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