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Step-by-Step Solution
Step 1: Represent the matrix A
Let the given symmetric matrix of order 2 be
$A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$,
where $a$, $b$, and $c$ are integers.
Step 2: Compute A²
The product $A^2 = A \times A$ is:
$A^2
= \begin{pmatrix} a & b \\ b & c \end{pmatrix}
\begin{pmatrix} a & b \\ b & c \end{pmatrix}
= \begin{pmatrix} a^2 + b^2 & ab + bc \\ ab + bc & b^2 + c^2 \end{pmatrix}$.
Step 3: Identify the given condition
The problem states that the sum of the diagonal elements of $A^2$ is 1. The sum of the diagonal elements (the trace) of $A^2$ is:
$(a^2 + b^2) + (b^2 + c^2) = a^2 + 2b^2 + c^2.$
Therefore, the condition becomes
$a^2 + 2b^2 + c^2 = 1.$
Step 4: Solve the Diophantine equation
We want integer solutions $(a, b, c)$ such that $a^2 + 2b^2 + c^2 = 1.$ Note that $a^2$, $2b^2$, and $c^2$ are all non-negative integers.
Because the total is 1, the possible cases are restricted. If $|b| = 1$, then $2b^2 = 2,$ which already exceeds 1. Hence $b$ must be 0.
With $b = 0,$ the equation reduces to:
$a^2 + c^2 = 1.$
The integer solutions to this equation are:
$a = 1, c = 0$
$a = -1, c = 0$
$a = 0, c = 1$
$a = 0, c = -1$
Step 5: Determine the possible matrices
For each of the above solutions for $(a, c)$ and $b = 0$, the matrix $A$ takes one of four forms:
$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$
$\begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix}$
$\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$
$\begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}$
Step 6: Conclusion
There are exactly 4 such symmetric matrices of order 2 (with integer entries) for which the sum of the diagonal elements of $A^2$ is 1.
Hence, the correct number of such matrices is 4.
