 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-1/1f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

(t - 1) (t - 2)^{2}

. Solving the two eigenvalue equations gives

E_{1} = span {(3, - 1, 3)^{T}}, E_{2} = span {(2, 1, 0)^{T}, (2, 0, 1)^{T}} .

For example, multiplying the matrix by these three vectors gives respectively

1

2

and

2

times the original vector. The eigenvectors form a basis: the matrix with these columns has nonzero determinant

- 1

. Thus an eigenbasis and its diagonal matrix are

B = ((3, - 1, 3)^{T}, (2, 1, 0)^{T}, (2, 0, 1)^{T}), [α]_{B} = diag (1, 2, 2) .

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