How do you solve the following system?: $7x+2y-z=3, 8x+3y+5z=9, 8x-8y-2z=4$

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Answer 1 · 2018-01-15T13:04:36

The solution is $((x),(y),(z))=((131/219),(-29/219),(202/219))$

Perform the Gauss Jordan elimination on the augmented matrix,

The last equation is first

$A=((8,-8,-2,|,4),(8,3,5,|,9),(7,2,-1,|,3))$

Perform the row operations

Make the first column the pivot , $R1larr(R1)/8$

$((1,-1,-1/4,|,1/2),(8,3,5,|,9),(7,2,-1,|,3))$

Eliminate the first column, $R2larr(R2-8R1)$ and $R3larr(R3-7R1)$

$((1,-1,-1/4,|,1/2),(0,11,7,|,5),(0,9,3/4,|,-1/2))$

Make the pivot in the second column, $R2larr(R2)/11$

$((1,-1,-1/4,|,1/2),(0,1,7/11,|,5/11),(0,9,3/4,|,-1/2))$

Eliminate the second column, $R3larr(R3-9R2)$

$((1,-1,-1/4,|,1/2),(0,1,7/11,|,5/11),(0,0,-219/44,|,-101/22))$

Make the pivot in the third column, $R3larr(R3xx-44/219)$

$((1,-1,-1/4,|,1/2),(0,1,7/11,|,5/11),(0,0,1,|,202/219))$

Eliminate the third column, $R2larr R2-7/11R3$

$((1,-1,-1/4,|,1/2),(0,1,0,|,-29/219),(0,0,1,|,202/219))$

$R1larr(R1+R2)$ and $R1larrR1+1/4R3$

$((1,0,0,|,131/219),(0,1,0,|,-29/219),(0,0,1,|,202/219))$

How do you solve the following system?: 7x+2y-z=3, 8x+3y+5z=9, 8x-8y-2z=4 7x+2y-z=3, 8x+3y+5z=9, 8x-8y-2z=4