 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ib/paper-1/5c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Expansion of the determinant along the first row gives

det M = 2 x^{2} - (3 - x) + 2 x (1 - x) = 3 (x - 1) .

Thus the matrix is invertible exactly when $x \neq = 1$ . The minor formed from the first two rows and the last two columns is

- 1

, independently of

x

, so the matrix rank is always at least two. Therefore

rank M = {3, 2, x \neq = 1, x = 1.

Taking cofactors and transposing them gives the adjugate matrix

adj M = 2 x x - 3 2 x (1 - x) 2 x - 1 x - 2 2 (1 - x^{2}) - 1 1 x - 1 .

M adj M = (det M) I

then gives

M^{- 1} = \frac{1}{3 ( x - 1 )} adj M (x \neq = 1) .

In particular, the singular case has rank two, rather than a further exceptional rank-one value.

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