Inverse of 2×2 matrix: Worked examples — Hinglish deep dive

2.6.10 · D3 · Maths › Matrices & Determinants — Introduction › Inverse of 2×2 matrix

Yaad karo woh do cheezein jo hum baar baar use karenge. Agar $A = (a c b d), det A = a d - b c, A^{- 1} = \frac{1}{a d - b c} (d - c - b a) .$

Yahan $det A$ (padho "determinant of $A$ ") ek single number hai $a d - b c$ — dekho Determinant of a 2×2 matrix. Yeh divisor hai, isliye sab kuch isi par depend karta hai.

Scenario matrix

Har $2 \times 2$ inverse problem inhi case classes mein se kisi ek mein aati hai. Har row ek alag trap ya skill hai; aakhri column batata hai kaun sa worked example use cover karta hai.

#	Case class	Kya special hai / kahan galti ho sakti hai	Covered by
C1	Saari entries positive, $det > 0$	"Clean" baseline case	Ex 1
C2	Negative entries, $det < 0$	Negative determinant — sign aakhir tak sahi rekhni hai	Ex 2
C3	Fractional / decimal entries	Determinant ek fraction hai, divide karna mushkil ho jaata hai	Ex 3
C4	$det = 0$ (singular)	Koi inverse exist nahi karta — detect karke rukna zaroori hai	Ex 4
C5	Ek zero entry ( $b = 0$ ya $c = 0$ , triangular)	Degenerate lagti hai lekin phir bhi invertible hai	Ex 5
C6	Linear system $A x = b$ solve karna	Application, sirf inversion nahi	Ex 6
C7	Real-world word problem	Words ko translate karo → matrix → solve karo	Ex 7
C8	Exam twist: aisa $k$ dhundho jisse $A$ singular ho jaaye	Logic ulta karo: $det = 0$ set karo	Ex 8

Do figures extreme rows ki geometry illustrate karti hain: C4 (area collapse ho jaata hai) aur C2 (ek flip).

Example 1 — C1: clean positive case

Worked example

$A = (3124)$ ka inverse nikalo Forecast: saari entries positive hain, toh andaaza lagao ki determinant ek modest positive number hoga aur inverse mein chhote fractions honge. Aage padhne se pehle guess karo.

Step 1 — determinant. $det A = (3) (4) - (2) (1) = 12 - 2 = 10.$ Yeh step kyun? Agar $det = 0$ toh hum turant ruk jaate hain — koi inverse nahi. Yahan $10 \neq = 0$ hai, toh hum aage badhte hain.

Step 2 — adjugate. Diagonal swap karo ( $3 \leftrightarrow 4$ ), off-diagonal negate karo ( $2, 1$ ): $adj A = (4 - 1 - 2 3) .$ Yeh step kyun? Yeh woh fixed pattern hai jo derivation ne produce kiya tha (dekho Adjugate and cofactors).

Step 3 — det se divide karo. $A^{- 1} = \frac{1}{10} (4 - 1 - 2 3) = (0.4 - 0.1 - 0.2 0.3) .$ Yeh step kyun? Adjugate akela $A \cdot adj A = (det A) I$ satisfy karta hai; $det A = 10$ se divide karne par woh $10 I$ exact $I$ ban jaata hai jो hum chahte hain, aur $A A^{- 1} = I$ complete ho jaata hai.

Verify: $A A^{- 1} = \frac{1}{10} (3124) (4 - 1 - 2 3) = \frac{1}{10} (100010) = I .$ ✓

Example 2 — C2: negative entries, negative determinant

Worked example

$A = (- 1 2 31)$ ka inverse nikalo Forecast: ek negative entry hai aur "backwards" product bada hai, toh guess karo ki $det$ negative aayega. Kya negative determinant inverse ko khatam kar deta hai? (Nahi — sirf $det = 0$ karta hai.)

Step 1 — determinant. $det A = (- 1) (1) - (3) (2) = - 1 - 6 = - 7.$ Yeh step kyun? $- 7 \neq = 0$ , toh inverse exist karta hai. Minus sign real hai aur use end tak carry karna hai.

Step 2 — adjugate. Diagonal swap karo ( $- 1 \leftrightarrow 1$ ), off-diagonal negate karo ( $3, 2$ ): $adj A = (1 - 2 - 3 - 1) .$ Yeh step kyun? Yeh fixed swap/negate pattern hai (dekho Adjugate and cofactors); diagonal par negative hone ki wajah se, swap karne par $- 1$ bottom-right corner mein aa jaata hai, toh uska sign dhyaan se dekho.

Step 3 — $- 7$ se divide karo. Negative se divide karne par adjugate ki har sign flip ho jaati hai: $A^{- 1} = \frac{1}{- 7} (1 - 2 - 3 - 1) = (- \frac{1}{7} \frac{2}{7} \frac{3}{7} \frac{1}{7}) .$ Yeh step kyun? Common galti: log bhool jaate hain ki $- 7$ adjugate entries ko bhi negate karta hai, isliye upar wale signs pe trust karo.

Verify: $A A^{- 1} = \frac{1}{- 7} (- 1 2 31) (1 - 2 - 3 - 1) = \frac{1}{- 7} (- 7 0 0 - 7) = I .$ ✓

Negative determinant ka matlab hai ki transformation orientation flip karta hai (jaisa ek mirror) aur saath mein rescale bhi. Neeche wali figure dekho: violet unit square counter-clockwise trace hota hai, lekin $A$ apply karne ke baad orange parallelogram ke corners clockwise jaate hain (magenta arrows follow karo). Yeh reversal — corners ka "galat direction mein" jaana — exactly wahi hai jo negative $det$ record karta hai; area magnitude $∣ - 7∣ = 7$ hai jitna square enlarge hua.

Example 3 — C3: fractional aur decimal entries

Worked example

$A = (\frac{1}{2} \frac{1}{4} 0.4 1)$ ka inverse nikalo Forecast: ek fraction ( $\frac{1}{2}$ ), ek decimal ( $0.4$ ), aur ek aur fraction ka mix — andaaza lagao ki determinant ek chhota fraction hoga, toh divide karne par numbers bade honge. Guess karo ki $det$ $\frac{1}{2}$ se upar hoga ya neeche.

Step 1 — determinant. Pehle decimal convert karo: $0.4 = \frac{2}{5}$ . Phir $det A = (\frac{1}{2}) (1) - (\frac{2}{5}) (\frac{1}{4}) = \frac{1}{2} - \frac{2}{20} = \frac{10}{20} - \frac{2}{20} = \frac{8}{20} = \frac{2}{5} .$ Yeh step kyun? $0.4$ ko $\frac{2}{5}$ mein convert karne par hum common denominator ( $20$ ) use kar sakte hain aur cleanly subtract kar sakte hain. $\frac{2}{5} \neq = 0$ ⟹ invertible.

Step 2 — adjugate. Diagonal swap karo ( $\frac{1}{2} \leftrightarrow 1$ ), off-diagonal negate karo ( $\frac{2}{5}$ aur $\frac{1}{4}$ ): $adj A = (1 - \frac{1}{4} - \frac{2}{5} \frac{1}{2}) .$ Yeh step kyun? Same fixed swap/negate pattern jaise hamesha (dekho Adjugate and cofactors) — fractions aur decimals recipe nahi badlaते, sirf arithmetic badal jaati hai.

Step 3 — $\frac{2}{5}$ se divide karo, yaani $\frac{5}{2}$ se multiply karo: $A^{- 1} = \frac{5}{2} (1 - \frac{1}{4} - \frac{2}{5} \frac{1}{2}) = (\frac{5}{2} - \frac{5}{8} - 1 \frac{5}{4}) .$ Yeh step kyun? Fraction se divide karna = uske reciprocal se multiply karna; isse $- \frac{2}{5}$ , $\frac{5}{2} \cdot (- \frac{2}{5}) = - 1$ ban jaata hai, wagera.

Verify: $A A^{- 1}$ ko $I$ dena chahiye. Top-left entry: $\frac{1}{2} \cdot \frac{5}{2} + \frac{2}{5} \cdot (- \frac{5}{8}) = \frac{5}{4} - \frac{1}{4} = 1$ ✓ (poora check VERIFY mein).

Example 4 — C4: singular case (koi inverse nahi)

Worked example

$A = (6496)$ ka inverse nikalne ki koshish karo Forecast: columns $(4 6)$ aur $(6 9)$ dekho. Kya doosra pehle ka multiple hai? Agar haan, toh area collapse ho jaata hai aur koi inverse nahi hoga. Step 1 se pehle decide karo.

Step 1 — determinant. $det A = (6) (6) - (9) (4) = 36 - 36 = 0.$ Yeh step kyun? Determinant hamesha pehle check karo. $0$ deal-breaker hai.

Step 2 — conclude. $det A = 0 \Rightarrow$ == $A$ singular hai, koi inverse exist nahi karta.== Yeh step kyun? Formula $0$ se divide karne ki demand karta (undefined). Dekho Singular vs non-singular matrices.

Geometrically kyun? Column 2 $= \frac{3}{2} \times$ column 1 (kyunki $9 = \frac{3}{2} \cdot 6$ aur $6 = \frac{3}{2} \cdot 4$ ). Dono columns ek hi direction mein point karte hain, toh $A$ poore plane ko ek single line par squash kar deta hai — area zero ho jaata hai aur information lost ho jaati hai.

Verify: columns parallel hain ⟺ unka cross-ratio $6 \cdot 6 - 9 \cdot 4 = 0$ , jo $det = 0$ se match karta hai. Consistent. ✓

Figure collapse ko concrete banata hai. Violet unit square (area $1$ ) ko $A$ mein daala jaata hai. Uska image koi parallelogram nahi hai — yeh sirf ek magenta line segment hai: plane ka har point us ek line par kahin na kahin land karta hai. Dono column vectors, orange $(6, 4)$ aur navy $(9, 6)$ , bilkul ek doosre ki direction ke upar laate hain (navy $= 1.5 \times$ orange), yahi wajah hai ki square ko spread hone ki jagah nahi milti. Zero area ke saath koi tarika nahi hai original square par "un-flatten" karne ka — isliye koi inverse nahi.

Example 5 — C5: zero entry (triangular matrix)

Worked example

$A = (2053)$ ka inverse nikalo Forecast: ek entry $0$ hai. Kya zero ka matlab degenerate hona hai? Columns dekho — kya woh parallel hain? Guess karo ki inverse exist karta hai ya nahi.

Step 1 — determinant. $det A = (2) (3) - (5) (0) = 6 - 0 = 6.$ Yeh step kyun? Zero $c$ position mein hai, toh woh $a d - b c$ se bahar ho jaata hai. $6 \neq = 0$ ⟹ invertible — zero entry ka matlab singular nahi hota.

Step 2 — adjugate. Diagonal swap karo, off-diagonal negate karo: $adj A = (30 - 5 2) .$ Yeh step kyun? $c = 0$ ko negate karne par bhi $0$ hi rehta hai — zero waise hi pada rehta hai.

Step 3 — $6$ se divide karo. $A^{- 1} = \frac{1}{6} (30 - 5 2) = (\frac{1}{2} 0 - \frac{5}{6} \frac{1}{3}) .$ Yeh step kyun? Adjugate ko $det A = 6$ se divide karne par $A \cdot adj A = 6 I$ exact $I$ tak rescale ho jaata hai, $A A^{- 1} = I$ complete karta hai — har doosre example jaisa hi finishing move.

Verify: $A A^{- 1} = \frac{1}{6} (2053) (30 - 5 2) = \frac{1}{6} (6006) = I .$ ✓ Triangular matrix ka inverse bhi usi shape mein triangular hota hai.

Example 6 — C6: linear system solve karna

Worked example

${4 x + 3 y = 18 2 x + 5 y = 16$ solve karo Forecast: yeh $A x = b$ hai jahan $A = (4235)$ , $b = (16 18)$ . Guess karo ki $x$ aur $y$ chhote whole numbers hain.

Step 1 — determinant. $det A = (4) (5) - (3) (2) = 20 - 6 = 14.$ Yeh step kyun? Unique solution tabhi exist karta hai jab $det \neq = 0$ ; $14$ kaam karta hai. (Dekho Solving linear systems.)

Step 2 — inverse. $A^{- 1} = \frac{1}{14} (5 - 2 - 3 4) .$ Yeh step kyun? Hum $A$ par poori recipe apply karte hain: diagonal swap karo ( $4 \leftrightarrow 5$ ), off-diagonal negate karo ( $3, 2$ ), phir $det A = 14$ se divide karo. Hum $A^{- 1}$ abhi banate hain kyunki agli step ko iske zaroorat hai.

Step 3 — $x = A^{- 1} b$ apply karo. Yeh step kyun? $A x = b$ ko left side se $A^{- 1}$ se multiply karo: kyunki $A^{- 1} A = I$ hai, left side $x$ ban jaata hai. $x = \frac{1}{14} (5 - 2 - 3 4) (16 18) = \frac{1}{14} (- 36 + 64 90 - 48) = \frac{1}{14} (28 42) = (2 3) .$

Verify: $4 (3) + 3 (2) = 12 + 6 = 18$ ✓ aur $2 (3) + 5 (2) = 6 + 10 = 16$ ✓.

Example 7 — C7: real-world word problem

Worked example Ek café muffins aur coffees bechta hai

Menu par do combos hain. Combo A = 2 muffins + 1 coffee = £7. Combo B = 1 muffin + 3 coffees = £11. Ek muffin ( $m$ ) aur ek coffee ( $c$ ) ki price nikalo. Forecast: whole-pound prices guess karo. Coffee shayad kuch pounds ki hogi.

Step 1 — matrix banao. Equations hain $2 m + 1 c = 7$ aur $1 m + 3 c = 11$ , yaani $A = (2113), b = (11 7), A (c m) = b .$ Yeh step kyun? Coefficients = har item ki quantities; right side = total prices.

Step 2 — determinant. $det A = (2) (3) - (1) (1) = 6 - 1 = 5.$ Yeh step kyun? Invert karne se pehle $det \neq = 0$ check karte hain — $5 \neq = 0$ guarantee karta hai ki prices uniquely determined hain (menu mein koi ambiguity nahi).

Step 3 — inverse aur solve. $A^{- 1} = \frac{1}{5} (3 - 1 - 1 2)$ , toh $(c m) = \frac{1}{5} (3 - 1 - 1 2) (11 7) = \frac{1}{5} (- 7 + 22 21 - 11) = \frac{1}{5} (15 10) = (3 2) .$ Yeh step kyun? $x = A^{- 1} b$ unknown prices recover karta hai.

Verify (units ke saath): muffin £2, coffee £3. Combo A: $2 (£2) + £3 = £7$ ✓. Combo B: $£2 + 3 (£3) = £11$ ✓. Prices positive hain aur sensible hain. ✓

Example 8 — C8: exam twist, $k$ nikalo jisse $A$ singular ho jaaye

Worked example Kis

$k$ ke liye $A = (k 3 4 k + 4)$ ka koi inverse nahi hoga? Forecast: "koi inverse nahi" ⟺ $det = 0$ . Yeh $k$ mein ek equation hai — guess karo ki do answers honge (yeh quadratic hai).

Step 1 — singularity condition set karo. Koi inverse nahi ⟺ $det A = 0$ . Yeh step kyun? Hum usual logic ulta karte hain: $det = 0$ avoid karne ki jagah, hum use demand karte hain.

Step 2 — determinant expand karo. $det A = k (k + 4) - (4) (3) = k^{2} + 4 k - 12.$ Yeh step kyun? $a d - b c$ multiply out karo taaki $k$ mein ek polynomial mile.

Step 3 — $k^{2} + 4 k - 12 = 0$ solve karo. Factor karo: $(k + 6) (k - 2) = 0$ , toh $k = - 6$ ya $k = 2$ . Yeh step kyun? Har factor ko zero set karne par woh values milti hain jahan area collapse hota hai.

Verify:

$k = 2$ : $A = (2346)$ , $det = 12 - 12 = 0$ ✓ (row 2 $= 1.5 \times$ row 1).
$k = - 6$ : $A = (- 6 3 4 - 2)$ , $det = 12 - 12 = 0$ ✓ (column 2 $= - \frac{2}{3} \times$ column 1). ✓

Active recall

Recall Kaun si case classes "invert karne ke liye safe" hain aur kaun si nahi?

Safe (invertible): C1, C2, C3, C5, C6, C7 — sab mein $det \neq = 0$ hai. Invertible nahi: C4 (aur C8 mein $k = - 6, 2$ par) — $det = 0$ .

Zero entry ka matlab matrix singular hai.

False — zero entry (Ex 5) ne phir bhi

det = 6 \neq = 0

diya. Sirf

det = a d - b c = 0

ka matlab singular hota hai.

Negative determinant ka matlab koi inverse nahi hai.

False —

det = - 7

(Ex 2) bilkul theek hai; yeh sirf orientation flip karta hai. Sirf exactly

det = 0

inverse block karta hai.

A

ko singular banane wala

k

nikalne ke liye, tum…

det A = 0

k

ke liye solve karo (Ex 8).

Inverse se

A x = b

solve karne ke liye, compute karo…

x = A^{- 1} b

(Ex 6).

Connections

Determinant of a 2×2 matrix — woh gatekeeper jo har example mein pehle check hota hai.
Adjugate and cofactors — woh swap/negate matrix jo har inversion mein use hoti hai.
Matrix multiplication — har "Verify" step mein use hoti hai.
Identity matrix — woh target $I$ jiske against hum check karte hain.
Solving linear systems — Examples 6 aur 7.
Singular vs non-singular matrices — Examples 4 aur 8.
Linear transformations and area — kyun $det = 0$ plane ko flatten karta hai (Ex 4).