How do you solve using gaussian elimination or gauss-jordan elimination, $3 x + y - 3 z = 3$ $3x + y - 3z =3$ , $x + 3 y - z = - 7$ $x + 3y - z = -7$ , $3 x + 3 y - z = - 1$ $3x + 3y - z = -1$ ?

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Answer 1 · 2017-12-13T21:58:27

$x = 3$ $x=3$ , $y = - 3$ $y=-3$ and $z = 1$ $z=1$

From second equation, $z = x + 3 y + 7$ $z=x+3y+7$

I took value of $z$ $z$ into third one,

$3 x + 3 y - (x + 3 y + 7) = - 1$ $3x+3y-(x+3y+7)=-1$

$2 x - 7 = - 1$ $2x-7=-1$

$2 x = 6$ $2x=6$ , so $x = 3$ $x=3$

After taking value of $z$ $z$ into first one,

$3 x + y - 3 \cdot (x + 3 y + 7) = 3$ $3x+y-3*(x+3y+7)=3$

$- 8 y - 21 = 3$ $-8y-21=3$

$- 8 y = 24$ $-8y=24$ , so $y = - 3$ $y=-3$

Thus $z = x + 3 y + 7 = 3 + 3 \cdot (- 3) + 7 = 3 - 9 + 7 = 1$ $z=x+3y+7=3+3*(-3)+7=3-9+7=1$

How do you solve using gaussian elimination or gauss-jordan elimination, 3x+y−3z=33x + y - 3z =3, x+3y−z=−7x + 3y - z = -7, 3x+3y−z=−13x + 3y - z = -1?