If \(R\) is a field and \(A, B \in M_n(R)\) are invertible then the matrix \(A + rB\) is invertible for all but a finite number of \(r\in R\).
Solution
Define \[ p(t)=\det(A+tB)\in R[t]. \]
1. The polynomial \(p(t)\) is nonzero.
Since \(B\) is invertible, \[ A+tB=B(B^{-1}A+tI_n). \] Therefore \[ p(t)=\det(B)\det(B^{-1}A+tI_n). \] The second determinant is a monic polynomial of degree \(n\) in \(t\), so the leading coefficient of \(p(t)\) is the nonzero element \(\det(B)\). Hence \(p\) is not the zero polynomial.
2. There are only finitely many \(r\in R\) for which \(p(r)=0\).
A nonzero polynomial of degree \(n\) over a field has at most \(n\) roots. Apply this to the polynomial \(p\) from 1.
3. The matrix \(A+rB\) is invertible for all but finitely many \(r\in R\).
Over a field, a square matrix is invertible exactly when its determinant is nonzero. Thus \[ A+rB\text{ is singular} \quad\Longleftrightarrow\quad p(r)=0. \] By 2 this happens for only finitely many scalars \(r\).