Invertibility Conditions for Symmetric Integer Matrix

Let $M$ be a $2 \times 2$ symmetric matrix with integer entries. Therefore, $M$ can be represented as:

$$M = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$$

where $a, b, c \in \mathbb{Z}$.

A matrix $M$ is invertible if and only if its determinant is non-zero, i.e., $\det(M) \neq 0$. The determinant of $M$ is given by: $\det(M) = ac - b^2$

Now, we analyze each option:

1. Analysis of Option (A): The first column of $M$ is $\begin{bmatrix} a \\ b \end{bmatrix}$ and the second row of $M$ is $\begin{bmatrix} b & c \end{bmatrix}$. The transpose of the second row is $\begin{bmatrix} b \\ c \end{bmatrix}$. Equating the first column to the transpose of the second row gives: $a = b \quad \text{and} \quad b = c \implies a = b = c$ Substituting this into the determinant expression: $\det(M) = a \cdot a - a^2 = 0$ Since $\det(M) = 0$, $M$ is not invertible. Thus, Option (A) is incorrect.

2. Analysis of Option (B): The second row of $M$ is $\begin{bmatrix} b & c \end{bmatrix}$ and the transpose of the first column is $\begin{bmatrix} a & b \end{bmatrix}$. Equating the second row to the transpose of the first column gives: $b = a \quad \text{and} \quad c = b \implies a = b = c$ As calculated in Option (A), this yields $\det(M) = 0$, meaning $M$ is not invertible. Thus, Option (B) is incorrect.

3. Analysis of Option (C): If $M$ is a diagonal matrix, then the off-diagonal entries are zero, so $b = 0$. We are given that the main diagonal entries $a$ and $c$ are non-zero ($a \neq 0, c \neq 0$). The determinant becomes: $\det(M) = ac - 0^2 = ac \neq 0$ Since $\det(M) \neq 0$, $M$ is invertible. Thus, Option (C) is correct.

4. Analysis of Option (D): The product of the main diagonal entries is $ac$. We are given that $ac$ is not the square of any integer, i.e., $ac \neq k^2$ for any $k \in \mathbb{Z}$. Since $b$ is an integer, $b^2$ is always the square of an integer. Because $ac$ is not an integer square, it can never be equal to $b^2$: $ac \neq b^2 \implies ac - b^2 \neq 0 \implies \det(M) \neq 0$ Since $\det(M) \neq 0$, $M$ is invertible. Thus, Option (D) is correct.

Conclusion: The conditions under which $M$ is invertible are given by options (C) and (D).