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] For unique solution, \operatorname{det}(A) \neq 0 \begin{gathered}\text{ Case I } \operatorname{det}( A)=\left|$\begin{array}{lll}1 & a & b \\ a & 1 & c \\ b & c & 1 \end{array}$\right| \\ =1-a²-b²-c²+2 a b c ≠ 0 \end{gathered} Here a, b, c$is selected from 1,0,0$. (No case is possible) Case II (i) \operatorname{det}(A)=\left|$\begin{array}{lll}1 & a & b \\ a & 0 & c \\ b & c & 0\end{array}$\right|=2 a b c-c^2 \neq 0 Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible) (ii) \operatorname{det}(A)=\left|$\begin{array}{lll}0 & a & b \\ a & 1 & c \\ b & c & 0\end{array}$\right|=2 a b c-b^2 \neq 0 Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible) (iii) \operatorname{det}(A)=\left|$\begin{array}{lll}0 & a & b \\ a & 0 & c \\ b & c & 1\end{array}$\right|=2 a b c-a^2 \neq 0 Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible) Hence, there are exactly 6 matrices for unique solution. Hence, option (b) is correct.