If A is a square matrix of order 3 × 3 whose elements are given by $a_{ij}=\frac{(i+2j)}{5}$ which of the following are not correct ?

(A) The value of $a_{12}+a_{23}-a_{32}=\frac{6}{5}$

(B) $\left(\left((A^T)^T\right)^T\right)^T=A^T$

(C) $a_{12}+a_{13}+a_{32}=a_{22}+a_{23}+a_{31}$

(D) Sum of all diagonal elements is $\frac{19}{5}$

(E) A is a symmetric matrix

Choose the correct answer from the options given below :

(B), (D), (E) Only

Given $a_{ij}=\frac{i+2j}{5}$ for $i,j=1,2,3$.

Compute required elements.

$a_{12}=\frac{1+4}{5}=1,\quad a_{23}=\frac{2+6}{5}=\frac{8}{5},\quad a_{32}=\frac{3+4}{5}=\frac{7}{5}$

(A) $a_{12}+a_{23}-a_{32}=1+\frac{8}{5}-\frac{7}{5}=1+\frac{1}{5}=\frac{6}{5}$ → correct.

(B) $((A^{T})^{T})^{T})^{T}=A^{T}$ since $(A^{T})^{T}=A$ → Incorrect.

(C)

$a_{13}=\frac{1+6}{5}=\frac{7}{5},\; a_{22}=\frac{2+4}{5}=\frac{6}{5},\; a_{31}=\frac{3+2}{5}=1$

LHS $=a_{12}+a_{13}+a_{32}=1+\frac{7}{5}+\frac{7}{5}=\frac{19}{5}$

RHS $=a_{22}+a_{23}+a_{31}=\frac{6}{5}+\frac{8}{5}+1=\frac{19}{5}$

So (C) is correct.

(D) Diagonal elements:

$a_{11}=\frac{3}{5},\; a_{22}=\frac{6}{5},\; a_{33}=\frac{9}{5}$

Sum $=\frac{18}{5}\ne\frac{19}{5}$ → incorrect.

(E) $a_{12}=1,\; a_{21}=\frac{4}{5}$ so $a_{12}\ne a_{21}$ → not symmetric → incorrect.

final answer: $\text{(B), (D) and (E)}$