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Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 1F / a / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2016 ib Paper 1 1F a
Created 2026-09-24 Updated 2026-10-06  0 By others on same topic  0 Discussions Create my own version
The characteristic polynomial of the given matrix is (t−1)(t−2)2. Solving the two eigenvalue equations gives
E1​=span{(3,−1,3)T},E2​=span{(2,1,0)T,(2,0,1)T}.
(1)
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).​
(2)

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