Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-2/8b/ii/solution

The three linearly independent eigenvectors still form a basis, so the same coefficient calculation applies to this diagonalizable matrix matrix. The general solution is
A repeated eigenvalue alone does not create a polynomial factor in time; that factor is needed only when a nontrivial Jordan block prevents a full eigenvector basis.

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