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

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Answer 1 · 2018-02-23T22:28:44

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

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 $t=1$

How do you solve the system 4r-4x+4t=-44r-4x+4t=-4, 4r+x-2t=54r+x-2t=5, and -3r-3x-4t=-16-3r-3x-4t=-16?