 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/8e/b/solution

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

A = 111 11 - 1 1 - 1 2,

because the off-diagonal entries contribute twice to

x^{T} A x

u_{1} = e_{1}, u_{2} = e_{3} - e_{1}, u_{3} = e_{2} - e_{1} + 2 u_{2} = - 3 e_{1} + e_{2} + 2 e_{3} .

These vectors are pairwise orthogonal for

B (u, v) = u^{T} A v

, and

B (u_{1}, u_{1}) = 1, B (u_{2}, u_{2}) = 1, B (u_{3}, u_{3}) = - 4.

Thus the form is diagonal in the basis

(u_{1}, u_{2}, u_{3})

, with matrix

diag (1, 1, - 4) .

3

and its signature is

2 - 1 = 1 .

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