D$ se divide karke hum legally $x = 0$ obtain karte hain. Final division ke bina, $D_{x} = 0$ akela meaningless hota.

Verify: eq.(1): $4 (0) + 5 (3) = 15.$ ✓ Consistency view se compare karo — ek unique point $(0, 3)$ dono lines par baithta hai.

Cell E — Degenerate: no solution (parallel distinct lines)

Worked example Ex 5 ·

$D = 0$ lekin $D_{x} \neq = 0$ $2 x - 6 y = 4, - x + 3 y = 5$ (eq.(1) upar, eq.(2) neeche.)

Forecast: Slopes: eq.(1) hai $y = \frac{x - 2}{3}$ , eq.(2) hai $y = \frac{x + 5}{3}$ — same slope $\frac{1}{3}$ , alag intercepts. Parallel aur distinct ⇒ kabhi nahi milenge ⇒ no solution. Dekho $D$ $0$ par hit karta hai.

$D$ . $2 - 1 - 6 3 = (2) (3) - (- 6) (- 1) = 6 - 6 = 0.$ Kyun: rows same direction mein point kar rahi hain ( $(- 1, 3) = - \frac{1}{2} (2, - 6)$ ), toh jo parallelogram woh span karti hain uska zero area hai — yahi hai $D = 0$ ka matlab. Yeh Figure s01 ka middle panel hai: do parallel lines jo kabhi cross nahi karti.
Kyunki $D = 0$ hai, ruko mat — $D_{x}$ test karo. $45 - 6 3 = (4) (3) - (- 6) (5) = 12 + 30 = 42 \neq = 0.$ Yeh step kyun: $D = 0$ ke saath koi bhi nonzero numerator matlab hai ratio $D_{x} / D$ hai $\frac{42}{0}$ — undefined. Koi point dono satisfy nahi karta. (Yahan division khud diagnosis hai: yeh hone se mana kar deta hai, jo exactly "no solution" hai.)
Conclude: $D = 0$ aur $D_{x} \neq = 0$ ⇒ no solution.

Verify: eq.(2) ko $- 2$ se multiply karo: $2 x - 6 y = - 10$ , lekin eq.(1) kehta hai $2 x - 6 y = 4$ . Kyunki $- 10 \neq = 4$ , equations contradict karti hain — confirmed inconsistent. ✓

Cell F — Degenerate: infinitely many (same line)

Worked example Ex 6 ·

$D = D_{x} = D_{y} = 0$ $3 x + 6 y = 9, x + 2 y = 3$ (eq.(1) upar, eq.(2) neeche.)

Forecast: eq.(1) exactly $3 \times$ eq.(2) hai. Ek hi line do baar draw ki gayi ⇒ us par har point kaam karta hai ⇒ infinitely many solutions. Expect karo ki teeno determinants $0$ honge.

$D$ . $3162 = (3) (2) - (6) (1) = 6 - 6 = 0.$
$D_{x}$ . $9362 = (9) (2) - (6) (3) = 18 - 18 = 0.$
$D_{y}$ . $3193 = (3) (3) - (9) (1) = 9 - 9 = 0.$ Yeh step kyun? Hum phir bhi sirf column 2 ko $y$ ke numerator ke liye swap karte hain; ise check karna (sirf $D_{x}$ nahi) matter karta hai kyunki lines ke sach mein coincide karne ke liye dono numerators vanish karne chahiye. Yeh Figure s01 ka right-hand panel hai: ek line do baar draw ki gayi.
Solution set describe karo. $x + 2 y = 3$ se: $x = 3 - 2 y$ , jahan $y$ free hai. Solutions: $(3 - 2 t, t)$ kisi bhi real $t$ ke liye. Yahan division kyun nahi? Har ratio $0/0$ hoga — machine ki final division undefined hai, jo exactly uska tarika hai yeh announce karne ka "ek answer nahi balki infinitely many." Toh divide karne ki jagah hum poori line describe karte hain.

Verify: $t = 1$ rakkho: $(1, 1)$ . eq.(1): $3 (1) + 6 (1) = 9$ ✓; eq.(2): $1 + 2 = 3$ ✓. $t = 0$ rakkho: $(3, 0)$ bhi kaam karta hai — infinitely many, forecast ke anusaar.

Cell G — Ek word problem (units ke saath)

Worked example Ex 7 · tickets aur paisa

Ek cinema adult tickets ₹120 mein aur child tickets ₹80 mein bechta hai. Ek shaam usne 50 tickets bechein total ₹5200 ke liye. Har tarah ke kitne? Maan lo $x =$ adult tickets, $y =$ child tickets. $x + y = 50 (count, eq.(1)), 120 x + 80 y = 5200 (rupees, eq.(2))$

Forecast: Agar saare 50 bachche hote toh hum ₹4000 collect karte; humne ₹1200 zyada collect kiya. Har adult ek child se ₹40 zyada deta hai, toh ≈ $1200/40 = 30$ adults. Guess $x = 30, y = 20$ ; Cramer se confirm karein.

$D$ . $1120180 = (1) (80) - (1) (120) = 80 - 120 = - 40.$ Kyun: $D \neq = 0$ ⇒ ek unique ticket split exist karta hai (physically toh hona hi chahiye).
$D_{x}$ (col 1 → $[505200]$ ): $505200180 = (50) (80) - (1) (5200) = 4000 - 5200 = - 1200.$
$D_{y}$ : $1120505200 = (1) (5200) - (50) (120) = 5200 - 6000 = - 800.$ Yeh step kyun? $y$ child tickets count karta hai aur uske coefficients column 2 mein baithte hain, toh hum column 2 swap karte hain (original $x$ -column rakhke) $y$ ka numerator banane ke liye.
Divide karo. $x = \frac{- 1200}{- 40} = 30$ adults, $y = \frac{- 800}{- 40} = 20$ children. Division last kyun? Determinants galat scale carry karte hain (woh "spread" areas hain, counts nahi); $D = - 40$ se divide karna unhe actual ticket numbers mein convert karta hai sahi units ke saath. Woh final ratio hi ek aisa step hai jo ek countable answer produce karta hai.

Verify (units bhi): tickets $30 + 20 = 50$ ✓; paisa $120 (30) + 80 (20) = 3600 + 1600 = 5200$ ₹ ✓. 30 adults ke forecast se match karta hai.

Cell H — Exam twist: letters se solve karo

Worked example Ex 8 · symbolic / literal coefficients

Constant $k$ ke terms mein $x, y$ ke liye solve karo (jahan $k \neq = 3$ ): $x + y = 2, k x + 3 y = 6$ (eq.(1) upar, eq.(2) neeche.)

Forecast: Answers $k$ mein formulas honge, numbers nahi. Kyunki hum $D$ se divide karte hain, jo bhi $D = 0$ banaye woh ek "forbidden" value hogi — use jaldi pakdo.

$D$ . $1 k 13 = (1) (3) - (1) (k) = 3 - k .$ Kyun: $D = 3 - k$ , toh $D = 0$ exactly $k = 3$ par — woh excluded value jiske baare mein problem ne warning di thi (wahan lines coincide ho jaati hain). Baaki har $k$ , including $k = 0$ , bilkul theek hai.
$D_{x}$ (col 1 → $[26]$ ): $2613 = (2) (3) - (1) (6) = 6 - 6 = 0.$ Yeh step kyun? $x$ column 1 ka owner hai, toh hum $x$ ka numerator banane ke liye woh column constants ke liye swap karte hain.
$D_{y}$ (col 2 → constants): $1 k 26 = (1) (6) - (2) (k) = 6 - 2 k .$ Yeh step kyun? Column 2 ( $y$ -column, yahan $[13]$ ) ko constants ke liye swap kiya jaata hai jabki column 1 $x$ -coefficients $[1 k]$ rakhta hai — woh placement jo ise $y$ ka numerator banati hai, $k$ ko wahan rakhte hue jahan woh belong karta hai.
Divide karo. $x = \frac{D _{x}}{D} = \frac{0}{3 - k} = 0$ (kisi bhi allowed $k$ ke liye), aur $y = \frac{D _{y}}{D} = \frac{6 - 2 k}{3 - k} = \frac{2 ( 3 - k )}{3 - k} = 2.$ Division last kyun? Sirf divide karne ke baad $k$ s cancel hote hain: $\frac{2 ( 3 - k )}{3 - k} = 2$ . Divide karne se pehle, numerator $6 - 2 k$ aur $D = 3 - k$ dono $k$ par depend karte hain; ratio hi woh surprising $k$ -independence reveal karta hai — aur yahi wajah hai ki $k = 3$ ( $D = 0$ banata hai) exclude karna zaroori hai, kyunki zero se divide nahi kar sakte.

Verify: $(x, y) = (0, 2)$ $k$ se independent hai. eq.(1): $0 + 2 = 2$ ✓; eq.(2): $k (0) + 3 (2) = 6$ ✓ har $k$ ke liye. Achha surprise — answer $k$ par depend nahi karta, phir bhi $k = 3$ $D$ ko break karta hai (wahan dono equations identical line $x + y = 2$ ban jaati hain, toh unique solution ki jagah infinitely many solutions).

Recall check

Recall Which cell is each situation?

$D \neq = 0$ , answer whole numbers ::: Cell A — clean unique. $D \neq = 0$ lekin $D_{x} = 0$ ::: Cell D — unknown $x$ genuinely $0$ hai, phir bhi unique solution. $D = 0$ aur $D_{x} \neq = 0$ ::: Cell E — no solution, parallel distinct lines. $D = D_{x} = D_{y} = 0$ ::: Cell F — infinitely many, same line. Ek coefficient $0$ hai (jaise $0 \cdot x + 3 y = 9$ ) ::: Cell C — allowed; sirf determinant mein ek $0$ jaata hai, $D$ phir bhi nonzero ho sakta hai. Coefficients letters hain ::: Cell H — answers formulas hain; jo value $D = 0$ banaye woh excluded hai.

Connections

Solving 2×2 systems using Cramer's rule — woh parent machine jise yeh examples exercise karti hain.
Determinant of a 2×2 matrix — woh criss-cross number jo har step mein compute hota hai.
Consistency of linear systems — Cells E aur F "inconsistent" aur "dependent" classes hain.
Matrix form of linear equations Ax=b — upar har example ek $A x = c$ hai.
Cramer's rule for 3×3 systems — same case-checklist, swap karne ke liye ek aur column.