find the system of equation.
x+y+w=0
y+z=0
x+y+z+w=0
x+y+2x=0
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Answer:
x+3y−2z=0
2x−y+4z=0
x−11y+14z=0
The given system of equations in the matrix form are written as below:
⎣
⎢
⎡
1
2
1
3
−1
−11
−2
4
14
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
0
0
0
⎦
⎥
⎥
⎤
AX=0...(1)
where
A=
⎣
⎢
⎢
⎡
1
2
1
3
−1
−11
−2
4
14
⎦
⎥
⎥
⎤
; X=
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
and 0=
⎣
⎢
⎢
⎡
0
0
0
⎦
⎥
⎥
⎤
∴∣A∣=1(−14+44)−3(28−4)−2(−22+1)=30−72+42=0
and therefore the system has a non-trivial solution. Now, we may write first two of the given equations
x+3y=2z and 2x−y=−4z
Solving these equations in terms of z, we get
x=−
7
10
z and y=
7
8
z
Now if z=7k, then x=−10k and y=8k
Hence, x=−10k, y=8k and z=7k where k is arbitrary, are the required solutions.
⇒2(∣y+z)/x∣=3
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