 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2/8e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

χ_{A} (t) = det (t I - A) = (t - 1)^{3} .

Moreover,

A - I = - 3 3 - 1 - 6 6 - 2 - 9 9 - 3

has rank one, so its kernel has dimension two, while

(A - I)^{2} = 0

and

A \neq = I

m_{A} (t) = (t - 1)^{2} .

There are two Jordan blocks, because the eigenspace has dimension two, and the largest has size two, because the minimal polynomial has exponent two. Thus

JNF (A) = J_{2} (1) \oplus J_{1} (1) .

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