Paragraph: Let $A$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1 …
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 $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ is is inconsistent, is
0
more than 2
2
1
Solution
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 $1,0,0$.
$\Rightarrow x+a y+b z=1 \Rightarrow a x+y+c z=0$ $b x+c y+z=0$
(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}
\Rightarrow x+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