Invertible matrix theorem — 12+ equivalent conditions: Worked examples

4.5.25 · D3 · Maths › Linear Algebra (Full) › Invertible matrix theorem — 12+ equivalent conditions

Definition Is page par use hone wale symbols aur words

$n \times n$ / square matrix: numbers ka ek grid jisme rows aur columns ki same sankhya ho ( $n$ dono ki). Theorem sirf square matrices ke baare mein baat karta hai; hum ek non-square matrix se purpose se milenge taaki dekh sakein ki theorem wahan chup rehta hai.
$I$ (identity matrix): woh square matrix jisme diagonal par $1$ hon aur baaki jagah $0$ ; e.g. $I_{2} = (1001)$ , aur $I_{3}$ iska $3 \times 3$ version hai. Yeh "kuch nahi karne wali" matrix hai: $A I = I A = A$ . Row operations se $I$ tak pahunchna hi "invertible" ka matlab hai.
$R^{n}$ : $n$ real numbers ki saari lists ka space, e.g. $R^{2}$ sabhi $(x_{1}, x_{2})$ ka flat plane hai, aur $R^{3}$ ordinary $3$ D space hai.
$A x = b$ : matrix $A$ ek input vector $x$ par act karke output vector $b$ deta hai. "Har $b$ ke liye solvable" ka matlab hai: kya yeh machine koi bhi target hit kar sakti hai?
$0$ (zero vector) aur null space: $0 = (0, \dots, 0)$ . Null space mein woh saare inputs $x$ hain jinhe machine crush karke $0$ bana deti hai. Agar sirf $x = 0$ hi crush hota hai, toh null space trivial hai — yeh ek accha sign hai.
pivot & RREF: jab tum ek matrix ko row reduction se tidy karte ho, har row ki pehli nonzero leading entry (clean up ke baad) ek pivot hoti hai. Poori tarah tidy ki gayi form hai Reduced Row Echelon Form (RREF). Ek square matrix jisme har row aur column mein pivot ho, uska RREF $I$ ke barabar hota hai — aur yahi exactly invertibility hai.
rank: pivots ki sankhya = rows ya columns mein genuinely independent directions ki sankhya. "Full rank" ka matlab hai rank $= n$ .

Formula Woh aath equivalent tests jo hum yahaan actually use karte hain

Parent 17 equivalent statements list karta hai. Is page par hum sirf inhi aath ko use karte hain — inme se sab ek hi same yes/no answer dete hain square $A$ ke liye. Ek baar is box ko skim karo; neechay har "yeh step kyun?" wahan naam se point karta hai, kisi bare number se nahi.

[det] $det A \neq = 0$ .
[RREF] $A$ row-reduce hokar $I$ ban jaata hai, matlab har row aur column mein ek pivot hai.
[null] $A x = 0$ ka sirf trivial solution $x = 0$ hai.
[indep] $A$ ke columns linearly independent hain.
[solve] $A x = b$ ka har $b$ ke liye ek solution hai (equivalently, columns pura $R^{n}$ span karte hain).
[rank] $rank A = n$ (full rank).
[eig] $0$ matrix $A$ ka koi eigenvalue nahi hai.
[side] ek one-sided inverse exist karta hai — $C A = I$ (left) ya $A D = I$ (right); square $A$ ke liye dono mein se koi ek bhi full inverse force kar deta hai.

Scenario matrix

Har square matrix exactly do camps mein se ek mein aati hai — invertible ya nahi — lekin wajah ki woh wahan kyun land karti hai, aur use dekhne ka sabse sasta tarika, alag alag hota hai. Yahaan un saare cases ka poora grid hai jo hum cover karenge. "Cheapest test" column upar ke box se test ka naam batata hai.

Cell	Scenario class	Yeh interesting kyun hai	Cheapest test	Example
A	Chhoti $2 \times 2$ , $det \neq = 0$	baseline invertible	[det]	Ex 1
B	Chhoti $2 \times 2$ , $det = 0$	baseline singular	[det]	Ex 2
C	Degenerate: ek zero row/column	ek puri row zero hai	zero dhundo → [RREF]	Ex 3
D	Degenerate: repeated / proportional rows	dependence seedhi nazar aati hai	[null] / [indep]	Ex 4
E	$3 \times 3$ triangular	determinant bahut aasaan	diagonal product → [det]	Ex 5
F	Eigenvalue angle — kya $0$ ek eigenvalue hai?	spectrum se connection	[eig]	Ex 6
G	Non-square trap	theorem laagu NAHI hota	pehle squareness check	Ex 7
H	One-sided inverse twist	right inverse $\Rightarrow$ full inverse	[side]	Ex 8
I	Word problem — mixing recipe	$A x = b$ ki real-world solvability	[solve]	Ex 9
J	Poori $3 \times 3$ ( $det$ by cofactors) + parameter twist	non-triangular determinant, phir "kis $t$ ke liye yeh toot jaata hai?"	[det] (cofactor / $t$ ka function)	Ex 10, Ex 11

Do bade camps aur unhe split karne wale tests neechay flowchart mein dikhaye gaye hain. Iske consequence nodes exactly wahi aath named tests hain jo formula box mein hain — toh picture aur box ek hi bhaasha bolte hain.

Intuition Yeh flowchart kaise padhein

Top node se shuru karo, "Square n x n matrix". Pehla fork ("det zero ?") single decisive test [det] hai — left arrow (labeled yes) agar $det A = 0$ , right arrow (labeled no) otherwise. Ek baar jab tum NOT invertible (left) ya invertible (right) par land kar jaao, toh neechay ke arrows consequences mein fan out ho jaate hain.

Har consequence node box ke unhi aath tests mein se ek hai, same naam ke saath:

Left branch (failure): [indep] fail (dependent columns), [null] fail (ek nonzero null vector), [eig] fail ( $0$ is ek eigenvalue hai), aur rank $n$ se neechay gir jaata hai ([rank] fail).
Right branch (success): [rank] pass (rank $= n$ ), [indep] pass (independent columns), aur [solve] pass (har $b$ reachable).

Yeh seekhne ki nayi baatein nahi hain — yeh wahi same tests hain jo tumne box mein already padhe hain, arrows ke roop mein re-drawn kiye gaye hain taaki tum dekh sako ki ek fork sab decide kar deta hai. Hum success side par sirf teen consequence nodes rakhte hain (har ek ki jagah) kyunki woh teen — full rank, independent columns, "har target hit karna" — wahi hain jo neechay ke worked examples actually use karte hain; baaki silently saath chale aate hain. Har example asliyat mein bas yahi picture hai: "fork [det] check karo, phir poori branch padho."

Worked examples

Cell A — baseline invertible $2 \times 2$

Worked example Example 1 —

$det \neq = 0$ , sab kuch kaam karta hai Kya $A = (3214)$ invertible hai?

Forecast: Padhne se pehle guess karo — kya ek aisi matrix jiske rows clearly alag alag directions mein point karti hain, reversible lagti hai?

Step 1. Determinant compute karo $det A = (3) (4) - (1) (2) = 12 - 2 = 10$ . Yeh step kyun? $2 \times 2$ ke liye, test [det] sabse sasta condition hai — ek multiplication, ek subtraction. Aur $det \neq = 0$ hote hi poora theorem trigger ho jaata hai.

Step 2. $det A = 10 \neq = 0 \Rightarrow$ invertible. Baaki sab free mein padh lo. Yeh step kyun? Kyunki box ke saare tests equivalent hain, ek confirm karne par ([det]) har doosra automatically confirm ho jaata hai: columns linearly independent, rank $= 2$ , null space $= {0}$ , $0$ koi eigenvalue nahi. Koi extra kaam nahi — theorem ka poora point yahi hai.

Step 3. Kyunki yeh hai invertible, hum $2 \times 2$ formula $A^{- 1} = \frac{1}{d e t A} (d - c - b a)$ use karke inverse likh sakte hain: $A^{- 1} = \frac{1}{10} (4 - 2 - 1 3) .$ Yeh step kyun? Invertibility decide karne ke baad, actual inverse dikhana worth hai taaki hum use directly agली line mein $I$ ke against sanity-check kar sakein.

Verify: $A A^{- 1} = \frac{1}{10} (3214) (4 - 2 - 1 3) = \frac{1}{10} (100010) = I$ . ✔

Cell B — baseline singular $2 \times 2$

Worked example Example 2 —

$det = 0$ , sab kuch saath mein fail Kya $A = (2163)$ invertible hai?

Forecast: Do rows dekho. Kya kuch compute karne se pehle koi hidden relationship nazar aati hai?

Step 1. Compute karo $det A = (2) (3) - (6) (1) = 6 - 6 = 0$ . Yeh step kyun? Test [det] phir — $2 \times 2$ ke liye sabse sasta, aur $det = 0$ poore failing side ke theorem ka alarm bell hai.

Step 2. $det A = 0 \Rightarrow$ not invertible. Har equivalent condition fail hoti hai: columns dependent, rank $= 1 < 2$ , ek nonzero null vector exist karta hai, $0$ is ek eigenvalue hai. Yeh step kyun? Equivalence dono taraf chalti hai: agar ek sasta test ([det]) fail ho, toh sab dusre bhi fail honge. Toh ek akela zero determinant poori list ko ek saath condemn kar deta hai.

Step 3. Dependence dhundo: row 1 $= 2 \times$ row 2. Toh column 2 $= 3 \times$ column 1 (kyunki $6 = 3 \cdot 2$ , $3 = 3 \cdot 1$ ). Null vector satisfy karta hai $x_{1} (col_{1}) + x_{2} (col_{2}) = 0$ , yaani $3 (col_{1}) - (col_{2}) = 0$ , toh $x = (3, - 1)^{T}$ . Yeh step kyun? Null vector produce karna directly prove karta hai ki test [null] fail hua — ek concrete nonzero input jise machine $0$ mein crush kar deti hai — toh hum sirf determinant par depend nahi karte.

Verify: $A (3 - 1) = (2 \cdot 3 + 6 \cdot (- 1) 1 \cdot 3 + 3 \cdot (- 1)) = (00)$ . ✔ Ek nonzero input zero mein squash ho gaya — machine ne information destroy kar di.

Cell C — degenerate: ek zero row

Worked example Example 3 — zeros ki puri row

Kya $A = 501703209$ invertible hai?

Forecast: Zeros ki ek row — kya aisi matrix kabhi reverse ho sakti hai?

Step 1. Zero row dhundo (row 2). Zero row ka matlab hai row reduction us row mein kabhi pivot (leading nonzero entry) nahi rakh sakti. Yeh step kyun? Test [RREF] ko har row mein pivot chahiye. Zero row immediately ise forbid kar deti hai — koi arithmetic nahi chahiye.

Step 2. $3$ se kam pivots hone par, RREF (poori tarah tidy ki gayi form) $I_{3}$ nahi ho sakti, toh not invertible. Yeh step kyun? [RREF] kehta hai invertible $⟺$ tidy form equals $I_{3}$ . Missing pivot ka matlab hai tidy form mein abhi bhi ek zero row hai, toh woh kabhi identity se match nahi kar sakti — invertibility condemn ho jaati hai.

Step 3. Determinant se cross-check karo: zero row ke along expand karne par $det A = 0$ milta hai (har term ek zero entry multiply karta hai). Yeh step kyun? Ek independent test ([det]) chalana jo [RREF] se agree karna chahiye ek slip se bachata hai — do equivalent conditions ka same verdict dena hamara confidence badhata hai.

Verify: $det A = 0$ (row 2 ke along cofactor expansion — saari terms vanish ho jaati hain). ✔ Dono tests agree karte hain.

Cell D — degenerate: ek $3 \times 3$ mein chupi proportional rows

Worked example Example 4 — dependence jo dhundhni padti hai

Kya $A = 120241365$ invertible hai?

Forecast: Inme se do rows secretly same direction mein hain. Kaunsi do?

Step 1. Notice karo ki row 2 $= 2 \times$ row 1 exactly. Toh rows dependent hain (ek doosri ki scaled copy hai). Yeh step kyun? Proportionality aankhon se dekhna sabse sasta possible route hai — na determinant, na reduction. Dependent rows rank ko $n$ se neechay gira deti hain.

Step 2. Dependent rows ka matlab hai rank $\leq 2 < 3$ , toh test [rank] (full rank $= 3$ ) fail, isliye not invertible. Yeh step kyun? Yahaan hum rank theory ka ek fact use karte hain: independent rows ki sankhya hamesha independent columns ki sankhya ke barabar hoti hai (row rank $=$ column rank). Toh dependent rows dependent columns bhi guarantee karti hain, jo exactly wahi hai jo [rank]/[indep] forbid karta hai. (Yeh equality Rank and Nullity note mein prove ki gayi hai; hum ise yahaan sirf cite karte hain.)

Step 3. Concretely, dependent columns ka matlab hai test [null] bhi fail hota hai — koi nonzero $x$ hai jiske liye $A x = 0$ . Yeh step kyun? Hum failing branch par specific consequence name karte hain taaki reader flowchart ka left side live dekhta hai, na ki bas ek abstract "rank $< 3$ ".

Verify: $det A = 0$ . (Do proportional rows hamesha $det = 0$ force karti hain, kyunki ek row ko scale karna determinant ko scale karta hai, aur do equal rows $det = 0$ deti hain.) ✔

Cell E — limiting-easy: ek triangular matrix

Worked example Example 5 — triangular determinant bas diagonal hai

Kya $A = 400950 - 1 73$ invertible hai?

Forecast: Diagonal ke neechay ki saari entries zero hain. Kya yeh invertibility ka sawaal trivially easy bana deta hai?

Step 1. Ek upper-triangular matrix ke liye, $det A =$ diagonal entries ka product $= 4 \cdot 5 \cdot 3 = 60$ . Yeh step kyun? Triangular matrices [det] ke liye limiting-easy case hain: determinant diagonal product mein collapse ho jaata hai, koi expansion nahi chahiye. Yeh sabse sasta possible determinant hai.

Step 2. $det A = 60 \neq = 0 \Rightarrow$ invertible, isliye full rank $3$ aur koi zero eigenvalue nahi. Yeh step kyun? Ek passing test ([det]) har equivalent condition ko apne saath le jaata hai — hum sirf wahi report karte hain jo name karne layak hain ([rank], [eig]).

Step 3. Bonus insight: ek triangular matrix ke eigenvalues wahi hote hain jo uski diagonal entries $4, 5, 3$ hain. Koi bhi $0$ nahi hai, toh test [eig] independently invertibility confirm karta hai. Yeh step kyun? Ek doosra, independent route ([eig]) jo same verdict tak pahunche ek free reliability check hai — aur yeh agले example ke eigenvalue angle ko preview karta hai.

Verify: $det A = 60$ aur eigenvalues $= {4, 5, 3}$ , koi zero nahi. ✔

Cell F — eigenvalue angle: kya $0$ ek eigenvalue hai?

Worked example Example 6 — invertibility as "no zero eigenvalue"

Kya $A = (2123)$ invertible hai? Sirf eigenvalues use karke decide karo.

Forecast: Ek eigenvalue ek number $λ$ hai jiske liye kisi nonzero $x$ ke saath $A x = λ x$ ho. Agar unme se ek $0$ hai, toh test [eig] kehta hai matrix singular hai. Guess karo: kya $0$ aayega?

Step 1. Eigenvalues $det (A - λ I) = 0$ solve karte hain. Characteristic polynomial compute karo: $det (2 - λ 1 2 3 - λ) = (2 - λ) (3 - λ) - 2 = λ^{2} - 5 λ + 4.$ Yeh step kyun? Test [eig] ek hi sawaal poochtha hai: "kya $0$ ek eigenvalue hai?" Definition ke hisaab se $λ$ tab ek eigenvalue hai jab $A - λ I$ singular ho, yaani jab $det (A - λ I) = 0$ . Toh zero eigenvalue dhundhne ka sabsa sasta tarika yeh hai ki $λ$ mein ek polynomial banao aur check karo ki $λ = 0$ uski roots mein hai ya nahi — yeh eigenvectors directly dhundne se kahin kam kaam hai.

Step 2. Factor karo: $λ^{2} - 5 λ + 4 = (λ - 1) (λ - 4)$ , toh eigenvalues hain $λ = 1$ aur $λ = 4$ . Yeh step kyun? Factoring polynomial ko ek aisa product bana deta hai jo exactly apni roots par zero hota hai, ek nazar mein dono eigenvalues read karne deta hai taaki hum unme $0$ check kar sakein.

Step 3. Koi bhi eigenvalue $0$ nahi hai $\Rightarrow$ invertible. (Equivalently, constant term $4$ equals $det A$ , aur $det A = 4 \neq = 0$ .) Yeh step kyun? Constant term ko [det] se connect karna dikhata hai ki do tests same coin hain — eigenvalues ka product hi determinant hai, toh "koi zero eigenvalue nahi" aur " $det \neq = 0$ " literally ek hi statement hain.

Verify: eigenvalues ${1, 4}$ , product $= 4 = det A \neq = 0$ . ✔ Parent note se product-of-eigenvalues rule hold karta hai.

Cell G — non-square trap

Worked example Example 7 — theorem chup rehta hai

Matrix $A = 101011$ ( $3 \times 2$ ) ke independent columns hain. Kya yeh invertible hai?

Forecast: Independent columns laga tha ki invertibility ki guarantee hai. Kya $A$ square hai?

Step 1. Shape count karo: $3$ rows, $2$ columns. Square nahi. Yeh step kyun? Parent note ki crucial caveat: theorem sirf $n \times n$ matrices par apply hota hai. Box ke kisi bhi test se pehle, squareness check karo — warna har conclusion void hai.

Step 2. Kyunki $A$ square nahi hai, "invertible" (two-sided inverse) impossible hai. Iska map $x \mapsto A x$ $R^{2} \to R^{3}$ bhejna hai; yeh injective ho sakta hai (columns independent) lekin $R^{3}$ ko kabhi hit nahi kar sakta. Yeh step kyun? Do columns zyaada se zyaada ek $2$ -dimensional slab span kar sakte hain, aur $3$ -dimensional space $R^{3}$ ke basis ke liye kam se kam $3$ vectors chahiye. Toh test [solve] koi bhi ho fail karna hi hai — independence aur invertibility yahaan alag ho jaate hain.

Step 3. Conclusion: columns independent hain, lekin matrix invertible nahi hai. Yeh step kyun? Paradox clearly state karna lesson cement karta hai — "independent columns $\Rightarrow$ invertible" sirf squareness gate ke baad true hai.

Verify: Columns independent (rank $= 2$ ) lekin rank $\neq = 3$ , toh columns $R^{3}$ span nahi kar sakte: $2 < 3$ . ✔

Cell H — one-sided inverse twist

Worked example Example 8 — exam-style: ek right inverse kaafi hai

Exam claim: " $A$ $2 \times 2$ hai aur ek aisa $D$ exist karta hai jiske saath $A D = I$ . Lekin humein kabhi nahi bataya gaya ki $D A = I$ . Kya hum conclude kar sakte hain ki $A$ invertible hai?"

Forecast: $A D = I$ $A D = D A = I$ se weaker lagta hai. Kya ek square matrix ke liye one-sided inverse ek real inverse hai?

Step 1. $A$ square hai, aur box mein test [side] kehta hai ki akela one-sided inverse kaafi hai. Yeh step kyun? Square matrices ke liye, $A D = I$ automatically $D A = I$ force karta hai — yeh parent ke "one-sided inverse" mistake box ka content hai, toh ek right inverse poora theorem trigger kar deta hai.

Step 2. Concrete instance: maano $A = (1011)$ aur propose karo $D = (10 - 1 1)$ . Yeh step kyun? Ek explicit pair se hum actually multiply karke claim ko action mein dekh sakte hain, na ki abstractly bharose par rehna.

Step 3. $A D = (1011) (10 - 1 1) = (1001) = I$ . Toh $D$ ek right inverse hai $\Rightarrow$ invertible, aur "missing" $D A = I$ bhi automatically hold karna chahiye. Yeh step kyun? $A D$ compute karna [side] ki hypothesis verify karta hai; theorem phir humein free mein $D A = I$ de deta hai, jo hum neechay confirm karte hain.

Verify: $D A = (10 - 1 1) (1011) = (1001) = I$ . ✔ Doosra side free mein aa gaya.

Cell I — word problem

Worked example Example 9 — paint-mixing shop

Ek shop do base paints, Red base aur Blue base, ko mix karke custom colours banati hai. $x_{1}$ litres Red base aur $x_{2}$ litres Blue base mix karne par ek aisi paint milti hai jiska (redness, blueness) reading hai $(redness blueness) = (2112) (x_{1} x_{2}) .$ Sawaal: Kya shop koi bhi target reading $b \in R^{2}$ hit kar sakti hai, aur kya recipe unique hai?

Forecast: Agar do bases "kaafi alag" directions mein point karte hain, toh unhe har colour span karna chahiye. Guess karo haan ya nahi.

Step 1. $A = (2112)$ lo. Compute karo $det A = (2) (2) - (1) (1) = 4 - 1 = 3$ . Yeh step kyun? "Har $b$ ke liye solvable" test [solve] hai, jo [det] ke equivalent hai. Ek number dono existence (kya hum har colour reach kar sakte hain?) aur uniqueness (kya recipe forced hai?) settle kar deta hai.

Step 2. $det A = 3 \neq = 0 \Rightarrow$ invertible. Toh columns $R^{2}$ span karte hain (har reading achievable hai) aur map injective hai (har target ke liye exactly ek recipe — do alag mixes same colour nahi de sakte). Yeh step kyun? [solve] pass hona bataata hai ki columns plane span karte hain, aur same invertibility map ko injective banati hai — isliye square matrix ke liye "har target hit karna" aur "unique recipe" saath aate hain.

Step 3. Target $b = (4, 5)^{T}$ hit karne ke liye, $x = A^{- 1} b$ solve karo jahan $A^{- 1} = \frac{1}{3} (2 - 1 - 1 2)$ : $x = \frac{1}{3} (2 - 1 - 1 2) (45) = \frac{1}{3} (36) = (12) .$ Yeh step kyun? Kyunki $A$ invertible hai, unique recipe literally $A^{- 1} b$ hai — inverse apply karna mixing ko undo kar deta hai aur zaroorat ke litres recover karta hai.

Verify: $A (1, 2)^{T} = (2 \cdot 1 + 1 \cdot 2, 1 \cdot 1 + 2 \cdot 2)^{T} = (4, 5)^{T} = b$ . ✔ Units: litres in, colour-reading out, recipe exactly target reproduce karta hai.

Cell J (part 1) — cofactors se poora $3 \times 3$ determinant

Worked example Example 10 — genuinely

$3 \times 3$ non-triangular determinant Kya $A = 211130122$ invertible hai? Koi bhi shortcut (zero row, proportional rows, triangular shape) apply nahi hota, toh humein actually $3 \times 3$ determinant compute karna hoga.

Forecast: Koi obvious dependence nahi dikhti. Kya tumhe lagta hai ki determinant $0$ par aayega ya nahi?

Step 1. $det A$ ko first row ke along cofactors use karke expand karo: $det A = 2 det (3022) - 1 det (1122) + 1 det (1130) .$ Yeh step kyun? Test [det] abhi bhi sabse sasta single verdict hai, lekin poore $3 \times 3$ ke liye hum $det$ diagonal se nahi padh sakte — hum ise teen $2 \times 2$ determinants mein todh dete hain jo hum already jaante hain kaise karna hai. $+, -, +$ sign pattern row 1 ke along standard cofactor checkerboard hai.

Step 2. Har $2 \times 2$ minor evaluate karo: $det (3022) = 6 - 0 = 6$ , $det (1122) = 2 - 2 = 0$ , $det (1130) = 0 - 3 = - 3$ . Yeh step kyun? Bade determinant ko chhhote mein reduce karna exactly wahi hai jo cofactor expansion kaam karta hai — har minor ek baseline $2 \times 2$ hai jo hum ek line mein kar sakte hain.

Step 3. Combine karo: $det A = 2 (6) - 1 (0) + 1 (- 3) = 12 - 0 - 3 = 9$ . Kyunki $det A = 9 \neq = 0$ , $A$ invertible hai — isliye full rank $3$ , independent columns, koi zero eigenvalue nahi. Yeh step kyun? Pieces ko unke signs ke saath wapas rakhne par woh ek number milta hai jo [det] ko chahiye; nonzero hona flowchart ki poori success branch on kar deta hai.

Verify: cofactor answer $det A = 9 \neq = 0$ . ✔

Cell J (part 2) — parameter / limiting case

Worked example Example 11 — kis

$t$ par invertibility toot ti hai? Maano $A (t) = (1 t t 1)$ . Kinkin real values of $t$ ke liye $A (t)$ not invertible hai?

Forecast: Jaise $t$ badhta hai, do rows ek doosre ki taraf tilt karti hain. Kis tilt par woh ek line mein aa jaati hain aur matrix collapse ho jaati hai?

Step 1. Determinant ko $t$ ka function maano: $det A (t) = (1) (1) - (t) (t) = 1 - t^{2}$ . Yeh step kyun? Test [det] poore sawaal ko "yahaan $1 - t^{2} = 0$ kahan hai?" mein badal deta hai — infinitely many matrices test karne ki jagah ek equation solve karo.

Step 2. $1 - t^{2} = 0 ⟺ t^{2} = 1 ⟺ t = 1 or t = - 1$ . Toh $A (t)$ singular exactly at $t = \pm 1$ hai, aur har doosre $t$ ke liye invertible. Yeh step kyun? Determinant function ki roots exactly woh parameter values hain jo [det] fail karti hain; unse door determinant nonzero hai, toh invertibility hold karta hai.

Step 3. Poori range mein geometry padho (neechay figure dekho):

$- 1 < t < 1$ : $det > 0$ , invertible — do rows clearly alag alag directions mein point karti hain.
$t = 1$ : $det = 0$ , dono rows $(1, 1)$ ban jaati hain — identical, toh woh ek line par collapse ho jaati hain; null vector $(1, - 1)^{T}$ appear hota hai kyunki $A (1) (1, - 1)^{T} = 0$ .
$t = - 1$ : $det = 0$ , rows $(1, - 1)$ aur $(- 1, 1)$ ban jaati hain — ek doosre ke negatives, phir ek hi line par; null vector $(1, 1)^{T}$ hai kyunki $A (- 1) (1, 1)^{T} = 0$ .
$∣ t ∣ > 1$ : $det < 0$ , phir se invertible — map ab flip karta hai orientation (negative determinant), lekin reversibility untouched hai, kyunki sirf $det \neq = 0$ matter karta hai, sign nahi. Yeh step kyun? Har interval aur determinant ke dono signs cover karna "all cases" requirement hai — reader ko koi bhi $t$ ki value nahi milti jise humne classify nahi kiya, aur woh dekhta hai ki collapse exactly $t = \pm 1$ par kyun hota hai (rows ek hi line par aa jaati hain).

Figure 1 — Determinant curve $det A (t) = 1 - t^{2}$ parameter $t$ ke against plot ki gayi hai. Curve zero cross karti hai (pink dots) exactly $t = - 1$ aur $t = + 1$ par, sirf do singular cases. Blue-shaded hump ( $- 1 < t < 1$ ) wahan hai jahan $det > 0$ aur matrix invertible hai; yellow-shaded tails ( $∣ t ∣ > 1$ ) wahan hain jahan $det < 0$ aur yeh abhi bhi invertible hai. Sirf do crossings singular hain.

Verify: $det A (1) = 0$ , $det A (- 1) = 0$ , aur $det A (0) = 1 \neq = 0$ , $det A (2) = - 3 \neq = 0$ . ✔ Singular exactly $t = \pm 1$ par.

Recall Quick self-test

$2 \times 2$ ke liye sabse sasta test kya hai? ::: Determinant $a d - b c$ (test [det]) — ek line poora theorem settle kar deta hai. Poore non-triangular $3 \times 3$ ka $det$ kaise nikaalte hain? ::: Row ke along cofactor expansion — $+, -, +$ sign pattern ke saath teen $2 \times 2$ minors. Ek $3 \times 3$ mein zeros ki ek puri row hai. Invertible? ::: Nahi — us row mein kabhi pivot nahi ho sakta, toh yeh [RREF]/[rank] fail karta hai. $3 \times 2$ matrix par independent columns — invertible? ::: Theorem apply nahi hota; non-square matrices kabhi invertible nahi hote. Square $A$ jiske saath sirf $A D = I$ ho — kya $A$ invertible hai? ::: Haan; [side] ke hisaab se, square matrices ke liye right inverse automatically full inverse hai. $(1 t t 1)$ ke liye, singular kab? ::: $t = \pm 1$ par, jahan $det = 1 - t^{2} = 0$ aur do rows ek line par aa jaati hain.