Let A = begin{bmatrix} 1 & 1 & 2 -2 & 0 & 1 | vedclass

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(D) First, calculate the determinant of $A$: $\det(A) = 1(0-3) - 1(-10-1) + 2(-6-0) = -3 + 11 - 12 = -4$.Using the property $	ext{adj}(	ext{adj}(M))

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-$3$

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(D) First, calculate the determinant of $A$: $\det(A) = 1(0-3) - 1(-10-1) + 2(-6-0) = -3 + 11 - 12 = -4$.Using the property $	ext{adj}(	ext{adj}(M)) = \det(M)^{n-2} M$ for an $n 	imes n$ matrix, here $n=3$, so $	ext{adj}(	ext{adj}(M)) = \det(M) M$.Let $M = 2(	ext{adj} A)^{-1}$. Since $	ext{adj} A = \det(A) A^{-1}$, we have $M = 2(\det(A) A^{-1})^{-1} = 2 \det(A)^{-1} A = 2(-4)^{-1} A = -\frac{1}{2} A$.Then $	ext{adj}(	ext{adj}(M)) = \det(M) M = \det(-\frac{1}{2} A) (-\frac{1}{2} A) = (-\frac{1}{2})^3 \det(A) (-\frac{1}{2} A) = \frac{1}{16} \det(A) A = \frac{1}{16} (-4) A = -\frac{1}{4} A$.The sum of all elements of $A$ is $1+1+2-2+0+1+1+3+5 = 12$.Therefore, the sum of all elements of $-\frac{1}{4} A$ is $-\frac{1}{4} 	imes 12 = -3$.

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

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