Let A be a symmetric matrix of order 2 with i | vedclass

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(A) Let $A = \begin{bmatrix} a & b \ b & c \end{bmatrix}$,where $a, b, c \in \mathbb{Z}$.Then $A^{2} = \begin{bmatrix} a & b \ b & c \end{bmatrix} \

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$4$

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(A) Let $A = \begin{bmatrix} a & b \ b & c \end{bmatrix}$,where $a, b, c \in \mathbb{Z}$.Then $A^{2} = \begin{bmatrix} a & b \ b & c \end{bmatrix} \begin{bmatrix} a & b \ b & c \end{bmatrix} = \begin{bmatrix} a^{2} + b^{2} & ab + bc \ ab + bc & b^{2} + c^{2} \end{bmatrix}$.The sum of the diagonal elements of $A^{2}$ is given by $	ext{tr}(A^{2}) = a^{2} + b^{2} + b^{2} + c^{2} = a^{2} + 2b^{2} + c^{2}$.We are given that $a^{2} + 2b^{2} + c^{2} = 1$,where $a, b, c \in \mathbb{Z}$.Since $a^{2}, b^{2}, c^{2} \ge 0$,for the sum to be $1$,we must have $b^{2} = 0$,which implies $b = 0$.Then the equation reduces to $a^{2} + c^{2} = 1$.Since $a, c \in \mathbb{Z}$,the possible integer solutions for $(a, c)$ are:$1$. If $a = 0$,then $c^{2} = 1 \Rightarrow c = \pm 1$. This gives two matrices: $\begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \ 0 & -1 \end{bmatrix}$.$2$. If $c = 0$,then $a^{2} = 1 \Rightarrow a = \pm 1$. This gives two matrices: $\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} -1 & 0 \ 0 & 0 \end{bmatrix}$.Thus,the total number of such matrices is $2 + 2 = 4$.

Let $A$ be a symmetric matrix of order $2$ with integer entries. If the sum of the diagonal elements of $A^{2}$ is $1,$ then the possible number of such matrices is

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