 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-1/8b/a/solution

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

det A = - 2 (μ - 1) (μ + 1) .

Thus for

μ \neq = \pm 1

, the kernel is

{0}

and the inhomogeneous system has exactly one solution.

μ = - 1

, row reduction gives

ker A = span {(1, - 2, 3)^{T}},

and the augmented matrix has the same rank two as

A

, so there are infinitely many solutions. At

μ = 1

ker A = span {(- 1, - 2, 3)^{T}},

but the augmented matrix has rank three while

A

has rank two, so there is no solution. Therefore the number of solutions is

⎩ ⎨ ⎧ 1, \infty, 0, μ \neq = \pm 1, μ = - 1, μ = 1.

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