Question

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 :

• A. $6$
• B. $1$
• C. $4$
• D. $12$

Answer 1

The correct option is C

$4$

Finding the possible number of matrices:

Given $A$ be a symmetric matrix of order $2$ i.e. $A=A^T$

$\left[\begin{array}{cc}a& b\\ b& c\end{array}\right]$

Finding $A^2$

$$\begin{gathered} A^2=\left[\begin{array}{cc}a& b\\ b& c\end{array}\right]\left[\begin{array}{cc}a& b\\ b& c\end{array}\right] \\ =\left[\begin{array}{cc}{a}^2+b^2& ab+bc\\ ba+cb& b^2+c^2\end{array}\right] \end{gathered}$$

Sum of the diagonal is$a^2+b^2+b^2+c^2=a^2+2b^2+c^2$

$a^2+2b^2+c^2=1$ [given sum of diagonal elements of $A^2$ is $1$]

$\Rightarrow$ $a^2+c^2=1\left[{\text{since the coefficient of b=2, it makes value of A}}^2\text{more than 1, hence}b=0\right]$

Since the coefficient of $\mathrm{b}=2$, it makes value of more than $1,$ hence $\mathrm{b}=0$

$\Rightarrow a=0\text{and}c=\pm 1\text{or}c=0\text{and}a=\pm 1$

The possibilities are

$$\begin{gathered} a=0,b=0,c=1 \\ a=0,b=0,c=-1 \\ a=1,b=0,c=0 \\ a=-1,b=0,c=0 \end{gathered}$$

Hence, the possible number of such matrices is $4$.

Hence, option (C) is the correct answer