 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-1/7a/iv/solution

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

A B = B A = 0 - 10 - 8 102210 - 8 - 10 0 .

Take

D = B

and

C = A

in the preceding result:

B

has the three distinct eigenvalues

2, 4, 6

and the matrices commute. The common eigenbasis from part (i) gives

S = 111 10 - 1 121, S^{- 1} A S = diag (1, 2, 2), S^{- 1} BS = diag (2, 4, 6) .

Indeed

A S = S diag (1, 2, 2)

and

BS = S diag (2, 4, 6)

, and

det S = 2 \neq = 0

. These equalities directly verify the simultaneous diagonalization.

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