the connection to the scalar nth-order case

You have met linear equations in two seemingly different guises: a single high-order equation like y'' + p y' + q y = 0, and a first-order system of several equations x' = A x. A natural worry is that you are learning two unrelated theories. You are not. A single nth-order linear equation is exactly a first-order system in disguise, and the matrix machinery contains the single-equation results as a special case.

The translation is mechanical. Given y^(n) + a_(n-1) y^(n-1) + ... + a_1 y' + a_0 y = 0, name the unknowns x_1 = y, x_2 = y', x_3 = y'', up to x_n = y^(n-1) — the function and its derivatives up to order n - 1. Then x_1' = x_2, x_2' = x_3, and so on, each new derivative simply being the next variable, while the last row x_n' = -a_0 x_1 - a_1 x_2 - ... - a_(n-1) x_n encodes the original equation solved for the top derivative. Writing this as x' = A x produces a special matrix A called the companion matrix: 1's on the superdiagonal, the negated coefficients along the bottom row. Now watch the gears mesh. The characteristic polynomial of this companion matrix is identical to the characteristic equation r^n + a_(n-1) r^(n-1) + ... + a_0 = 0 of the original equation — same roots, same multiplicities. Its eigenvalues are the equation's characteristic roots.

Every system result then specializes correctly. A distinct eigenvalue lambda gives a solution e^(lambda t), matching the single-equation rule. A repeated eigenvalue makes the companion matrix defective, and the Jordan structure manufactures the t e^(lambda t), t^2 e^(lambda t), ... that you learned as the repeated-root case. Complex eigenvalues become sines and cosines. So the matrix exponential of the companion matrix repackages exactly the nth-order theory — and conversely, you can always reduce a system to scalar equations, confirming the two pictures are one. The first component of e^(At) x0 even reproduces the scalar solution.