Question

3x+y+z=3,2x+2y+5z+1=0,x=3y+4z+2

Answer 1

Answer:

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MATHS

The number of solutions of the equation 3x+3y−z=5,x+y+z=3,2x+2y−z=3 is:

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ANSWER

The coefficient Matrix A is

A=

∣

∣

∣

∣

∣

∣

∣

∣

3

1

2

3

1

2

−1

1

−1

∣

∣

∣

∣

∣

∣

∣

∣

∣A∣=3[−1−2]−3[−1−2]−1[2−2]

=−9+9=0

Put x=0 in eq (1) eq (2)

3y−z=5

y+z=3.

4y=8

y=2,z=1

Now put this in equation (3)

x=0,

y=2

z=1.

⇒3=3

∴ Infinite solution

Answer 2

Step-by-step explanation:

so this question's answer is

x=1,y=1,and z=-1