Paragraph:
Let \(A\) be the set of all \(3 \times 3\) symmetric matrices all of whose entries are either 0 or 1 . Five of these entries are 1 and four of them are 0 .Question:
The number of matrices \(A\) in \(A\) for which the system of linear equations \(A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\) has a unique solution, is

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, \(\operatorname{det}(A) \neq 0\)
\[
\begin{gathered}
\text { Case I } \operatorname{det}(A)=\left|\begin{array}{lll}
1 & a & b \\
a & 1 & c \\
b & c & 1
\end{array}\right| \\
=1-a^2-b^2-c^2+2 a b c \neq 0
\end{gathered}
\]
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.