Inverse of 2×2 matrix: Question bank — Hinglish deep dive

2.6.10 · D5 · Maths › Matrices & Determinants — Introduction › Inverse of 2×2 matrix

Sach ya jhooth — justify karo

Sach ya jhooth: Har square matrix ka inverse hota hai.

Jhooth. Sirf non-singular matrices ka hota hai. Agar

det A = 0

toh map area ko zero kar deta hai, information destroy ho jaati hai, isliye koi undo machine exist nahi kar sakti.

Sach ya jhooth: Agar

det A = 0

toh

A

har vector ko zero vector par bhejta hai.

Jhooth. Yeh plane ko ek line par collapse karta hai (ya sirf extreme case

A = 0

mein ek point par). Zyaadatar inputs us line par nonzero vectors ke roop mein land karte hain — lekin bahut saare inputs ek hi output share karte hain, isliye ise undo nahi kiya ja sakta.

Sach ya jhooth: Agar

A

invertible hai, toh

A^{- 1}

bhi hai, aur

(A^{- 1})^{- 1} = A

Sach. Undo ka undo original machine hai. Kyunki

A^{- 1} A = I

hai,

A

woh defining property satisfy karta hai jo

A^{- 1}

ke inverse ki hoti hai.

Sach ya jhooth:

det (A^{- 1}) = \frac{1}{det A}

Sach.

A A^{- 1} = I

se aur is fact se ki determinants multiply hote hain,

det A \cdot det A^{- 1} = det I = 1

, isliye

det A^{- 1} = 1/ det A

(jiske liye

det A \neq = 0

chahiye — jo

A

ke invertible hone ke saath consistent hai).

Sach ya jhooth: Agar

A

aur

B

dono invertible hain, toh

(A B)^{- 1} = A^{- 1} B^{- 1}

Jhooth. Order reverse hota hai:

(A B)^{- 1} = B^{- 1} A^{- 1}

. "Pehle

A

lagao phir

B

" ko undo karne ke liye pehle

B

ko undo karna padega, phir

A

ko — bilkul jaise shoes pehle utaarne padte hain socks se pehle.

Sach ya jhooth: Identity matrix apna khud ka inverse hai.

Sach.

I \cdot I = I

, isliye

I^{- 1} = I

. Jo machine kuch nahi karti, usse kuch na karne se hi undo kiya jaata hai.

Sach ya jhooth: Agar

det A

negative hai, toh

A

ka koi inverse nahi.

Jhooth. Koi bhi nonzero determinant inverse deta hai. Negative determinant ka matlab sirf yeh hai ki map orientation bhi flip karta hai (ek mirror-flip), jo bilkul reversible hai.

Sach ya jhooth:

A

ki dono rows swap karne se kabhi yeh nahi badalta ki woh invertible hai ya nahi.

Sach. Row-swap sirf determinant ki sign flip karta hai, kabhi nonzero value ko zero nahi banata. Invertibility

det A \neq = 0

par depend karti hai, jo sign change ke baad bhi bachi rehti hai.

Sach ya jhooth: Agar

A

ke do columns identical hain, toh

A

singular hai.

Sach. Identical columns parallel hote hain, isliye unse bana parallelogram ka area zero hota hai, yaani

det A = 0

. Dekho Singular vs non-singular matrices.

Unhone har entry ka reciprocal le liya. Matrix inversion entrywise nahi hoti; Matrix multiplication rows aur columns ko mix karta hai, isliye ise undo karne ke liye swap-diagonal, negate-off-diagonal, divide by det chahiye, har slot par

1/ x

nahi.

Ek student

(a c b d)

ka adjugate

(- a c b - d)

compute karta hai. Unhone kaun sa rule ulta kiya?

Unhone diagonal ko negate kiya aur off-diagonal ko waise hi choda — bilkul ulta. Rule yeh hai: main diagonal par

a \leftrightarrow d

swap karo aur off-diagonal

b, c

ko negate karo:

(d - c - b a)

Ek student kehta hai "

det A = 3

, isliye

A^{- 1} = 3 (d - c - b a)

." Theek karo.

Unhone det se divide karne ki jagah multiply kar diya. Formula hai

A^{- 1} = \frac{1}{d e t A} adj A

, isliye yeh hona chahiye tha

\frac{1}{3} (d - c - b a)

Ek student

A x = b

x = b A^{- 1}

likh kar solve karta hai. Kya galti hai?

Multiplication order (aur isliye dimensions) galat hai. Left-multiply karna zaroori hai:

A^{- 1} A x = A^{- 1} b

x = A^{- 1} b

milta hai, matrix column vector ke left mein hoti hai. Dekho Solving linear systems.

Ek student kehta hai "

A^{2} = I

, isliye

A

zaroor

I

- I

ke barabar hai." Iska jawab do.

Zaroorat nahi. Koi bhi reflection square hoke identity deta hai, jaise

A = (10 0 - 1)

satisfy karta hai

A^{2} = I

lekin na

I

hai na

- I

A^{2} = I

sirf yeh kehta hai ki

A

apna khud ka inverse hai.

Ek student likhta hai "

det A = a d + b c

." Kahan slip hui?

Unhone off-diagonal product ko subtract karne ki jagah add kar diya. Determinant hai

a d - b c

; minus hi signed area encode karta hai, isliye ise kabhi drop nahi kiya ja sakta.

Ek student kehta hai "

A^{- 1}

exist karta hai, isliye

A^{- 1}

mein

A

ke wahi entries hain bas rearrange hue hain." Kya yeh generally sach hai?

Nahi. Zyaadatar matrices ke liye entries

1/ det A

se scale bhi hoti hain, jo unki numeric values badal deta hai, sirf positions nahi. Sirf jab

det A = 1

(ya

- 1

) ho tab

A^{- 1}

A

ki entries ka pure rearrangement/sign-flip hota hai.

Why ke sawaal

Formula mein

det A

se divide kyun kiya jaata hai, na ki, maan lo,

a

se?

Kyunki

det A = a d - b c

wahi common factor hai jo saari four defining equations

A A^{- 1} = I

solve karne par nikla tha. Yeh woh akela quantity hai jo answer ki har entry share karti hai.

det A = 0

invertible aur not ke beech ki exact boundary kyun hai, na ki koi aisi cheez jaise

det A < 1

Kyunki

det A

area measure karta hai, aur inverse tab exist karta hai jab koi bhi area destroy na ho. Koi bhi nonzero area — chahe kitna bhi tiny ho — rescale hoke wapas aa sakta hai; sirf zero area information irreversibly kho deta hai.

Negative determinant physically phir bhi valid inverse kyun deta hai?

Negative det ka matlab hai map orientation flip karta hai (plane ko page ki tarah palat deta hai). Flipping reversible hai — bas wapas flip karo — isliye orientation reversal kabhi inverse ko nahi rokta; sirf area collapse rokta hai.

Hum inverse recipe apply karne se pehle determinant kyun compute karte hain?

Agar woh zero hai, toh recipe zero se divide karne ki demand karti hai aur koi inverse exist nahi karta — isliye pehle check karna tumhe undefined entries wala meaningless answer banane se bachata hai.

(A B)^{- 1} = B^{- 1} A^{- 1}

kyun hai,

A^{- 1} B^{- 1}

nahi?

Kyunki sequence ko undo karna uska order reverse karta hai.

A B

kisi vector par act karte waqt pehle

B

ka effect apply karta hai; reverse karne ke liye, last-applied cheez pehle undo karni padti hai, yaani

A

ko last undo karo, isliye product mein

B^{- 1}

pehle aata hai.

Geometric picture (columns se parallelogram banana) yeh guarantee kyun karta hai ki parallel columns ka matlab no inverse hai?

Parallel columns ek degenerate parallelogram span karte hain jiska area zero hai, isliye

det A = 0

. Zero area ka matlab hai poori plane ek line par squash ho jaati hai — bahut saare inputs ek output share karte hain — aur jo function one-to-one nahi hai usse undo nahi kiya ja sakta.

Edge cases

Kya zero matrix

(0000)

invertible hai?

Nahi. Uska determinant

0

hai, aur yeh har vector ko

0

par bhejti hai, saari information erase kar deti hai — sabse extreme singular case possible.

Agar

A

ki ek row poori tarah zero hai, toh kya keh sakte hain?

Yeh singular hai. Ek zero row

det A = 0

force karta hai (jaise

(a 0 b 0)

deta hai

a \cdot 0 - b \cdot 0 = 0

), isliye koi inverse exist nahi karta.

Kya diagonal matrix

(a 0 0 d)

ka hamesha inverse hota hai?

Sirf jab dono

a \neq = 0

aur

d \neq = 0

hon. Tab

det = a d \neq = 0

aur

A^{- 1} = (1/ a 0 0 1/ d)

; agar koi bhi diagonal entry

0

hai, toh woh axis collapse ho jaati hai aur matrix singular hai.

Identity-scaled matrix

k I = (k 0 0 k)

ka inverse kya hai

k \neq = 0

ke liye?

\frac{1}{k} I

k

se uniform scaling ko

1/ k

se uniform scaling se undo kiya jaata hai; yeh sirf

k = 0

par fail karta hai, jahan sab kuch origin par collapse ho jaata hai.

Kya koi matrix apne khud ke adjugate ke barabar ho sakti hai?

Haan — jaise

(a 0 0 a)

ka adjugate

(a 0 0 a)

hai (swap equal diagonal par kuch nahi karta, zeros zero rehte hain). Yeh tab hota hai jab

a = d

aur

b = c = 0

ho.

Agar

det A = 1

hai, toh

A^{- 1}

A

ke adjugate se kaise related hai?

Dono equal hain:

A^{- 1} = \frac{1}{1} adj A = adj A

1

se divide karne par kuch nahi badalta, isliye inverse sirf swap/negate matrix hai bina kisi rescaling ke.

Recall Yahan har trap ki ek-line summary

Inverse exist karta hai $⟺ det A \neq = 0$ (koi bhi nonzero value, positive ya negative); recipe hai swap diagonal, negate off-diagonal, det se divide karo; aur order reverse hota hai $(A B)^{- 1} = B^{- 1} A^{- 1}$ ke under. Baaki sab inme se kisi ek ko bhool jaane ka variation hai.

Connections

Inverse of 2×2 matrix — parent note poori derivation aur formula ke saath.
Determinant of a 2×2 matrix — woh $a d - b c$ gatekeeper jo upar har trap ke peeche hai.
Singular vs non-singular matrices — invertible/not dividing line jo edge cases mein probe ki gayi.
Matrix multiplication — kyun order reverse hota hai aur kyun entrywise reciprocals fail karte hain.
Identity matrix — woh $I$ jo self-invert karta hai aur inverse define karta hai.
Solving linear systems — "Spot the error" mein left-multiply application.
Adjugate and cofactors — woh swap/negate object jise kai sawaal test karte hain.
Linear transformations and area — "why" sawaalon ka geometric core.

Question bank — Inverse of 2×2 matrix

Sach ya jhooth — justify karo

Galti dhundho

Why ke sawaal

Edge cases

Connections