How do you solve $x + 2 y + z = 6$ $x + 2y + z = 6$ , $2 x - y - z = 0$ $2x - y - z = 0$ , and $3 x + 2 y + z = 10$ $3x + 2y +z = 10$ using matrices?

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Answer 1 · 2017-12-17T05:03:18

$x = 2$ $x=2$
$y = 0$ $y=0$
$z = 4$ $z=4$

Given -

$x + 2 y + z = 6$ $x+2y+z=6$
$2 x - y - z = 0$ $2x-y-z=0$
$3 x + 2 y + z = 10$ $3x+2y+z=10$
Answer is developed from the template I created in Excel

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Answer 2 · 2017-12-17T21:06:59

$x = 2$ $x=2$ , $y = 0$ $y=0$ and $z = 4$ $z=4$

Perform the Gauss Jordan elimination on the augmented matrix

$A=((1,2,1,|,6),(2,-1,-1,|,0),(3,2,1,|,10))$

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

$R2larrR2-2R1$ ; $R3larrR3-3R1$

$A=((1,2,1,|,6),(0,-5,-3,|,-12),(0,-4,-2,|,-8))$

$R1larrR1-(R3)/2$ ; $R2larrR2-R3$

$A=((1,0,0,|,2),(0,-1,-1,|,-4),(0,-4,-2,|,-8))$

$R3larrR3-4R2$

$A=((1,0,0,|,2),(0,-1,-1,|,-4),(0,0,2,|,8))$

$R2larr(R2)/(-1)$ ; $R3larr(R3)/2$

$A=((1,0,0,|,2),(0,1,1,|,4),(0,0,1,|,4))$

$R2larrR2-R3$

$A=((1,0,0,|,2),(0,1,0,|,0),(0,0,1,|,4))$

Thus $x=2$ , $y=0$ and $z=4$

How do you solve x + 2y + z = 6x+2y+z=6x + 2y + z = 6, 2x - y - z = 02x−y−z=02x - y - z = 0, and 3x + 2y +z = 103x+2y+z=103x + 2y +z = 10 using matrices?