Number of 3 × 3 symmetric matrices which can be formed by three '0', three '1' & three '-1' only, is
Step-by-Step Solution
Key Concept: A 3x3 symmetric matrix has 6 independent entries (a11, a22, a33, a12, a13, a23). We need to distribute three 0s, three 1s, and three -1s among these 6 positions such that the total count of each digit is satisfied.
A 3x3 symmetric matrix A = [aij] has a12=a21, a13=a31, a23=a32. The independent entries are a11, a22, a33, a12, a13, a23. Let these be x1, x2, x3, x4, x5, x6. We have three 0s, three 1s, and three -1s. Let n0, n1, n-1 be the number of 0s, 1s, and -1s in the diagonal (x1, x2, x3). Let m0, m1, m-1 be the number of 0s, 1s, and -1s in the off-diagonal (x4, x5, x6). Then n0+m0=3, n1+m1=3, n-1+m-1=3. Also, the number of ways to arrange the diagonal is 3!/(n0!n1!n-1!) and the off-diagonal is 3!/(m0!m1!m-1!). Summing over all valid partitions gives 36.
Correct Answer: 36