Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 2 / b

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- Solution b

Solution

Writing the unknown vector as

(x, y, z)^{T}

, the first two linear equations give

x = z - 1

and

y = z - β

. Substitution into the third gives

- (z - 1) - (z - β) + α z = 1, (α - 2) z = - β .

Thus, when

α \neq = 2

, put

t = - β / (α - 2)

. There is exactly one solution:

(x, y, z) = (t - 1, t - β, t), t = - \frac{β}{α - 2} .

When

α = 2

, the remaining equation is

0 = - β

. Hence

α = 2, β \neq = 0 α = 2, β = 0 : \emptyset, : {(- 1, 0, 0) + t (1, 1, 1) : t \in R} .

The second case is an affine line in a vector space, so it contains infinitely many solutions. These cases exhaust all real numbers

α, β

.

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