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

This is a companion drill page for the parent matrix-method note. The parent showed you the machinery. Here we make sure you have fired that machinery at every kind of situation it can meet — every sign of eigenvalue, every geometry, the degenerate and the disguised.

Before we start, one honest reminder of the vocabulary, in plain words, so nothing below is a symbol you have not been handed:

Definition The five objects, in plain English

$A$ is a fixed grid of numbers (a matrix) that "mixes" the coordinates of a moving point.
$x (t)$ is the position of that point at time $t$ — a little arrow from the origin.
$x^{'}$ is its velocity arrow: which way and how fast it is drifting right now.
$A x$ is the velocity the matrix assigns to a point sitting at $x$ . So $x^{'} = A x$ reads: "at every point, the arrow of motion is whatever $A$ turns your position into."
$I$ is the identity matrix — the "do-nothing" matrix. In 2-D, $I = (1001)$ ; multiplying any vector by $I$ leaves it unchanged. We write $λ I = (λ 0 0 λ)$ so that $A - λ I$ means "subtract $λ$ from each diagonal entry of $A$ " — that is what appears inside $det (A - λ I)$ . An eigenvector $v$ is a special direction on which $A$ acts as pure stretching: $A v = λ v$ . The stretch factor $λ$ is the eigenvalue. (Prerequisites live in Eigenvalues and Eigenvectors.)

The scenario matrix

Every $2 \times 2$ system falls into exactly one of the cells below. The eigenvalues decide the geometry — that is the whole classification, and it is the subject of Phase Portraits and Stability.

Cell	Eigenvalue situation	Geometry of trajectories	Stability
C1	Real, opposite signs $λ_{1} < 0 < λ_{2}$	Saddle	unstable
C2	Real, both $< 0$	Node, arrows inward	stable
C3	Real, both $> 0$	Node, arrows outward	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, only one eigenvector (defective)	Improper node (shear)	sign of $λ$
C7★	Repeated eigenvalue, two independent eigenvectors	Proper / star node (straight rays)	sign of $λ$
C8	One eigenvalue $= 0$	Line of equilibria (degenerate)	non-isolated
C9	Word problem (mixing tanks) → lands in C2	physical decay	stable
C10	Exam twist: given a solution, reconstruct $A$	reverse-engineering	—

We now hit every cell.

Forecast: eigenvalues will come out one positive, one negative → this is a saddle, so the point should escape to infinity along one direction while being pulled toward the origin along another. Guess the escape direction before reading on.

Characteristic equation. $det (A - λ I) = (1 - λ)^{2} - 9 = 0$ . Why this step? A nonzero eigenvector exists only when $A - λ I$ is singular, i.e. its determinant vanishes. $(1 - λ)^{2} = 9 \Rightarrow 1 - λ = \pm 3 \Rightarrow λ = 4, - 2.$ Opposite signs → saddle, C1. ✓ matches forecast.
Eigenvector for $λ = 4$ . $(A - 4 I) = (- 3 3 3 - 3)$ , so $- 3 v_{1} + 3 v_{2} = 0 \Rightarrow v_{1} = v_{2}$ , giving $v_{+} = (1 1)$ . Why this step? We solve the singular system for its null direction — the line $A$ merely stretches.
Eigenvector for $λ = - 2$ . $(A + 2 I) = (3333)$ , so $v_{1} = - v_{2}$ , giving $v_{-} = (- 1 1)$ . Why this step? Same null-direction hunt, now for the second eigenvalue — this gives the other special line, the one along which motion decays.
General solution + IC. $x (t) = c_{1} e^{4 t} (1 1) + c_{2} e^{- 2 t} (- 1 1) .$ At $t = 0$ : $c_{1} + c_{2} = 1, c_{1} - c_{2} = 0 \Rightarrow c_{1} = c_{2} = \frac{1}{2}$ . Why this step? Superposition (system is linear) gives the two-parameter family; the two initial numbers pin the constants. $x (t) = \frac{1}{2} e^{4 t} (1 1) + \frac{1}{2} e^{- 2 t} (- 1 1) .$

Look at the figure: the $(1 1)$ direction (red) blows up, the $(- 1 1)$ direction (blue) dies — the classic saddle cross.

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

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

Forecast: if both eigenvalues are negative, everything is dragged to the origin — a stable node. The slower-decaying direction dominates late motion. Guess which one.

Eigenvalues. $det (A - λ I) = (- 3 - λ)^{2} - 1 = 0 \Rightarrow (- 3 - λ) = \pm 1 \Rightarrow λ = - 2, - 4.$ Both negative → C2. ✓ Why this step? Same singularity condition; the signs alone already classify the picture.
Eigenvectors. For $λ = - 2$ : $(- 1 1 1 - 1)$ , $v_{1} = v_{2} \Rightarrow (1 1)$ . For $λ = - 4$ : $(1111)$ , $v_{1} = - v_{2} \Rightarrow (- 1 1)$ . Why this step? Each singular matrix $A - λ I$ is solved for its null direction — the two lines along which decay is pure.
Solution + IC. $x = c_{1} e^{- 2 t} (1 1) + c_{2} e^{- 4 t} (- 1 1)$ . At $t = 0$ : $c_{1} + c_{2} = 2, c_{1} - c_{2} = 4 \Rightarrow c_{1} = 3, c_{2} = - 1$ . $x (t) = 3 e^{- 2 t} (1 1) - e^{- 4 t} (- 1 1) .$ Why this step? $e^{- 4 t}$ vanishes faster, so for large $t$ the trajectory hugs the $(1 1)$ line — the slow eigendirection wins. Forecast confirmed.

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

Forecast: diagonal matrix means the axes are already the eigendirections — no mixing. Both entries positive → outward node. Which axis flees faster?

Eigenvalues/vectors read off directly. A diagonal $A$ has $λ_{1} = 2$ with $e_{1} = (0 1)$ , and $λ_{2} = 5$ with $e_{2} = (1 0)$ . Both $> 0$ → C3. Why this step? On a diagonal matrix each coordinate obeys its own scalar ODE $x_{i}^{'} = λ_{i} x_{i}$ — no coupling to untangle.
Solution. $x_{1} = e^{2 t}$ , $x_{2} = e^{5 t}$ (IC gives $c_{1} = c_{2} = 1$ ): $x (t) = (e ^{5 t} e ^{2 t}) .$ Why this step? Each decoupled scalar equation $x_{i}^{'} = λ_{i} x_{i}$ solves to $x_{i} = x_{i} (0) e^{λ_{i} t}$ . The $y$ -axis (rate 5) outruns the $x$ -axis (rate 2).

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

Forecast: eigenvalues will be pure imaginary ( $α = 0$ ). No growth, no decay → the point should orbit forever on a closed loop. Guess: circle or ellipse?

Eigenvalues. $det (A - λ I) = λ^{2} + 9 = 0 \Rightarrow λ = \pm 3 i$ . So $α = 0, β = 3$ → center, C4. ✓ Why this step? Euler's Formula turns $e^{λ t} = e^{(α + i β) t} = e^{α t} (cos β t + i sin β t)$ ; here $e^{α t} = e^{0} = 1$ , so no amplitude change — pure rotation.
Complex eigenvector for $λ = 3 i$ . $(- 3 i 3 - 3 - 3 i) v = 0 \Rightarrow - 3 i v_{1} - 3 v_{2} = 0 \Rightarrow v_{2} = - i v_{1}$ . Take $v = (- i 1) = p (0 1) + i q (- 1 0)$ . Why this step? One complex branch carries both real solutions; splitting $v = p + i q$ separates them cleanly.
Two real solutions ( $α = 0, β = 3$ ). $x_{1} = p cos 3 t - q sin 3 t = (s i n 3 t c o s 3 t), x_{2} = p sin 3 t + q cos 3 t = (- c o s 3 t s i n 3 t) .$ Why this step? A complex solution $e^{λ t} v$ has a real and an imaginary part, and for a real matrix each part is itself a real solution. Expanding $e^{i β t} (p + i q) = (p cos β t - q sin β t) + i (p sin β t + q cos β t)$ via Euler's Formula and reading off the two brackets is exactly the complex-to-real conversion recipe — that is why $x_{1}, x_{2}$ take this form.
Apply IC. $x = c_{1} (s i n 3 t c o s 3 t) + c_{2} (- c o s 3 t s i n 3 t)$ . At $t = 0$ : $c_{1} (0 1) + c_{2} (- 1 0) = (0 2) \Rightarrow c_{1} = 2, c_{2} = 0$ . $x (t) = (2 sin 3 t 2 cos 3 t) .$ This traces a circle of radius 2 — forecast confirmed.

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 a negative real part → spiral that winds inward to the origin. Guess the winding direction.

Eigenvalues. $det (A - λ I) = (- 1 - λ)^{2} + 4 = 0 \Rightarrow (- 1 - λ)^{2} = - 4 \Rightarrow λ = - 1 \pm 2 i$ . So $α = - 1 < 0, β = 2$ → stable spiral, C5. ✓ Why this step? $α < 0$ makes $e^{α t}$ shrink; $β \neq = 0$ makes it rotate. Shrink + rotate = inward spiral.
Eigenvector for $λ = - 1 + 2 i$ . $(A - λ I) = (- 2 i 2 - 2 - 2 i)$ ; first row: $- 2 i v_{1} - 2 v_{2} = 0 \Rightarrow v_{2} = - i v_{1}$ . So $v = (- i 1) = (0 1) + i (- 1 0)$ , giving $p = (0 1), q = (- 1 0)$ . Why this step? We solve the (complex) singular system for its null direction, then split $v = p + i q$ so the real-part/imag-part recipe can build two real solutions.
Real solutions ( $α = - 1, β = 2$ ). $x_{1} = e^{- t} (p cos 2 t - q sin 2 t) = e^{- t} (s i n 2 t c o s 2 t), x_{2} = e^{- t} (- c o s 2 t s i n 2 t) .$ Why this step? Same real-part/imag-part recipe as C4, but now the $e^{- t}$ envelope shrinks the radius each turn.
Apply IC. $c_{1} (0 1) + c_{2} (- 1 0) = (0 1) \Rightarrow c_{1} = 1, c_{2} = 0$ . $x (t) = e^{- t} (sin 2 t cos 2 t) .$

Verify: $∣ x (t) ∣^{2} = e^{- 2 t} (cos^{2} 2 t + sin^{2} 2 t) = e^{- 2 t} \to 0$ — radius shrinks, confirming inward spiral ✓. $x (0) = (0 1)$ ✓.

Forecast: same as C5 but the real part is now positive → spiral flies outward. No IC needed; we just classify and give the general solution.

Eigenvalues. $det (A - λ I) = (1 - λ)^{2} + 4 = 0 \Rightarrow λ = 1 \pm 2 i$ . $α = 1 > 0, β = 2$ → unstable spiral, C6. ✓ Why this step? $α > 0$ makes $e^{α t}$ grow — the envelope expands as it rotates.
Eigenvector for $λ = 1 + 2 i$ . $(A - λ I) = (- 2 i 1 - 4 - 2 i)$ ; second row: $v_{1} - 2 i v_{2} = 0 \Rightarrow v_{1} = 2 i v_{2}$ . Take $v = (1 2 i) = (1 0) + i (0 2)$ , so $p = (1 0), q = (0 2)$ . Why this step? We solve the singular complex system for its null direction and split $v = p + i q$ , because the real and imaginary parts of the complex branch are precisely what we feed into the two-real-solution recipe.
General real solution ( $α = 1, β = 2$ ). $x (t) = e^{t} [c_{1} (cos 2 t - 2 sin 2 t) + c_{2} (sin 2 t 2 cos 2 t)] .$ Why this step? $p cos 2 t - q sin 2 t = (c o s 2 t - 2 s i n 2 t)$ and $p sin 2 t + q cos 2 t = (s i n 2 t 2 c o s 2 t)$ ; the $e^{t}$ front makes both blow up.

Verify: $tr (A) = 1 + 1 = 2$ equals the sum of eigenvalues $(1 + 2 i) + (1 - 2 i) = 2$ ✓ and $det (A) = 1 + 4 = 5$ equals the product $(1 + 2 i) (1 - 2 i) = 1 + 4 = 5$ ✓ (trace = sum, det = product of eigenvalues).

Forecast: a repeated eigenvalue with only one eigenvector. The extra solution must carry a factor of $t$ . Since $λ < 0$ , everything still decays — but along a sheared path, not straight.

Eigenvalue. $det (A - λ I) = (- 2 - λ)^{2} = 0 \Rightarrow λ = - 2$ (double) → C7. Why this step? An upper-triangular matrix has its eigenvalues sitting on the diagonal; here both are $- 2$ , so we must check whether the repeated root has enough eigenvectors.
Eigenvector. $(A + 2 I) = (0010)$ , so $v_{2} = 0 \Rightarrow v = (0 1)$ . Only one eigenvector → defective. Why this step? We solve the singular system $(A + 2 I) v = 0$ for its null direction; the null space turns out to be one-dimensional, which is exactly the warning sign of a defective (shear) node.
Generalized eigenvector. Solve $(A + 2 I) w = v$ : $(0010) w = (0 1) \Rightarrow w_{2} = 1$ , take $w = (1 0)$ . Why this step? The parent note's rule: $x_{2} = e^{λ t} (t v + w)$ works exactly when $(A - λ I) w = v$ . The $t$ supplies the missing degree of freedom.
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: $x^{'} = e^{- 2 t} (- 2 - 2 ( 1 + t ) + 1) = e^{- 2 t} (- 2 - 1 - 2 t)$ , and $A x = e^{- 2 t} (- 2 - 2 ( 1 + t ) + 1) = e^{- 2 t} (- 2 - 1 - 2 t)$ ✓.

Forecast: here $λ = - 2$ is again doubled, but the matrix is a pure scalar multiple of $I$ . Every direction gets stretched by the same factor, so every vector is an eigenvector — trajectories should be perfectly straight rays into the origin, not curved. Contrast this with C7's shear.

Eigenvalue. $det (A - λ I) = (- 2 - λ)^{2} = 0 \Rightarrow λ = - 2$ (double) → repeated, like C7. Why this step? We must first find the repeated root before we can count how many eigenvectors it carries — the count is what separates C7 from C7★.
Count eigenvectors. $(A + 2 I) = (0000)$ — the zero matrix. Every nonzero vector satisfies $(A + 2 I) v = 0$ , so we have two independent eigenvectors (e.g. $(0 1), (1 0)$ ) → not defective → C7★. ✓ Why this step? The number of independent eigenvectors is exactly the dimension of the null space of $A - λ I$ ; here that null space is all of the plane, so no generalized eigenvector is needed.
Solution + IC. With two eigenvectors no $t$ -factor is needed; the general solution is $x = e^{- 2 t} (c_{1} (0 1) + c_{2} (1 0))$ . IC gives $c_{1} = 3, c_{2} = 1$ : $x (t) = e^{- 2 t} (1 3) .$ Why this step? Because $A = - 2 I$ , the whole system is just $x^{'} = - 2 x$ — every component decays at the same rate, so the direction never changes: a straight ray.

Verify: $x (0) = (1 3)$ ✓. $x^{'} = - 2 e^{- 2 t} (1 3) = A x$ ✓. Direction $(1 3)$ is constant for all $t$ (straight ray) ✓.

Forecast: if $det A = 0$ then one eigenvalue is $0$ . An eigendirection with $λ = 0$ means $A v = 0$ — points there never move. So there is a whole line of equilibria, not just the origin.

Eigenvalues. $det (A - λ I) = (1 - λ) (2 - λ) - 2 = λ^{2} - 3 λ = λ (λ - 3) = 0 \Rightarrow λ = 0, 3$ . A zero eigenvalue → C8. Why this step? $det A = 0$ warns us immediately that $A$ is singular, so $λ = 0$ is forced.
Eigenvectors. For $λ = 0$ : $A v = 0 \Rightarrow v_{1} + v_{2} = 0 \Rightarrow v_{0} = (- 1 1)$ — a line of rest points. For $λ = 3$ : $(- 2 2 1 - 1)$ , $2 v_{1} = v_{2} \Rightarrow v_{3} = (2 1)$ . Why this step? Each eigenvalue's null direction is found the usual way; the $λ = 0$ direction is special because motion there is frozen ( $A v_{0} = 0$ ).
Solution + IC. $x = c_{1} (- 1 1) + c_{2} e^{3 t} (2 1)$ (the $λ = 0$ term has $e^{0 t} = 1$ , a constant part). At $t = 0$ : $c_{1} (- 1 1) + c_{2} (2 1) = (0 3)$ . Row-solve: $c_{1} + c_{2} = 3, - c_{1} + 2 c_{2} = 0 \Rightarrow c_{2} = 1, c_{1} = 2$ . $x (t) = 2 (- 1 1) + e^{3 t} (2 1) .$ Why this step? The constant part is the frozen contribution from the zero eigenvalue; only the $λ = 3$ part evolves.

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

Forecast: physically, salt should redistribute and (since water is flowing out) decay toward zero — expect both eigenvalues negative → stable node (C2).

Write the matrix. $A = (- 2 2 1 - 3)$ . Why this step? Reading the coupling coefficients off the two rate equations turns words into $x^{'} = A x$ .
Eigenvalues. $det (A - λ I) = (- 2 - λ) (- 3 - λ) - 2 = λ^{2} + 5 λ + 4 = (λ + 1) (λ + 4) = 0 \Rightarrow λ = - 1, - 4.$ Both negative → C2, stable. ✓ physical sense. Why this step? The signs of the eigenvalues decide whether salt decays or grows; both negative confirms the physical expectation of draining tanks.
Eigenvectors. For $λ = - 1$ : $(- 1 2 1 - 2)$ , $v_{1} = v_{2} \Rightarrow (1 1)$ . For $λ = - 4$ : $(2211)$ , $2 v_{1} = - v_{2} \Rightarrow (- 2 1)$ . Why this step? Standard null-direction solve for each eigenvalue, giving the two decay modes.
Solution + IC. $x = c_{1} e^{- t} (1 1) + c_{2} e^{- 4 t} (- 2 1)$ . At $t = 0$ : $c_{1} + c_{2} = 10, c_{1} - 2 c_{2} = 0 \Rightarrow c_{1} = \frac{20}{3}, c_{2} = \frac{10}{3}$ . $x (t) = \frac{20}{3} e^{- t} (1 1) + \frac{10}{3} e^{- 4 t} (- 2 1) .$ Why this step? Superposition + the two initial salt amounts fix the mixture.

Verify (units + physics): $x (0) = \frac{20}{3} (1 1) + \frac{10}{3} (- 2 1) = (0 10)$ kg ✓. As $t \to \infty$ , both $\to 0$ kg (tanks drain) ✓. Every $c_{i} e^{λ t} v$ carries units of kg ✓.

Forecast: the exponents are the eigenvalues, the constant vectors are the eigenvectors. We reverse the diagonalization $A = P D P^{- 1}$ — see Diagonalization.

Read off eigenpairs. $λ_{1} = 2, v_{1} = (1 1)$ ; $λ_{2} = - 1, v_{2} = (- 1 1)$ . Why this step? Each basic solution has the form $e^{λ t} v$ , so the data hands us $(λ, v)$ directly.
Assemble $P$ and $D$ . $P = (11 1 - 1)$ , $D = (20 0 - 1)$ . Why this step? Put the eigenvectors as the columns of $P$ and the eigenvalues on the diagonal of $D$ ; this is exactly the recipe $A P = P D$ .
Invert $P$ . $det P = - 2$ , so $P^{- 1} = \frac{1}{- 2} (- 1 - 1 - 1 1) = (1/2 1/2 1/2 - 1/2)$ . Why this step? $A = P D P^{- 1}$ needs the inverse to "undo" the change of basis.
Multiply. $A = P D P^{- 1} = (11 1 - 1) (20 0 - 1) (1/2 1/2 1/2 - 1/2) = (\frac{1}{2} \frac{3}{2} \frac{3}{2} \frac{1}{2}) .$ $A = (1/2 3/2 3/2 1/2) .$

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

Recall Self-test: match each system to its cell

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

Related building blocks: Matrix Exponential packages all of this as $x (t) = e^{A t} x_{0}$ , and every 2×2 case above is the ODE-mirror of the Second-Order Linear ODEs characteristic-root story.