solve by cramer's rule 3x-4y+5z=4,x-y+10z=10,4x-7y+6z=3
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Answered by
0
Answer:
Given equations are 3x+4y+5z=18,2x−y+8z=13,5x−2y+7z=20
D=
∣
∣
∣
∣
∣
∣
∣
∣
3
2
5
4
−1
−2
5
8
7
∣
∣
∣
∣
∣
∣
∣
∣
=3(−7+16)−4(14−40)+5(−4+5)
=3(9)−4(−26)+5(1)=27+104+5=136
D
1
=
∣
∣
∣
∣
∣
∣
∣
∣
18
13
20
4
−1
−2
5
8
7
∣
∣
∣
∣
∣
∣
∣
∣
=18(−7+16)−4(91−160)+5(−26+20)
=18(9)−4(−69)+5(−6)=162+276−30=408
D
2
=
∣
∣
∣
∣
∣
∣
∣
∣
3
2
5
18
13
20
5
8
7
∣
∣
∣
∣
∣
∣
∣
∣
=3(91−160)−18(14−40)+5(40−65)
=3(−69)−18(−26)+5(−25)=−207+468−125=136
D
3
=
∣
∣
∣
∣
∣
∣
∣
∣
3
2
5
4
−1
−2
18
13
20
∣
∣
∣
∣
∣
∣
∣
∣
=3(−20+26)−4(40−65)+18(−4+5)
=3(6)−4(−25)+18(1)=18+100+18=136
Now, x=
D
D
1
=
136
408
=3
y=
D
D
2
=
136
136
=1
z=
D
D
3
=
136
136
=1
Hence, the solution for the system is (3,1,1)
Answered by
0
Answer:
{x,y,z} = {-8/7,-9/7,-3/7}
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