Check if a Matrix is Symmetric

Symmetry compares swapped indices, not reversed rows

A matrix is symmetric when reflecting it across its main diagonal leaves every value unchanged. For each row index i and column index j, the required condition is A[i][j] == A[j][i]. Matrix A is square because it has n = 3 rows and 3 columns, so every swapped position exists inside the matrix.

The rows of A are [1, 2, 1], [2, 4, 2], and [5, 6, 5]. Each row reads the same from left to right and right to left, but that only compares cells within one row. Matrix symmetry compares a cell in one row with a cell in a different row after swapping its two indices. Palindromic rows do not prove symmetry.

The first unequal mirrored pair fixes the answer

Start with the cells above the main diagonal and compare each one with its swapped-index partner. The first comparison is A[0][1] = 2 against A[1][0] = 2, so this mirrored pair matches. The next comparison is A[0][2] = 1 against A[2][0] = 5, so the matrix is not symmetric.

Once one mirrored pair has different values, the entire matrix fails the test. The remaining pair A[1][2] = 2 and A[2][1] = 6 also differs, but checking it cannot change the answer from false to true. A symmetric matrix must pass every mirrored-pair comparison.

The upper triangle checks every mirrored pair exactly once

For each row i, start the column index j at i + 1 and continue while j < n. This visits only cells above the main diagonal. Compare A[i][j] with A[j][i], and return false when they differ.

for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
        if (A[i][j] != A[j][i]) {
            return false;
        }
    }
}
return true;

A diagonal cell such as A[1][1] maps to itself, so comparing it with itself cannot reveal a mismatch. Starting j at i + 1 skips these self-comparisons. The lower triangle would only repeat the same pairs in reverse order: after checking A[0][2] against A[2][0], checking A[2][0] against A[0][2] adds no information.

for (j = ___; j < n; j++)

Early return saves comparisons without changing the test

For A, the algorithm returns false after comparing only two off-diagonal pairs. It first sees that 2 equals 2, then sees that 1 does not equal 5, so it stops immediately. A successful check cannot stop early, because it must confirm that every mirrored pair matches.

A square matrix with n rows has n(n - 1) / 2 off-diagonal mirrored pairs in its upper triangle. A full check therefore takes O(n^2) time. The algorithm stores only the loop indices and a few temporary values, so its extra space is O(1).