In C++ (2D array filled by column-major style indexing), what sequence does the program output? #include<iostream.h> class CuriousTabArray { int Matrix[3][3]; public: CuriousTabArray() { for(int i = 0; i < 3; i++) for(int j = 0; j < 3; j++) Matrix[j][i] = i + j; } void Display(void) { for(int i = 0; i < 3; i++) for(int j = 0; j < 3; j++) cout << Matrix[j][i] << " "; } }; int main() { CuriousTabArray objCuriousTab; objCuriousTab.Display(); return 0; }

Introduction / Context:
The code writes and then reads a 3x3 matrix using the same Matrix[j][i] indexing order inside nested loops. Although C++ stores arrays in row-major order, the explicit indices determine which element is updated. Because both the constructor and the display use the same indexing, values are read exactly as written.

Given Data / Assumptions:

• Initialization: Matrix[j][i] = i + j for i, j = 0..2.
• Display: prints Matrix[j][i] in the same i-outer, j-inner order.

Concept / Approach:
Compute each entry: For i=0, the row of outputs is 0,1,2; for i=1, outputs are 1,2,3; for i=2, outputs are 2,3,4. Since display mirrors initialization order and indexing, no uninitialized values are printed.

Step-by-Step Solution:

Verification / Alternative check:
Swapping indices in Display to Matrix[i][j] would produce a different sequence because the matrix is not symmetric in general for arbitrary formulas.

Why Other Options Are Wrong:

• 4 3 2 ...: Reversed order, not what the loops produce.
• Garbage / Compilation error: All elements are assigned; headers and syntax are fine.

Common Pitfalls:
Confusing row-major storage with the iteration order used for printing; storage order does not affect correctness of this deterministic computation.

Final Answer:
The program will display the output 0 1 2 1 2 3 2 3 4.