let D1=∣∣
∣∣aba+bcdc+daba−b∣∣
∣∣ and D2=∣∣
∣∣aca+cbdb+daca+b+c∣∣
∣∣ then the value of D1D2 where b≠0 and ad≠bc ,is
D1∣∣
∣∣aba+bcdc+baba−b∣∣
∣∣
=∣∣
∣∣aba+bcdc+doo−2d∣∣
∣∣R3→R3−R1
=−2b(ad−bc)
D2=∣∣
∣∣aca+cbdb+daca+b+c∣∣
∣∣
=∣∣
∣∣aca+cbdb+doob∣∣
∣∣R3→R3−R1
=b(aa−bc)
Now D1D2=−2b(ad−bc)b(ad−bc)
=−2