Question

How do you solve the system `4r-4x+4t=-4`, `4r+x-2t=5`, and `-3r-3x-4t=-16`?

Answer 1

`r=1`, `x=3` and `t=1`

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((4,-4,4,|,-4),(4,1,-2,|,5),(-3,-3,-4,|,-16))`

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

`R1larr(R1)/4`

`A=((1,-1,1,|,-1),(4,1,-2,|,5),(-3,-3,-4,|,-16))`

`R2larrR1-4R2`; `R3larrR3+3R1`

`A=((1,-1,1,|,-1),(0,5,-6,|,9),(0,-6,-1,|,-19))`

`R2larrR2*6`; `R3larrR3*5`

`A=((1,-1,1,|,-1),(0,30,-36,|,54),(0,-30,-5,|,-95))`

`R3larrR3+R2`

`A=((1,-1,1,|,-1),(0,30,-36,|,54),(0,0,-41,|,-41))`

`R3larr(R3)/(-41)`

`A=((1,-1,1,|,-1),(0,30,-36,|,54),(0,0,1,|,1))`

`R1larrR1-R2; R2larrR2+36R3`

`A=((1,-1,0,|,-2),(0,30,0,|,90),(0,0,1,|,1))`

`R2larr(R2)/30`

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

`R1larrR1+R2`

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

Thus, `r=1`, `x=3` and `t=1`