you can solve this question quickly without any matrix multiplication.
Since, given matrix 'A' is a lower triangular matrix.
So, |A| = -6 (multiplication all principal diagonals)
and eigen values of A will be -1 , 2 , 3 (i.e. all principal diagonals)
Now , since , Adj(A) = |A| * A^{-1} and eigen values of A^{-1 }are the reverse of eigen values of A.
So, eigenvalues of A^{-1} will be $\frac{1}{\lambda _{i}}$ for i = 1,2,3...
So, Eigen values of Adj(A) = $\frac{\left | A \right |}{Eigen \;values\; of \; matrix\; A}$
So, here , Eigen values of Adj(A) will be :- $\frac{-6}{-1}$ , $\frac{-6}{2}$ , $\frac{-6}{3}$ . ie. 6,-3,-2.
Since , AX = $\lambda$X and A^{2}X = $\lambda$^{2}X
So, if $\lambda$ is an eigen value of A then $\lambda$^{2} will be an eigen value of A^{2}
So, Eigen values of A^{2} will be :- 1, 4, 9
Now , Eigen values fro matrix 3A^{2 }+ adj (A) will be :-
1) 3*1 + 6 = 9
2) 3*4 + (-3) = 9
3) 3*9 + (-2) = 25
Since , Trace of a matrix = Summation of its eigen values
So here , Trace = 9 + 9 + 25 = 43