 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/17/b/ii/solution

det L = 4 - λ^{2}

, so there is exactly one solution whenever

λ \neq = \pm 2

. A zero determinant does not alone distinguish inconsistency from infinitely many solutions, so check those values separately.

Eliminate

x

and

z

using

x = 1 - λ y

and

z = 1 - 4 y

. The remaining equation is

(4 - λ^{2}) y = 3 - λ .

For

λ = 2

this reads

0 = 1

, and for

λ = - 2

it reads

0 = 5

. Both are inconsistent. Consequently

Solutions exist exactly for λ \in R ∖ {- 2, 2},

and each permitted value has a unique solution.

 New to topics? Read the docs here!