Question

solve the system of equations x+3y+2z=0,2x-y+3z,3x-5y+4z=0,x+17y+4z=0​

Answer 1

Answer:

ANSWER

We have

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

Answer 2

A + 3 + 2 = 0

2 − + 3 = 0

3 − 5 + 4 = 0

+ 17 + 4 = 0

Step-by-step explanation: