Let $$A = \begin{bmatrix} 1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5 \end{bmatrix}$$. Then the sum of all elements of the matrix $$\text{adj}\left(\text{adj}\left(2(\text{adj}\,A)^{-1}\right)\right)$$ is equal to :

For a square matrix of order $$n=3$$ we use the two standard identities
    (i) $$\det\!\bigl(\operatorname{adj}M\bigr)=\bigl(\det M\bigr)^{\,n-1}=\bigl(\det M\bigr)^2$$
    (ii) $$\operatorname{adj}\!\bigl(\operatorname{adj}M\bigr)=\bigl(\det M\bigr)^{\,n-2}\,M=\det(M)\,M$$.

Given $$A=\begin{bmatrix}1 & 1 & 2\\ -2 & 0 & 1\\ 1 & 3 & 5\end{bmatrix},$$ first find its determinant.

Expanding about the first row:
$$\det A = 1\,(0\cdot5-1\cdot3)\;-\;1\,(-2\cdot5-1\cdot1)\;+\;2\,(-2\cdot3-0\cdot1)$$
$$=1(-3)-1(-11)+2(-6)=-3+11-12=-4.$$

Hence $$\det A=-4.$$ Therefore $$\det\!\bigl(\operatorname{adj}A\bigr)=(\det A)^2=(-4)^2=16.$$ Also, by (ii), $$\operatorname{adj}\!\bigl(\operatorname{adj}A\bigr)=\det(A)\,A=-4\,A.$$

Using $$\operatorname{adj}A=\det(A)\,A^{-1}=-4\,A^{-1},$$
invert both sides to get
$$(\operatorname{adj}A)^{-1}=(-4)^{-1}\,(A^{-1})^{-1}=(-\tfrac14)\,A.$$

Now define $$B=2\,(\operatorname{adj}A)^{-1}.$$
Substituting the expression obtained above,
$$B=2\left(-\tfrac14\,A\right)=-\tfrac12\,A.$$

The required matrix is $$\operatorname{adj}\!\bigl(\operatorname{adj}B\bigr).$$
For a $$3\times3$$ matrix, identity (ii) gives
$$\operatorname{adj}\!\bigl(\operatorname{adj}B\bigr)=\det(B)\,B.$$

Compute $$\det(B).$$ Since $$B=-\tfrac12 A,$$ and scaling a $$3\times3$$ matrix by a factor $$k$$ multiplies its determinant by $$k^3,$$
$$\det(B)=\left(-\tfrac12\right)^3\det A=-\tfrac18\,(-4)=\tfrac12.$$

Next, find the sum of all elements of $$B.$$ Sum of elements of $$A$$: row-wise
$$1+1+2=4,\quad -2+0+1=-1,\quad 1+3+5=9.$$
Total $$\Sigma_A = 4-1+9 = 12.$$ Hence
$$\Sigma_B = -\tfrac12\,\Sigma_A = -\tfrac12\,(12) = -6.$$

Finally, the sum of elements of $$\operatorname{adj}\!\bigl(\operatorname{adj}B\bigr)=\det(B)\,B$$ is
$$\det(B)\;\Sigma_B = \left(\tfrac12\right)(-6) = -3.$$

Option D which is: $$-3$$