Let $M$ be a $2 \times 2$ symmetric matrix with integer entries. Then $M$ is invertible if

Let M be a 2 × 2 symmetric matrix with integer entries. Then M is invertible if $M = \begin{pmatrix} a & b \\ b & d \end{pmatrix},\qquad a,b,d \in \mathbb{Z}$ $\text{To find: which conditions imply } \det(M) \neq 0$ $\text{Plan: } \det(M) = ad - b^2,\quad \text{test each condition} \Rightarrow \det \neq 0?$ $\det(M) = \begin{vmatrix} a & b \\ b & d \end{vmatrix} = ad - b^2$ $(1)\ (a,b) = (b,d)\ \text{(transpose of 2nd row)} \Rightarrow a=b=d$ $\det = a\cdot a - a^2 = 0 \Rightarrow (1)\ \text{FALSE}$ $(2)\ (b,d) = (a,b)\ \text{(transpose of 1st column)} \Rightarrow a=b=d$ $\det = 0 \Rightarrow (2)\ \text{FALSE}$ $(3)\ b=0,\ a\neq 0,\ d\neq 0 \Rightarrow \det = ad \neq 0$ $\Rightarrow (3)\ \text{TRUE}$ $(4)\ ad \neq \text{(perfect square)}$ $\det = 0 \Rightarrow ad = b^2 \text{ — a perfect square, contradiction}$ $\therefore \det \neq 0 \Rightarrow (4)\ \text{TRUE}$ $(3)\ \text{and}\ (4)\ \text{imply } \det(M) \neq 0$ $\boxed{\text{Answer: } (3)\ \text{and}\ (4)}$