Let Mn×n be the set of all n-square symmetric matrices and the characteristics polynomial of each A∈Mn×n is...

Question

Answer 1

This question is wrong, it could be a typo mistake. First I will explain why this question is wrong and then I will correct the question and explain the solution. Just so you know, trace of a matrix is the sum of the diagonal elements of a matrix, and also the sum of eigen values of that matrix.

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Here Matrix M is a symmetric matrix with characteristic equation as given in the question. For a characteristic equation we know that coefficient of t^n-1 is the trace of Matrix (check eqn (6) of this link). Here in question the coefficient of t^n-1 is 0, which means trace of M is 0 (Tr(M) = 0).

Now coefficient of t^n-2 can be represented as (Tr(M)² - Tr(M²))/2 = -Tr(M²)/2 (as Tr(M) = 0) (check eqn(17) in this class note. Look for eqn c_{2 = ...}).
Coefficient of t^n-2 given in question is 1, that means Tr(M²) = -2, but it could never be negative for a symmetric matrix. The reason for Tr(M²) to be non-negative can be explained in different ways. One way is as follows:
A symmetric matrix A has all real eigen values, and eigen values of A² is the square of eigen values of A, hence trace of A² is sum of squares of real numbers, which makes it non-negative. Hence such a matrix doesn't exist.

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My guess the correct characteristic equation is: 
tⁿ + a_n-2t^n-2 + a_n-3t^n-3 + ...+a₁t + a₀
Now, this characteristic equation puts the constraint that Tr(M) = 0.

Now jumping into main question. The dimension of the matrix. All square matrices of size n can be considered as a linear space of dimension n², as every element corresponds to a matrix with that element 1 and all others 0. We could also think dimension as the freedom of filling this matrix. We can we fill whole of n² with no restrictions. But in this question we have two restrictions:
1) Matrix should be symmetric.
2) Matrix should have trace 0.
So for symmetric matrix, while filling one side of diagonal (say upper triangle), corresponding elements in the other side is already determined. Which makes it (n² - n)/2. Diagonals can be anything so dimension is updated with +n. But now to have trace 0, last element in the diagonal has to be the -1 times sum of rest of elements in the diagonal, making the updation +(n-1). So the dimension at the end is:
(n²-n)/2   +n-1 = (n-1)(n+2)/2. which is option (c)

Answer 2

by using the rank-nullity theorem, dim M is n^2 and the nullity for the symmetric matrix is n(n+1)/2 so if we mines n(n+1)/2 from n^2 we will have n(n-1)/2 so option a is right.