If $A = [a_{ij}]$ be square matrix of order 3, such that $a_{ij} = i + j$, $\forall i, j$ then which of the following are correct?

(A) A is a skew-symmetric matrix.

(B) A is a non-singular matrix.

(C) The inverse of A does not exist.

(D) A is a symmetric matrix.

Choose the correct answer from the options given below:

Given $a_{ij} = i + j$ for a 3×3 matrix.

Each element is obtained by adding its row number and column number:

$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$

$A = \begin{bmatrix} 1+1 & 1+2 & 1+3 \\ 2+1 & 2+2 & 2+3 \\ 3+1 & 3+2 & 3+3 \end{bmatrix}$

$A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}$

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For statement (A), a matrix is skew-symmetric if $a_{ij} = -a_{ji}$.

Here $a_{12} = 3$ and $a_{21} = 3$, so $a_{12} \neq -a_{21}$.

Therefore, statement (A) is false.

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For statement (D), a matrix is symmetric if $a_{ij} = a_{ji}$.

Since $a_{ij} = i + j$ and $a_{ji} = j + i$, and addition is commutative:

$a_{ij} = i + j = j + i = a_{ji}$

Therefore, statement (D) is true.

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For statement (B), a matrix is non-singular if its determinant is non-zero.

$\det(A) = 2\begin{vmatrix} 4 & 5 \\ 5 & 6 \end{vmatrix} - 3\begin{vmatrix} 3 & 5 \\ 4 & 6 \end{vmatrix} + 4\begin{vmatrix} 3 & 4 \\ 4 & 5 \end{vmatrix}$

$\det(A) = 2(4 \times 6 - 5 \times 5) - 3(3 \times 6 - 5 \times 4) + 4(3 \times 5 - 4 \times 4)$

$\det(A) = 2(24 - 25) - 3(18 - 20) + 4(15 - 16)$

$\det(A) = 2(-1) - 3(-2) + 4(-1)$

$\det(A) = -2 + 6 - 4$

$\det(A) = 0$

Since $\det(A) = 0$, the matrix is singular.

Therefore, statement (B) is false.

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For statement (C), a matrix has an inverse only if its determinant is non-zero.

Since $\det(A) = 0$, the inverse does not exist.

Therefore, statement (C) is true.

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Statements (C) and (D) are correct.