Let D1=a b a+bc d c+da b a−b and D2=aca+cbdb+daca+b+c then the value of D1D2, Where b≠0 and ad≠bc, is ____________.

D1=aba+bcdc+daba−bC2→C1+C2C3→C2−C3+C1D1=aa+b0cc+d0aa+b2b=2b(ac+ad−ac−cb)=2b(ad−bc)D2=aa+c0bb+d0aa+c−b=−b(ab+ad−ab−ac)=−b(ad−bc)we have D1D2=−2b(ad−bc)b(ad−bc)=−2

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Let $D_{1} = |\begin{array}{lll} a b a + b \\ c d c + d \\ a b a - b \end{array}| and D_{2} = |\begin{array}{ccc} a & c & a + c \\ b & d & b + d \\ a & c & a + b + c \end{array}|$ then the value of $|\frac{D_{1}}{D_{2}}|,$ Where $b \neq 0 and ad \neq bc$ , is ____________.

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Detailed Solution

$\begin{aligned} D_{1} = |\begin{array}{ccc} a & b & a + b \\ c & d & c + d \\ a & b & a - b \end{array}| \\ C_{2} \to C_{1} + C_{2} \end{aligned}$

$\begin{aligned} C_{3} \to C_{2} - C_{3} + C_{1} \\ D_{1} = |\begin{array}{ccc} a & a + b & 0 \\ c & c + d & 0 \\ a & a + b & 2 b \end{array}| \end{aligned}$

$\begin{aligned} = 2 b (ac + ad - ac - cb) \\ = 2 b (ad - bc) \end{aligned}$

$\begin{aligned} D_{2} = |\begin{array}{ccc} a & a + c & 0 \\ b & b + d & 0 \\ a & a + c & - b \end{array}| \\ = - b (ab + ad - ab - ac) \\ = - b (ad - bc) \end{aligned}$