x+y+z= 6, x-y+z = 2, x+2y=z= 2
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Answer:
Given equations are
x + y + z = 6,
x – y + z = 2,
x + 2y – z = 2
D = |1111-1112-1|
= 1(1 – 2) – 1(–1 – 1) + 1(2 + 1)
= 1 (–1) –1 (–2) + 1(3)
= –1 + 2 + 3
= 4 ≠ 0
Dx = |6112-1122-1|
= 6(1 – 2) – 1(–2 – 2) + 1(4 + 2)
= 6(– 4) – 1 (– 4) + 1(6)
= –6 + 4 + 6
= 4
Dy = |16112112-1|
= 1(–2 – 2) – 6(–1 – 1) + 1(2 – 2)
= 1(–4) – 6(–2) + 1(0)
= –4 + 12 + 0
= 8
Dz = |1161-12122|
= 1(–2 – 4) – 1(2 – 2) + 6(2 + 1)
= 1(– 6) – 1(0) + 6(3)
= –6 + 0 + 18
= 12
By Cramer’s Rule,
x = DxD=44=1, y=DyD=84 = 2 and
z = DzD=1214 = 3
∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.
Step-by-step explanation:
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