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The number of ordered pairs $(\mathrm{x}, \mathrm{y})$ for which $\mathrm{A}=$ $\left(\begin{array}{lll}1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2\end{array}\right)$ is a singular and symmetric matrix is
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Verified Answer
The correct answer is:
$0$
Given $A=\left[\begin{array}{lll}1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2\end{array}\right]$
Since $A$ is singular i.e. $|A|=0$
$$
\begin{aligned}
\Rightarrow & \left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 2 & x \\
y & 1 & 2
\end{array}\right|=0 \\
\Rightarrow & 1(4-x)-2(4-x y)+1(2-2 y)=0 \\
\Rightarrow & 4-x-8+2 x y+2-2 y=0
\end{aligned}
$$
$\Rightarrow \quad-2-x-2 y+2 x y=0$ ...(i)
$$
\Rightarrow \quad-2-x-2 y+2 x y=0
$$
Since $A$ is symmetric
$$
\begin{aligned}
& \Rightarrow A^T=A \\
& \Rightarrow\left[\begin{array}{lll}
1 & 2 & \mathrm{y} \\
2 & 2 & 1 \\
1 & \mathrm{x} & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 2 & \mathrm{x} \\
\mathrm{y} & 1 & 2
\end{array}\right]
\end{aligned}
$$
Comparing above, we get
$y=1, x=1$ ...(i)
Since $A$ is singular i.e. $|A|=0$
$$
\begin{aligned}
\Rightarrow & \left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 2 & x \\
y & 1 & 2
\end{array}\right|=0 \\
\Rightarrow & 1(4-x)-2(4-x y)+1(2-2 y)=0 \\
\Rightarrow & 4-x-8+2 x y+2-2 y=0
\end{aligned}
$$
$\Rightarrow \quad-2-x-2 y+2 x y=0$ ...(i)
$$
\Rightarrow \quad-2-x-2 y+2 x y=0
$$
Since $A$ is symmetric
$$
\begin{aligned}
& \Rightarrow A^T=A \\
& \Rightarrow\left[\begin{array}{lll}
1 & 2 & \mathrm{y} \\
2 & 2 & 1 \\
1 & \mathrm{x} & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 2 & \mathrm{x} \\
\mathrm{y} & 1 & 2
\end{array}\right]
\end{aligned}
$$
Comparing above, we get
$y=1, x=1$ ...(i)
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