Systems of first-order linear ODEs — matrix method: Worked examples

4.6.21 · D3 · Maths › Ordinary Differential Equations › Systems of first-order linear ODEs — matrix method

Yeh parent matrix-method note ka ek companion drill page hai. Parent ne tumhe machinery dikhaayi. Yahan hum ensure karte hain ki tumne woh machinery har tarah ki situation par chalayi ho — eigenvalue ka har sign, har geometry, degenerate aur disguised cases bhi.

Shuru karne se pehle, vocabulary ka ek honest reminder, seedhi zubaan mein, taaki neeche koi bhi symbol aisa na ho jo tumhe diya na gaya ho:

Definition Paanch objects, seedhi English mein

$A$ ek fixed grid of numbers hai (ek matrix) jo ek moving point ke coordinates ko "mix" karta hai.
$x (t)$ us point ki position hai time $t$ par — origin se ek chhota sa arrow.
$x^{'}$ uska velocity arrow hai: woh abhi kis direction mein aur kitni tezi se drift kar raha hai.
$A x$ woh velocity hai jo matrix $x$ par baithe point ko assign karta hai. Toh $x^{'} = A x$ padhta hai: "har point par, motion ka arrow wahi hai jo $A$ tumhari position ko banaa deta hai."
$I$ identity matrix hai — "kuch na karne waali" matrix. 2-D mein, $I = (1001)$ ; kisi bhi vector ko $I$ se multiply karne par woh unchanged rehta hai. Hum $λ I = (λ 0 0 λ)$ likhte hain taaki $A - λ I$ ka matlab ho " $A$ ki har diagonal entry mein se $λ$ ghataao" — yahi $det (A - λ I)$ ke andar aata hai. Ek eigenvector $v$ ek special direction hai jis par $A$ sirf stretching ki tarah kaam karta hai: $A v = λ v$ . Stretch factor $λ$ eigenvalue hai. (Prerequisites Eigenvalues and Eigenvectors mein hain.)

The scenario matrix

Har $2 \times 2$ system neeche diye gaye cells mein se bilkul ek mein aata hai. Eigenvalues geometry decide karte hain — yahi poora classification hai, aur yeh Phase Portraits and Stability ka subject hai.

Cell	Eigenvalue situation	Trajectories ki geometry	Stability
C1	Real, opposite signs $λ_{1} < 0 < λ_{2}$	Saddle	unstable
C2	Real, dono $< 0$	Node, arrows andar ki taraf	stable
C3	Real, dono $> 0$	Node, arrows bahar ki taraf	unstable
C4	Complex, real part $= 0$ ( $α = 0$ )	Center — closed circles/ellipses	neutral
C5	Complex, real part $< 0$ ( $α < 0$ )	Spiral inward	stable
C6	Complex, real part $> 0$ ( $α > 0$ )	Spiral outward	unstable
C7	Repeated eigenvalue, sirf ek eigenvector (defective)	Improper node (shear)	sign of $λ$
C7★	Repeated eigenvalue, do independent eigenvectors	Proper / star node (seedhi rays)	sign of $λ$
C8	Ek eigenvalue $= 0$	Line of equilibria (degenerate)	non-isolated
C9	Word problem (mixing tanks) → C2 mein aata hai	physical decay	stable
C10	Exam twist: ek solution diya, $A$ reconstruct karo	reverse-engineering	—

Ab hum har cell ko hit karte hain.

Forecast: eigenvalues ek positive, ek negative aayenge → yeh ek saddle hai, toh point ek direction mein infinity ki taraf escape karega jabki doosri direction mein origin ki taraf kheencha jaayega. Padhne se pehle escape direction guess karo.

Characteristic equation. $det (A - λ I) = (1 - λ)^{2} - 9 = 0$ . Yeh step kyun? Ek nonzero eigenvector tabhi exist karta hai jab $A - λ I$ singular ho, yaani uska determinant zero ho. $(1 - λ)^{2} = 9 \Rightarrow 1 - λ = \pm 3 \Rightarrow λ = 4, - 2.$ Opposite signs → saddle, C1. ✓ forecast se match karta hai.
Eigenvector for $λ = 4$ . $(A - 4 I) = (- 3 3 3 - 3)$ , toh $- 3 v_{1} + 3 v_{2} = 0 \Rightarrow v_{1} = v_{2}$ , giving $v_{+} = (1 1)$ . Yeh step kyun? Hum singular system ko uski null direction ke liye solve karte hain — woh line jise $A$ sirf stretch karta hai.
Eigenvector for $λ = - 2$ . $(A + 2 I) = (3333)$ , toh $v_{1} = - v_{2}$ , giving $v_{-} = (- 1 1)$ . Yeh step kyun? Wahi null-direction hunt, ab doosre eigenvalue ke liye — yeh doosri special line deta hai, woh line jiske saath motion decay karta hai.
General solution + IC. $x (t) = c_{1} e^{4 t} (1 1) + c_{2} e^{- 2 t} (- 1 1) .$ At $t = 0$ : $c_{1} + c_{2} = 1, c_{1} - c_{2} = 0 \Rightarrow c_{1} = c_{2} = \frac{1}{2}$ . Yeh step kyun? Superposition (system linear hai) do-parameter family deta hai; do initial numbers constants pin karte hain. $x (t) = \frac{1}{2} e^{4 t} (1 1) + \frac{1}{2} e^{- 2 t} (- 1 1) .$

Figure dekho: $(1 1)$ direction (red) blow up karti hai, $(- 1 1)$ direction (blue) mitti hai — classic saddle cross.

Figure — Systems of first-order linear ODEs — matrix method

Verify: $x (0) = \frac{1}{2} (1 1) + \frac{1}{2} (- 1 1) = (0 1)$ ✓. Aur $A v_{+} = (3 \cdot 1 + 1 \cdot 1 1 \cdot 1 + 3 \cdot 1) = (4 4) = 4 v_{+}$ ✓.

Forecast: agar dono eigenvalues negative hain, toh sab kuch origin ki taraf kheencha jaata hai — ek stable node. Slow-decaying direction late motion dominate karti hai. Guess karo kaun si.

Eigenvalues. $det (A - λ I) = (- 3 - λ)^{2} - 1 = 0 \Rightarrow (- 3 - λ) = \pm 1 \Rightarrow λ = - 2, - 4.$ Dono negative → C2. ✓ Yeh step kyun? Wahi singularity condition; sirf signs already picture classify kar dete hain.
Eigenvectors. For $λ = - 2$ : $(- 1 1 1 - 1)$ , $v_{1} = v_{2} \Rightarrow (1 1)$ . For $λ = - 4$ : $(1111)$ , $v_{1} = - v_{2} \Rightarrow (- 1 1)$ . Yeh step kyun? Har singular matrix $A - λ I$ ko uski null direction ke liye solve karte hain — woh do lines jin par decay pure hoti hai.
Solution + IC. $x = c_{1} e^{- 2 t} (1 1) + c_{2} e^{- 4 t} (- 1 1)$ . At $t = 0$ : $c_{1} + c_{2} = 2, c_{1} - c_{2} = 4 \Rightarrow c_{1} = 3, c_{2} = - 1$ . $x (t) = 3 e^{- 2 t} (1 1) - e^{- 4 t} (- 1 1) .$ Yeh step kyun? $e^{- 4 t}$ tezi se vanish hota hai, toh large $t$ ke liye trajectory $(1 1)$ line ke saath chipak jaati hai — slow eigendirection jeetta hai. Forecast confirm hua.

Verify: $x (0) = 3 (1 1) - (- 1 1) = (4 2)$ ✓. Jab $t \to \infty$ , $x \to 0$ (stable) ✓.

Forecast: diagonal matrix ka matlab hai axes already eigendirections hain — koi mixing nahi. Dono entries positive → outward node. Kaun sa axis tezi se bhaagega?

Eigenvalues/vectors seedhe padhte hain. Ek diagonal $A$ mein $λ_{1} = 2$ with $e_{1} = (0 1)$ , aur $λ_{2} = 5$ with $e_{2} = (1 0)$ . Dono $> 0$ → C3. Yeh step kyun? Ek diagonal matrix mein har coordinate apni khud ki scalar ODE $x_{i}^{'} = λ_{i} x_{i}$ follow karta hai — koi coupling nahi sulajhaani.
Solution. $x_{1} = e^{2 t}$ , $x_{2} = e^{5 t}$ (IC deta hai $c_{1} = c_{2} = 1$ ): $x (t) = (e ^{5 t} e ^{2 t}) .$ Yeh step kyun? Har decoupled scalar equation $x_{i}^{'} = λ_{i} x_{i}$ solve hoti hai $x_{i} = x_{i} (0) e^{λ_{i} t}$ mein. $y$ -axis (rate 5) $x$ -axis (rate 2) se aage nikal jaata hai.

Verify: $x (0) = (e ^{0} e ^{0}) = (1 1)$ ✓. $x^{'} = (5 e ^{5 t} 2 e ^{2 t}) = A x$ ✓.

Forecast: eigenvalues pure imaginary honge ( $α = 0$ ). Koi growth nahi, koi decay nahi → point hamesha ek closed loop par orbit karta rahega. Guess karo: circle ya ellipse?

Eigenvalues. $det (A - λ I) = λ^{2} + 9 = 0 \Rightarrow λ = \pm 3 i$ . Toh $α = 0, β = 3$ → center, C4. ✓ Yeh step kyun? Euler's Formula $e^{λ t} = e^{(α + i β) t} = e^{α t} (cos β t + i sin β t)$ banata hai; yahan $e^{α t} = e^{0} = 1$ , toh koi amplitude change nahi — pure rotation.
Complex eigenvector for $λ = 3 i$ . $(- 3 i 3 - 3 - 3 i) v = 0 \Rightarrow - 3 i v_{1} - 3 v_{2} = 0 \Rightarrow v_{2} = - i v_{1}$ . Lo $v = (- i 1) = p (0 1) + i q (- 1 0)$ . Yeh step kyun? Ek complex branch dono real solutions carry karta hai; $v = p + i q$ split karne se woh saaf ho jaate hain.
Do real solutions ( $α = 0, β = 3$ ). $x_{1} = p cos 3 t - q sin 3 t = (s i n 3 t c o s 3 t), x_{2} = p sin 3 t + q cos 3 t = (- c o s 3 t s i n 3 t) .$ Yeh step kyun? Ek complex solution $e^{λ t} v$ ka ek real aur ek imaginary part hota hai, aur ek real matrix ke liye har part khud ek real solution hai. $e^{i β t} (p + i q) = (p cos β t - q sin β t) + i (p sin β t + q cos β t)$ ko Euler's Formula se expand karna aur do brackets padhna bilkul wahi hai complex-to-real conversion recipe — isliye $x_{1}, x_{2}$ yeh form lete hain.
IC lagao. $x = c_{1} (s i n 3 t c o s 3 t) + c_{2} (- c o s 3 t s i n 3 t)$ . At $t = 0$ : $c_{1} (0 1) + c_{2} (- 1 0) = (0 2) \Rightarrow c_{1} = 2, c_{2} = 0$ . $x (t) = (2 sin 3 t 2 cos 3 t) .$ Yeh radius 2 ka ek circle trace karta hai — forecast confirm hua.

Verify: $x_{1}^{2} + x_{2}^{2} = 4 cos^{2} 3 t + 4 sin^{2} 3 t = 4$ (constant radius = closed orbit) ✓. $x (0) = (0 2)$ ✓.

Forecast: complex eigenvalues with negative real part → spiral jo origin ki taraf andar wind karta hai. Winding direction guess karo.

Eigenvalues. $det (A - λ I) = (- 1 - λ)^{2} + 4 = 0 \Rightarrow (- 1 - λ)^{2} = - 4 \Rightarrow λ = - 1 \pm 2 i$ . Toh $α = - 1 < 0, β = 2$ → stable spiral, C5. ✓ Yeh step kyun? $α < 0$ $e^{α t}$ ko shrink karta hai; $β \neq = 0$ use rotate karta hai. Shrink + rotate = inward spiral.
Eigenvector for $λ = - 1 + 2 i$ . $(A - λ I) = (- 2 i 2 - 2 - 2 i)$ ; pehli row: $- 2 i v_{1} - 2 v_{2} = 0 \Rightarrow v_{2} = - i v_{1}$ . Toh $v = (- i 1) = (0 1) + i (- 1 0)$ , giving $p = (0 1), q = (- 1 0)$ . Yeh step kyun? Hum (complex) singular system ko uski null direction ke liye solve karte hain, phir $v = p + i q$ split karte hain taaki real-part/imag-part recipe do real solutions bana sake.
Real solutions ( $α = - 1, β = 2$ ). $x_{1} = e^{- t} (p cos 2 t - q sin 2 t) = e^{- t} (s i n 2 t c o s 2 t), x_{2} = e^{- t} (- c o s 2 t s i n 2 t) .$ Yeh step kyun? Wahi real-part/imag-part recipe jaise C4, lekin ab $e^{- t}$ envelope har turn mein radius ko shrink karta hai.
IC lagao. $c_{1} (0 1) + c_{2} (- 1 0) = (0 1) \Rightarrow c_{1} = 1, c_{2} = 0$ . $x (t) = e^{- t} (sin 2 t cos 2 t) .$

Verify: $∣ x (t) ∣^{2} = e^{- 2 t} (cos^{2} 2 t + sin^{2} 2 t) = e^{- 2 t} \to 0$ — radius shrink ho raha hai, inward spiral confirm karta hai ✓. $x (0) = (0 1)$ ✓.

Forecast: C5 jaisa hi lekin real part ab positive hai → spiral bahar ki taraf udta hai. IC ki zaroorat nahi; hum sirf classify karte hain aur general solution dete hain.

Eigenvalues. $det (A - λ I) = (1 - λ)^{2} + 4 = 0 \Rightarrow λ = 1 \pm 2 i$ . $α = 1 > 0, β = 2$ → unstable spiral, C6. ✓ Yeh step kyun? $α > 0$ $e^{α t}$ ko grow karta hai — envelope rotate hote hue expand karta hai.
Eigenvector for $λ = 1 + 2 i$ . $(A - λ I) = (- 2 i 1 - 4 - 2 i)$ ; doosri row: $v_{1} - 2 i v_{2} = 0 \Rightarrow v_{1} = 2 i v_{2}$ . Lo $v = (1 2 i) = (1 0) + i (0 2)$ , toh $p = (1 0), q = (0 2)$ . Yeh step kyun? Hum singular complex system ko uski null direction ke liye solve karte hain aur $v = p + i q$ split karte hain, kyunki complex branch ke real aur imaginary parts bilkul wahi hain jo hum two-real-solution recipe mein feed karte hain.
General real solution ( $α = 1, β = 2$ ). $x (t) = e^{t} [c_{1} (cos 2 t - 2 sin 2 t) + c_{2} (sin 2 t 2 cos 2 t)] .$ Yeh step kyun? $p cos 2 t - q sin 2 t = (c o s 2 t - 2 s i n 2 t)$ aur $p sin 2 t + q cos 2 t = (s i n 2 t 2 c o s 2 t)$ ; saame ka $e^{t}$ dono ko blow up karta hai.

Verify: $tr (A) = 1 + 1 = 2$ eigenvalues ke sum ke barabar hai $(1 + 2 i) + (1 - 2 i) = 2$ ✓ aur $det (A) = 1 + 4 = 5$ product ke barabar hai $(1 + 2 i) (1 - 2 i) = 1 + 4 = 5$ ✓ (trace = sum, det = product of eigenvalues).

Forecast: ek repeated eigenvalue sirf ek eigenvector ke saath. Extra solution mein $t$ ka factor hona chahiye. Kyunki $λ < 0$ , sab kuch phir bhi decay karta hai — lekin ek sheared path par, seedha nahi.

Eigenvalue. $det (A - λ I) = (- 2 - λ)^{2} = 0 \Rightarrow λ = - 2$ (double) → C7. Yeh step kyun? Ek upper-triangular matrix ke eigenvalues uske diagonal par baithe hote hain; yahan dono $- 2$ hain, isliye hume check karna hai ki repeated root ke paas kaafi eigenvectors hain ya nahi.
Eigenvector. $(A + 2 I) = (0010)$ , toh $v_{2} = 0 \Rightarrow v = (0 1)$ . Sirf ek eigenvector → defective. Yeh step kyun? Hum singular system $(A + 2 I) v = 0$ ko uski null direction ke liye solve karte hain; null space ek-dimensional nikalta hai, jo bilkul defective (shear) node ki warning sign hai.
Generalized eigenvector. Solve karo $(A + 2 I) w = v$ : $(0010) w = (0 1) \Rightarrow w_{2} = 1$ , lo $w = (1 0)$ . Yeh step kyun? Parent note ka rule: $x_{2} = e^{λ t} (t v + w)$ bilkul tab kaam karta hai jab $(A - λ I) w = v$ ho. $t$ missing degree of freedom supply karta hai.
General solution + IC. $x = c_{1} e^{- 2 t} (0 1) + c_{2} e^{- 2 t} (t (0 1) + (1 0)) .$ At $t = 0$ : $c_{1} (0 1) + c_{2} (1 0) = (1 1) \Rightarrow c_{1} = 1, c_{2} = 1$ . $x (t) = e^{- 2 t} (1 1 + t) .$

Verify: $x (0) = (1 1)$ ✓. Plug back karo: $x^{'} = e^{- 2 t} (- 2 - 2 ( 1 + t ) + 1) = e^{- 2 t} (- 2 - 1 - 2 t)$ , aur $A x = e^{- 2 t} (- 2 - 2 ( 1 + t ) + 1) = e^{- 2 t} (- 2 - 1 - 2 t)$ ✓.

Forecast: yahan $λ = - 2$ phir se double hai, lekin matrix identity ka pure scalar multiple hai. Har direction ko same factor se stretch kiya jaata hai, isliye har vector ek eigenvector hai — trajectories bilkul seedhi rays honi chahiye origin mein, curved nahi. Ise C7 ke shear se compare karo.

Eigenvalue. $det (A - λ I) = (- 2 - λ)^{2} = 0 \Rightarrow λ = - 2$ (double) → repeated, C7 jaisa. Yeh step kyun? Repeated root pehle dhundhna zaroori hai, uske baad count karo ki kitne eigenvectors hain — count hi C7 aur C7★ ko alag karta hai.
Eigenvectors count karo. $(A + 2 I) = (0000)$ — zero matrix. Har nonzero vector $(A + 2 I) v = 0$ satisfy karta hai, toh hamare paas do independent eigenvectors hain (e.g. $(0 1), (1 0)$ ) → not defective → C7★. ✓ Yeh step kyun? Independent eigenvectors ki sankhya bilkul $A - λ I$ ke null space ki dimension ke barabar hai; yahan woh null space poora plane hai, isliye koi generalized eigenvector ki zaroorat nahi.
Solution + IC. Do eigenvectors ke saath koi $t$ -factor ki zaroorat nahi; general solution hai $x = e^{- 2 t} (c_{1} (0 1) + c_{2} (1 0))$ . IC deta hai $c_{1} = 3, c_{2} = 1$ : $x (t) = e^{- 2 t} (1 3) .$ Yeh step kyun? Kyunki $A = - 2 I$ , poora system sirf $x^{'} = - 2 x$ hai — har component same rate se decay karta hai, toh direction kabhi nahi badlti: ek seedhi ray.

Verify: $x (0) = (1 3)$ ✓. $x^{'} = - 2 e^{- 2 t} (1 3) = A x$ ✓. Direction $(1 3)$ har $t$ ke liye constant hai (seedhi ray) ✓.

Forecast: agar $det A = 0$ toh ek eigenvalue $0$ hai. $λ = 0$ wali eigendirection ka matlab hai $A v = 0$ — wahan ke points kabhi nahi hilte. Toh sirf origin nahi, equilibria ki poori line hai.

Eigenvalues. $det (A - λ I) = (1 - λ) (2 - λ) - 2 = λ^{2} - 3 λ = λ (λ - 3) = 0 \Rightarrow λ = 0, 3$ . Ek zero eigenvalue → C8. Yeh step kyun? $det A = 0$ turant warn karta hai ki $A$ singular hai, isliye $λ = 0$ forced hai.
Eigenvectors. For $λ = 0$ : $A v = 0 \Rightarrow v_{1} + v_{2} = 0 \Rightarrow v_{0} = (- 1 1)$ — rest points ki ek line. For $λ = 3$ : $(- 2 2 1 - 1)$ , $2 v_{1} = v_{2} \Rightarrow v_{3} = (2 1)$ . Yeh step kyun? Har eigenvalue ki null direction usual tarike se dhundhi jaati hai; $λ = 0$ direction special hai kyunki wahan motion frozen hai ( $A v_{0} = 0$ ).
Solution + IC. $x = c_{1} (- 1 1) + c_{2} e^{3 t} (2 1)$ ( $λ = 0$ term mein $e^{0 t} = 1$ hai, ek constant part). At $t = 0$ : $c_{1} (- 1 1) + c_{2} (2 1) = (0 3)$ . Row-solve: $c_{1} + c_{2} = 3, - c_{1} + 2 c_{2} = 0 \Rightarrow c_{2} = 1, c_{1} = 2$ . $x (t) = 2 (- 1 1) + e^{3 t} (2 1) .$ Yeh step kyun? Constant part zero eigenvalue ka frozen contribution hai; sirf $λ = 3$ wala part evolve karta hai.

Verify: $x (0) = 2 (- 1 1) + (2 1) = (0 3)$ ✓. Aur $A (- 1 1) = (2 - 2 1 - 1) = (0 0) = 0 \cdot (- 1 1)$ ✓ (frozen direction).

Forecast: physically, salt redistribute hogi aur (kyunki paani baah raha hai) zero ki taraf decay hogi — umeed hai dono eigenvalues negative honge → stable node (C2).

Matrix likho. $A = (- 2 2 1 - 3)$ . Yeh step kyun? Do rate equations se coupling coefficients padhna words ko $x^{'} = A x$ mein badal deta hai.
Eigenvalues. $det (A - λ I) = (- 2 - λ) (- 3 - λ) - 2 = λ^{2} + 5 λ + 4 = (λ + 1) (λ + 4) = 0 \Rightarrow λ = - 1, - 4.$ Dono negative → C2, stable. ✓ physical sense. Yeh step kyun? Eigenvalues ke signs decide karte hain ki salt decay hogi ya badhegi; dono negative draining tanks ki physical expectation confirm karte hain.
Eigenvectors. For $λ = - 1$ : $(- 1 2 1 - 2)$ , $v_{1} = v_{2} \Rightarrow (1 1)$ . For $λ = - 4$ : $(2211)$ , $2 v_{1} = - v_{2} \Rightarrow (- 2 1)$ . Yeh step kyun? Har eigenvalue ke liye standard null-direction solve, do decay modes deta hai.
Solution + IC. $x = c_{1} e^{- t} (1 1) + c_{2} e^{- 4 t} (- 2 1)$ . At $t = 0$ : $c_{1} + c_{2} = 10, c_{1} - 2 c_{2} = 0 \Rightarrow c_{1} = \frac{20}{3}, c_{2} = \frac{10}{3}$ . $x (t) = \frac{20}{3} e^{- t} (1 1) + \frac{10}{3} e^{- 4 t} (- 2 1) .$ Yeh step kyun? Superposition + do initial salt amounts constants fix karte hain.

Verify (units + physics): $x (0) = \frac{20}{3} (1 1) + \frac{10}{3} (- 2 1) = (0 10)$ kg ✓. Jab $t \to \infty$ , dono $\to 0$ kg (tanks drain) ✓. Har $c_{i} e^{λ t} v$ kg ke units carry karta hai ✓.

Forecast: exponents eigenvalues hain, constant vectors eigenvectors hain. Hum diagonalization $A = P D P^{- 1}$ reverse karte hain — dekho Diagonalization.

Eigenpairs padhte hain. $λ_{1} = 2, v_{1} = (1 1)$ ; $λ_{2} = - 1, v_{2} = (- 1 1)$ . Yeh step kyun? Har basic solution $e^{λ t} v$ form mein hai, toh data seedha $(λ, v)$ hamare haath mein deta hai.
$P$ aur $D$ assemble karo. $P = (11 1 - 1)$ , $D = (20 0 - 1)$ . Yeh step kyun? Eigenvectors ko $P$ ke columns ke roop mein rakho aur eigenvalues ko $D$ ke diagonal par; yahi bilkul $A P = P D$ ki recipe hai.
$P$ invert karo. $det P = - 2$ , toh $P^{- 1} = \frac{1}{- 2} (- 1 - 1 - 1 1) = (1/2 1/2 1/2 - 1/2)$ . Yeh step kyun? $A = P D P^{- 1}$ ko basis ke change ko "undo" karne ke liye inverse chahiye.
Multiply karo. $A = P D P^{- 1} = (11 1 - 1) (20 0 - 1) (1/2 1/2 1/2 - 1/2) = (\frac{1}{2} \frac{3}{2} \frac{3}{2} \frac{1}{2}) .$ $A = (1/2 3/2 3/2 1/2) .$

Verify: $A (1 1) = (3/2 + 1/2 1/2 + 3/2) = (2 2) = 2 (1 1)$ ✓ aur $A (- 1 1) = (3/2 - 1/2 1/2 - 3/2) = (1 - 1) = - 1 \cdot (- 1 1)$ ✓.

Recall Self-test: har system ko uske cell se match karo

Eigenvalues $- 3, + 3$ → kaun sa cell? ::: C1 (saddle) Eigenvalues $- 1 \pm 4 i$ → kaun sa cell? ::: C5 (stable spiral, $α < 0$ ) Eigenvalues $\pm 5 i$ → kaun sa cell? ::: C4 (center) Eigenvalue $4$ repeated with ek eigenvector → kaun sa cell? ::: C7 (defective / improper node) $A = - 3 I$ (eigenvalue $- 3$ repeated, do eigenvectors) → kaun sa cell? ::: C7★ (star / proper node, stable) Eigenvalues $0$ aur $2$ → kaun sa cell? ::: C8 (line of equilibria, degenerate)

Related building blocks: Matrix Exponential yeh sab $x (t) = e^{A t} x_{0}$ mein package karta hai, aur upar ke har 2×2 case Second-Order Linear ODEs characteristic-root story ka ODE-mirror hai.