Solving 2×2 systems using Cramer's rule: Visual walkthrough

2.6.11 · D2 · Maths › Matrices & Determinants — Introduction › Solving 2×2 systems using Cramer's rule

Step 1 — Do secret numbers ke baare mein do rules

KYA. Hum do chhupe hue numbers dhundh rahe hain. Pehle ko $x$ aur doosre ko $y$ kaho. Do "rules" unhe aapas mein bandhte hain:

$how much x counts a_{1} x + how much y counts b_{1} y = the total we’re told c_{1}$ $a_{2} x + b_{2} y = c_{2}$

Yahan $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}$ sirf known numbers hain jo problem humein deti hai. Letters $x, y$ unknowns hain. Symbol $=$ ka matlab hai "left side aur right side same amount hain."

KYON. Akela ek rule kaafi nahi — wo kaafi saare pairs $(x, y)$ allow karta hai. Sirf dono milkar ek answer fix karte hain — isliye dono use karne padenge.

PICTURE. Har rule flat plane pe ek seedhi line khiinchta hai jiske across-axis pe $x$ hai aur up-axis pe $y$ . Ek point $(x, y)$ kisi line pe tab hota hai jab wo us line ka rule maanta ho. Humara jawaab dono lines pe hota hai — unka crossing point.

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

Step 2 — Numbers ko ek grid (matrix) mein pack karo

KYA. Coefficients ko $2 \times 2$ grid mein stack karo aur unknowns/constants ko columns mein:

$A = coefficient matrix [a_{1} a_{2} b_{1} b_{2}] the unknowns [x y] = the constants [c_{1} c_{2}]$

Ek matrix bas numbers ka ek rectangle hota hai. Yeh wala $2 \times 2$ hai: do rows, do columns.

KYON. $A x = c$ likhne se hum poore system ko ek object ki tarah refer kar sakte hain — aur, sabse zaroori baat, isse columns clearly dikhte hain. $A$ ka left column " $x$ ka ghar" hai; right column " $y$ ka ghar" hai. Yeh yaad rakho — poori trick ek column swap karna hi hai.

PICTURE. Pehle column ( $x$ -column) ko cyan colour karo, doosre ( $y$ -column) ko amber.

Dekho Matrix form of linear equations Ax=b ki yeh multiplication do original rules kaise rebuild karti hai.

Step 3 — $y$ ko hatao taaki $x$ akela khade

KYA. $x$ ko isolate karne ke liye, hum $y$ -terms ko cancel karte hain. Pehle rule ko $b_{2}$ se aur doosre ko $b_{1}$ se multiply karo:

$a_{1} b_{2} x + now equal b_{1} b_{2} y = c_{1} b_{2}$ $a_{2} b_{1} x + now equal b_{1} b_{2} y = c_{2} b_{1}$

Yeh multipliers kyun? Hum chahte hain ki dono $y$ -coefficients same number, yaani $b_{1} b_{2}$ , ban jaayein. Rule 1 ko $b_{2}$ se multiply karne par uska $b_{1}$ ban jaata hai $b_{1} b_{2}$ ; rule 2 ko $b_{1}$ se multiply karne par uska $b_{2}$ ban jaata hai $b_{1} b_{2}$ . Dono pe same number — subtract karne ke liye taiyaar.

PICTURE. Kisi rule ko ek constant se scale karne se uski line hilt nahi (dono sides equally bade ho gaye); sirf label change hota hai. Toh crossing point wahi rehta hai — hum bas convenient labels choose kar rahe hain.

Step 4 — Subtract karo, aur determinant se milo

KYA. Pehle scaled rule mein se doosra subtract karo. Same $b_{1} b_{2} y$ terms gayab ho jaate hain:

$call this D (a_{1} b_{2} - a_{2} b_{1}) x = call this D_{x} (c_{1} b_{2} - c_{2} b_{1})$

KYON. $y$ ke jaane ke baad, $x$ finally akela hai — divide karo aur mil gaya. Lekin in do clumps ko dekho: dono mein main-diagonal product minus anti-diagonal product ka pattern hai. Yahi pattern exactly ek $2 \times 2$ determinant hai.

Toh: $D = a_{1} a_{2} b_{1} b_{2}, D_{x} = c_{1} c_{2} b_{1} b_{2}$

PICTURE. $D_{x}$ dekho: yeh matrix $A$ hai jisme uska pehla column ( $x$ -column) constant column $c$ se swap ho gaya hai. $b$ 's wahin rahe; $a$ 's ko hataakar $c$ 's aaye.

Step 5 — $y$ ke liye same dance karo

KYA. Dobarao karo, lekin ab $x$ cancel karo. Rule 1 ko $a_{2}$ se, rule 2 ko $a_{1}$ se multiply karo, subtract karo:

$same D (a_{1} b_{2} - a_{2} b_{1}) y = D_{y} (a_{1} c_{2} - a_{2} c_{1}), D_{y} = a_{1} a_{2} c_{1} c_{2}$

KYON. Symmetry se, neeche wala wahi same $D$ hai (jo geometry decide karti hai "kya lines cross hoti hain" use koi fark nahi ki hum kaunsa unknown dhundh rahe hain). Upar wala $D_{y}$ , $A$ hai jisme doosra column ( $y$ -column) $c$ se swap hua hai.

PICTURE. Step 4 ka mirror: is baar amber $y$ -column ko nikala gaya hai aur constants aaye hain; cyan $x$ -column wahin hai.

Step 6 — Ek real example mein chalao

KYA. Solve karo $2 x + 3 y = 8, 4 x - y = 6.$

$D = 24 3 - 1 = (2) (- 1) - (3) (4) = - 14$ $D_{x} = 86 3 - 1 = - 8 - 18 = - 26, D_{y} = 2486 = 12 - 32 = - 20$ $x = \frac{- 26}{- 14} = \frac{13}{7}, y = \frac{- 20}{- 14} = \frac{10}{7}$

$D$ pehle kyun compute karein? Agar $D$ $0$ hota toh hum ruk jaate — koi unique answer dhundne ki zaroorat nahi.

PICTURE. Dono lines drawn hain; amber dot $(\frac{13}{7}, \frac{10}{7})$ pe unique crossing hai.

Step 7 — Degenerate case: jab $D = 0$ ho

KYA. $D = 0$ ka matlab hai $a_{1} b_{2} = a_{2} b_{1}$ , yaani do coefficient rows proportional hain — dono lines parallel hain. Tab Step 4 ne humein $0 \cdot x = D_{x}$ diya, jise divide nahi kar sakte. Do sub-cases hain:

$D = 0$ aur $D_{x}$ ya $D_{y} \neq = 0$ → parallel distinct lines → no solution. Example: $x + 2 y = 3, 2 x + 4 y = 10$ se $D = 0, D_{x} = 12 - 20 = - 8 \neq = 0$ milta hai.
$D = D_{x} = D_{y} = 0$ → dono lines identical hain → infinitely many solutions. Example: $x + 2 y = 3, 2 x + 4 y = 6$ se $D = D_{x} = D_{y} = 0$ milta hai; $x + 2 y = 3$ pe har point kaam karta hai.

KYON. $D$ coefficient rows se bane parallelogram ka signed area hai. Parallel rows us parallelogram ko flat kar deti hain — area $0$ — toh "lines kitni sharply cross hoti hain" ka koi jawab nahi hota.

PICTURE. Left: parallel-distinct (ek visible gap, koi meeting nahi). Right: coincident (ek hi line do baar draw ki gayi, har jagah milt hai).

Dekho Consistency of linear systems full classification ke liye.

Ek picture mein poora summary

Sab kuch ek saath: do lines $(x, y)$ pe milti hain; side mein coefficient grid; teen swap-grids $D$ , $D_{x}$ , $D_{y}$ ; aur do chhote fractions jo answers bahar nikalte hain. Arrows follow karo.

Recall Feynman retelling (ek 12-saal ke bachche ko batao)

Do dost tumhe do secret numbers ke baare mein ek straight-line rule dete hain. Paper pe draw karo toh dono lines usually ek dot pe milti hain — us dot ki across-value $x$ hai, upar-value $y$ hai. $x$ find karne ke liye bina draw kiye, ek criss-cross game khelo: coefficient grid lo, $x$ ka column dhako, answer numbers slide karo, aur "down-diagonal times, minus up-diagonal times" karo. Yahi top number $D_{x}$ hai. Usi criss-cross ko untouched grid pe karo — woh bottom number $D$ hai. Divide karo: $x = D_{x} / D$ . Doosra column dhak ke $y$ ke liye repeat karo. Agar bottom number $D$ $0$ aaye, toh dono lines secretly parallel hain — ya toh milti hi nahi (koi answer nahi) ya same line hain (endless answers). Kisi bhi case mein koi single dot nahi hai, toh division hone se mana kar deta hai.

Recall checks

Dono fractions ke neeche hamesha kaunsa determinant hota hai?

D

, yaani coefficient determinant.

D_{x}

paane ke liye kaunsa column constants se replace karte hain?

Column 1 — yaani

x

-column.

2 \times 2

determinant ka rule kya hai?

Main-diagonal product minus anti-diagonal product,

p s - q r

D = 0

ka geometrically kya matlab hai?

Coefficient rows parallel hain, toh unka parallelogram zero area ka hai — lines uniquely cross nahi karti.

D = 0

aur

D_{x} \neq = 0

No solution — parallel distinct lines.

D = D_{x} = D_{y} = 0

Infinitely many solutions — ek hi line do baar draw ki gayi.

Connections

Solving 2×2 systems using Cramer's rule — parent note jise yeh page visually explain karta hai.
Determinant of a 2×2 matrix — har step mein use hone wala down-minus-up engine.
Matrix form of linear equations Ax=b — Step 2 ki packaging.
Inverse of a 2×2 matrix — isme bhi same $1/ D$ factor chhupa hai.
Consistency of linear systems — Step 7 ki classification.
Cramer's rule for 3×3 systems — same column-swap idea, ek dimension upar.