 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013-resit/5/solution

Subtracting the first equation from the third gives

4 x - 4 z = 6

, so

z = x - 3/2

. The second equation then becomes

(α - 2) x = β - 3

. The third fixes

y = 7 (1 - x) /5

. Therefore the three possibilities are

α \neq = 2 α = 2, β = 3 α = 2, β \neq = 3 one solution, x = \frac{β - 3}{α - 2}, y = \frac{7}{5} (1 - x), z = x - \frac{3}{2} infinitely many, (x, y, z) = (t, \frac{7}{5} (1 - t), t - \frac{3}{2}) no solution .

20 (α - 2) \neq = 0

, confirming uniqueness. At

α = 2

its matrix rank is two. The augmented matrix has the same matrix rank only for

β = 3

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