How do you solve using gaussian elimination or gauss-jordan elimination, $2 x + 4 y - 6 z = 48$ $2x+4y-6z=48$ , $x + 2 y + 3 z = - 6$ $x+2y+3z=-6$ , $3 x - 4 y + 4 z = - 23$ $3x-4y+4z=-23$ ?

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Answer 1 · 2017-05-19T20:41:11

Using Gauss elimination gives $x = 3$ $x=3$ , $y = 3$ $y=3$ , and $z = - 5$ $z=-5$

Step 1. Rewrite the equations in matrix form
$((2,4,-6,|,48),(1,2,3,|,-6),(3,-4,4,|,-23))$

Step 2. Switch $R_2$ with $R_1$
$((1,2,3,|,-6),(2,4,-6,|,48),(3,-4,4,|,-23))$

Step 3. $-2R_1+R_2->R_2$
$((1,2,3,|,-6),(0,0,-12,|,60),(3,-4,4,|,-23))$

Step 4. $-3R_1+R_3->R_3$
$((1,2,3,|,-6),(0,0,-12,|,60),(0,-10,-5,|,-5))$

Step 5. Switch $R_2$ with $R_3$
$((1,2,3,|,-6),(0,-10,-5,|,-5),(0,0,-12,|,60))$

Step 6. $-1/10R_2 -> R_2$
$((1,2,3,|,-6),(0,1,1/2,|,1/2),(0,0,-12,|,60))$

Step 7. $-1/12R_3->R_3$
$((1,2,3,|,-6),(0,1,1/2,|,1/2),(0,0,1,|,-5))$

Translating back into the values $x$ , $y$ , and $z$
$z=-5$
$y+1/2(-5)=1/2=>y=3$
$x+2(3)+3(-5)=-6=>x=3$

How do you solve using gaussian elimination or gauss-jordan elimination, 2x+4y-6z=482x+4y−6z=482x+4y-6z=48, x+2y+3z=-6x+2y+3z=−6x+2y+3z=-6, 3x-4y+4z=-233x−4y+4z=−233x-4y+4z=-23?