Question

How do you solve the system `x+y+z=1`, `2x-y+2z=-1`, and `-x-3y+z=1`?

Answer 1

The solution is `((x),(y),(z))=((-2),(1),(2))`

Explanation:

Perform the Gauss- Jordan Elimination on the Augmented Matrix

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

The pivot is in the first column and first row

Eliminate the first column

`R_2larr(R2-2R1)` and `R3larr(R3+R1)`

`((1,1,1,|,1),(0,-3,0,|,-3),(0,-2,2,|,2))`

Make the pivot in the second row of the second column

`R2larr(R2)/(-3)`

`((1,1,1,|,1),(0,1,0,|,1),(0,-2,2,|,2))`

Eliminate the second column

`R1larr(R1-R2)`

`((1,0,1,|,0),(0,1,0,|,1),(0,-2,2,|,2))`

`R3larrR3+2R2`

`((1,0,1,|,0),(0,1,0,|,1),(0,0,2,|,4))`

Make the pivot in the third column

`R3larr(R3/2)`

`((1,0,1,|,0),(0,1,0,|,1),(0,0,1,|,2))`

Eliminate the third column

`R1larr(R1-R3)`

`((1,0,0,|,-2),(0,1,0,|,1),(0,0,1,|,2))`

The solution is

`((x),(y),(z))=((-2),(1),(2))`

Answer 2

`x=-2, y=1, z=2`

Explanation:

Given equations:

`x+y+z=1\ ...........(1)`

`2x-y+2z=-1\ ...........(2)`

`-x-3y+z=1\ ...........(3)`

Adding (1) & (2), we get

`x+y+z+2x-y+2z=1+(-1)`

`3x+3z=0`

`x+z=0\ ........(4)`

Multiplying (1) by `3` & adding to (3) as follows

`-x-3y+z+3(x+y+z)=1+3\cdot 1`

`2x+4z=4`

`x+2z=2\ .........(5)`

subtracting (4) from (5) as follows

`x+2z-(x+z)=2-0`

`z=2`

Setting `z=2` in (4), we get

`x+2=0`

`x=-2`

Now, setting `x=-2` & `z=2` in (1), we get

`-2+y+2=1`

`y=1`

Hence, we get

`x=-2, y=1, z=2`