Sum of Elements of Adjoint Matrix Expression

To find the sum of all elements of the matrix $\text{adj}(\text{adj}(2(\text{adj}A)^{-1}))$, we start by evaluating the properties of the given matrix $A$.

The matrix $A$ is given as: $A = \begin{bmatrix} 1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5 \end{bmatrix}$

Step 1: Compute the determinant of $A$ $|A| = 1(0 - 3) - 1(-10 - 1) + 2(-6 - 0)$ $|A| = -3 + 11 - 12 = -4$

Step 2: Simplify $(\text{adj}A)^{-1}$ We know the identity $A \cdot \text{adj}A = |A| I_3$. Taking inverse on both sides, we get: $(\text{adj}A)^{-1} = \left(|A| A^{-1}\right)^{-1} = \frac{1}{|A|} A$

Substituting $|A| = -4$: $(\text{adj}A)^{-1} = -\frac{1}{4} A$

Step 3: Simplify $2(\text{adj}A)^{-1}$ Let $B = 2(\text{adj}A)^{-1}$. Then: $B = 2 \left( -\frac{1}{4} A \right) = -\frac{1}{2} A$

Step 4: Compute the determinant of $B$ Since $B = -\frac{1}{2} A$ and $A$ is a $3 \times 3$ matrix: $|B| = \left( -\frac{1}{2} \right)^3 |A| = -\frac{1}{8} (-4) = \frac{1}{2}$

Step 5: Simplify $\text{adj}(\text{adj} B)$ For any $n \times n$ matrix $B$, the property of iterated adjoint states that: $\text{adj}(\text{adj} B) = |B|^{n-2} B$

For $n = 3$: $\text{adj}(\text{adj} B) = |B|^{3-2} B = |B| B$

Substituting $|B| = \frac{1}{2}$ and $B = -\frac{1}{2} A$: $\text{adj}(\text{adj}(2(\text{adj}A)^{-1})) = \frac{1}{2} \left( -\frac{1}{2} A \right) = -\frac{1}{4} A$

Step 6: Find the sum of all elements of $-\frac{1}{4} A$ First, find the sum of elements of matrix $A$: $\text{Sum}(A) = 1 + 1 + 2 + (-2) + 0 + 1 + 1 + 3 + 5 = 12$

Thus, the sum of all elements of the given matrix is: $\text{Sum} \left( -\frac{1}{4} A \right) = -\frac{1}{4} \times \text{Sum}(A) = -\frac{1}{4} \times 12 = -3$

Hence, the correct option is D.