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(B) Given $A = \begin{bmatrix} 1 & 2 & 1 \ 2 & 2 & x \ y & 1 & 2 \end{bmatrix}$.Since $A$ is a symmetric matrix,$A^T = A$.Comparing the elements of

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(B) Given $A = \begin{bmatrix} 1 & 2 & 1 \ 2 & 2 & x \ y & 1 & 2 \end{bmatrix}$.Since $A$ is a symmetric matrix,$A^T = A$.Comparing the elements of $A$ and $A^T$:$\begin{bmatrix} 1 & 2 & y \ 2 & 2 & 1 \ 1 & x & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 1 \ 2 & 2 & x \ y & 1 & 2 \end{bmatrix}$.By comparing corresponding elements,we get $y = 1$ and $x = 1$.Now,check if $A$ is singular for these values,i.e.,$|A| = 0$.$|A| = \begin{vmatrix} 1 & 2 & 1 \ 2 & 2 & 1 \ 1 & 1 & 2 \end{vmatrix} = 1(4 - 1) - 2(4 - 1) + 1(2 - 2) = 1(3) - 2(3) + 0 = 3 - 6 = -3$.Since $|A| = -3 
eq 0$,there are no values of $(x, y)$ that satisfy both conditions simultaneously.Therefore,the number of such ordered pairs is $0$.

The number of ordered pairs $(x, y)$ for which $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2 \end{bmatrix}$ is a singular and symmetric matrix is

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