Question

How can I solve this equation system? `x + 5y + 2z = 19" "3x + 2y+ z = 13" "2x - y - z = 0` How should I think about this?

Answer 1

Solution: `x=2 , y=3 , z=1`

Explanation:

`x+5 y+2 z=19;(1), 3 x+2 y+ z=13; (2)`

`2 x- y- z=0 ; (3) :. 2 x = y+z` . Multiplying equation (1)

by `2` we get , `2 x + 10 y + 4 z = 38 ;` or

`y +z + 10 y + 4 z = 38 ; or 11 y +5 z =38; (4)`

Multiplying equation (2) by `2` we get ,

`6 x + 4 y + 2 z = 26 ;` or

`3(y +z) + 4 y + 2 z = 26 ; or 7 y +5 z =26; (5)`

Subtracting equation (5) from equation (4) we get,

`4 y = 12 or y =3 :. 5 z= 38 - 11 y` or

`5 z = 38 - 11*3 or 5 z= 5 or z=1`

`:. 2 x= y+ z or 2 x= 3 +1 or 2 x =4 or x=2`

Solution:`x=2 , y=3 , z=1` [Ans]

Answer 2

Thus `x=2`, `y=3` and `z=1`

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((1,5,2,|,19),(3,2,1,|,13),(2,-1,-1,|,0))`

I have written the equations not in the sequence as in the question in order to get `1` as pivot.

Perform the folowing operations on the rows of the matrix

`R2larrR2-3R1`; `R3larrR3-2R1`

`A=((1,5,2,|,19),(0,-13,-5,|,-44),(0,-11,-5,|,-38))`

`R2larrR2-R3`

`A=((1,5,2,|,19),(0,-2,0,|,-6),(0,-11,-5,|,-38))`

`R2larr(R2)/(-2)`

`A=((1,5,2,|,19),(0,1,0,|,3),(0,-11,-5,|,-38))`

`R1larrR1-5R2`; `R3larrR3+11R2`

`A=((1,0,2,|,4),(0,1,0,|,3),(0,0,-5,|,-5))`

`R3larr(R3)/(-5)`

`A=((1,0,2,|,4),(0,1,0,|,3),(0,0,1,|,1))`

`R1larrR1-2R3`

`A=((1,0,0,|,2),(0,1,0,|,3),(0,0,1,|,1))`

Thus `x=2`, `y=3` and `z=1`