Paragraph: Let A be the set of all 3 × 3 symmetric matrices all of...

Given, $A$\left[$$\begin{array}{l}x \\ y \\ z\end{array}$$\right]=\left[$\begin{array}{l}1 \\ 0 \\ 0\end{array}$\right] Case I \left[$\begin{array}{lll}1 & a & b \\ a & 1 & c \\ b & c & 1\end{array}$\right]\left[$\begin{array}{l}x \\ y \\ z\end{array}$\right]=\left[$\begin{array}{l}1 \\ 0 \\ 0\end{array}$\right]a, b, c are selected from $1,0,0$. $\Rightarrow x+a y+b z=1 \Rightarrow a x+y+c z=0b x+c y+z=0$ (i)$If a=1, b=c=0$, then $x+y=1$ Inconsistent system of equation $x+y=0$ (ii)$If a=0=c, b=1$, then $x+z=1$, $y=0$ Inconsistent system of equation x+z=0 (iii)$If c=1, a=b=0$, then $x=1, z=0$, $y=0$ Case II (i) \left[$\begin{array}{lll}1 & a & b \\ a & 0 & c \\ b & c & 0\end{array}$\right]\left[$\begin{array}{l}x \\ y \\ z\end{array}$\right]=\left[$\begin{array}{l}1 \\ 0 \\ 0\end{array}$\right]a, b, c are selected from $1,1,0$. $$\begin{aligned} \Rightarrowx+a y+b z & =1 \Rightarrow a x+c z=0 \\ b x+c y & =0 \end{aligned}$$Clearly, in all three cases, solutions are possible, so system is consistent. \begin{gathered}\text{ (ii) }\left[$\begin{array}{lll}0 & a & b \\ a & 1 & c \\ b & c & 0 \end{array}$\right]\left[$\begin{array}{l}x \\ y \\ z \end{array}$\right]=\left[$\begin{array}{l}1 \\ 0 \\ 0 \end{array}$\right] \\ \Rightarrow a y+b z=1 \Rightarrow a x+y+c z=0 \\ b x+c y=0 \end{gathered} Clearly,b=0, a=c=1$gives y=1 ; x+y+z=0 Inconsistent system$y=0$ More than 2 matrices are possible. Hence, option (b) is correct.