Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 1 / 5C / Solution

Expansion of the determinant along the first row givesThus the matrix is invertible exactly when $x\ne 1$. 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. ThereforeTaking cofactors and transposing them gives the adjugate matrixThe adjugate identity $M\mathrm{adj}M=(\det M)I$ then givesIn particular, the singular case has rank two, rather than a further exceptional rank-one value.