How do you solve $3 x + 2 y + z = 8$ $3x + 2y + z = 8$ , $2 x - 3 y + 2 z = - 16$ $2x -3y + 2z = -16$ , and $x + 4 y - z = 20$ $x + 4y -z = 20$ using matrices?

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Answer 1 · 2016-03-20T08:31:54

$x = 1$ $x=1$ and $y = 4$ $y=4$ and $z = - 3$ $z=-3$

from the given
$3 x + 2 y + z = 8$ $3x+2y+z=8$
$2 x - 3 y + 2 z = - 16$ $2x-3y+2z=-16$
$x + 4 y - z = 20$ $x+4y-z=20$

For the matrix

$D=((a_11, a_12,a_13),(a_21,a_22,a_23),(a_31,a_32, a_33))=((3,2,1),(2, -3, 2), (1, 4, -1))$

$C=((c_1),(c_2),(c_3))=((8),(-16),(20))$

We will use the formula

$D=(a_11* a_22*a_33+a_12* a_23*a_31+a_13* a_21*a_32$
$-a_31* a_22*a_13-a_32* a_23*a_11-a_33* a_21*a_12)$

For x:

$x=(((c_1, a_12,a_13),(c_2,a_22,a_23),(c_3,a_32, a_33)))/D$

For y:

$y=(((a_11, c_1,a_13),(a_21,c_2,a_23),(a_31,c_3, a_33)))/D$

For z:

$z=(((a_11, a_12,c_1),(a_21,a_22,c_2),(a_31,a_32, c_3)))/D$

We form the matrix

$D=((3,2,1),(2, -3, 2), (1, 4, -1))=4$

For x:

$x=(((c_1, a_12,a_13),(c_2,a_22,a_23),(c_3,a_32, a_33)))/D=(((8, 2,1),(-16,-3,2),(20,4, -1)))/4=1$

For y:

$y=(((a_11, c_1,a_13),(a_21,c_2,a_23),(a_31,c_3, a_33)))/D=(((3, 8,1),(2,-16,2),(1,20, -1)))/4=4$

For z:

$z=(((a_11, a_12,c_1),(a_21,a_22,c_2),(a_31,a_32, c_3)))/D=(((3,2,8),(2,-3,-16),(1,4, 20)))/4=-3$

solution set $x=1$ and $y=4$ and $z=-3$

God bless...I hope the explanation is useful.

How do you solve 3x + 2y + z = 83x+2y+z=83x + 2y + z = 8, 2x -3y + 2z = -162x−3y+2z=−162x -3y + 2z = -16, and x + 4y -z = 20x+4y−z=20x + 4y -z = 20 using matrices?