Analysis: Because the element cij=ai in row I and column J of the matrix? aj+ai+aj =(2i- 1)(2j- 1)+2i- 1+2j- 1 = 2i+j- 1(I = 1，2，…，7； J = 1, 2, …, 12), so aij=amn(i, m = 1, 2, …, 7; j，n= 1，2，…， 12)。

Then satisfy 2i+j- 1=2m+n- 1, and i+j=m+n can be obtained from the monotonicity of the exponential function: when i+j≠m+n, aij≠amn, so the different values that this matrix element can get are all for I+J.

Solution: the element cij=ai in row I and column J of the matrix? aj+ai+aj =(2i- 1)(2j- 1)+2i- 1+2j- 1 = 2i+j- 1(I = 1，2，…，7； j= 1，2，…， 12)，

If and only if: i+j=m+n, aij=amn(i, M = 1, 2, …, 7; j，n= 1，2，…， 12)，

So the different values that matrix elements can get are the different sums of i+j, and the sums are 2, 3, …, 19, *** 18.

So choose a.

Please adopt O(∩_∩)O~ Oh if you are satisfied!