Let D_1 = begin{vmatrix} a & b & a+b c & d & | vedclass

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(A) For $D_1$,apply the column operation $C_3 ightarrow C_3 - (C_1 + C_2)$:$D_1 = \begin{vmatrix} a & b & a+b-(a+b) \ c & d & c+d-(c+d) \ a & b &

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$-2$

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(A) For $D_1$,apply the column operation $C_3 ightarrow C_3 - (C_1 + C_2)$:$D_1 = \begin{vmatrix} a & b & a+b-(a+b) \ c & d & c+d-(c+d) \ a & b & a-b-(a+b) \end{vmatrix} = \begin{vmatrix} a & b & 0 \ c & d & 0 \ a & b & -2b \end{vmatrix}$.Expanding along $C_3$,we get $D_1 = -2b(ad - bc)$.For $D_2$,apply the column operation $C_3 ightarrow C_3 - (C_1 + C_2)$:$D_2 = \begin{vmatrix} a & c & a+c-(a+c) \ b & d & b+d-(b+d) \ a & c & a+b+c-(a+c) \end{vmatrix} = \begin{vmatrix} a & c & 0 \ b & d & 0 \ a & c & b \end{vmatrix}$.Expanding along $C_3$,we get $D_2 = b(ad - bc)$.Therefore,$\frac{D_1}{D_2} = \frac{-2b(ad - bc)}{b(ad - bc)} = -2$.

Let $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$ and $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$. Then the value of $\frac{D_1}{D_2}$,where $b \neq 0$ and $ad \neq bc$,is:

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