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tion:GIVEN equations are 3x+4y+5z=18,2x−y+8z=13,5x−2y+7z=20D= ∣∣∣∣∣∣∣∣ 325 4−1−2 587 ∣∣∣∣∣∣∣∣ =3(−7+16)−4(14−40)+5(−4+5)=3(9)−4(−26)+5(1)=27+104+5=136D 1 = ∣∣∣∣∣∣∣∣ 181320 4−1−2 587 ∣∣∣∣∣∣∣∣ =18(−7+16)−4(91−160)+5(−26+20)=18(9)−4(−69)+5(−6)=162+276−30=408D 2 = ∣∣∣∣∣∣∣∣ 325 181320 587 ∣∣∣∣∣∣∣∣ =3(91−160)−18(14−40)+5(40−65)=3(−69)−18(−26)+5(−25)=−207+468−125=136D 3 = ∣∣∣∣∣∣∣∣ 325 4−1−2 181320 ∣∣∣∣∣∣∣∣ =3(−20+26)−4(40−65)+18(−4+5)=3(6)−4(−25)+18(1)=18+100+18=136Now, x= DD 1 = 136408 =3Y= DD 2 = 136136 =1z= DD 3 = 136136 =1Hence, the solution for the SYSTEM is (3,1,1).


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