# `x + 2y - 3z = -28, 4y + 2z = 0, -x + y - z = -5` Use matricies to solve the system of equations. Use Gaussian elimination with back-substitution. `x+2y-3z=-28`

`4y+2z=0`

`-x+y-z=-5`

The equations in the matrix form can be written as,

`[[1,2,-3,-28],[0,4,2,0],[-1,1,-1,-5]]`

Add Row 1 and Row 3

`[[1,2,-3,-28],[0,4,2,0],[0,3,-4,-33]]`

Multiply Row 2 by 2 and Add it to Row 3

`[[1,2,-3,-28],[0,4,2,0],[0,11,0,-33]]`

Now the equations can be written as,

`x+2y-3z=-28`     ----- equation 1

`4y+2z=0`         ...

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`x+2y-3z=-28`

`4y+2z=0`

`-x+y-z=-5`

The equations in the matrix form can be written as,

`[[1,2,-3,-28],[0,4,2,0],[-1,1,-1,-5]]`

Add Row 1 and Row 3

`[[1,2,-3,-28],[0,4,2,0],[0,3,-4,-33]]`

Multiply Row 2 by 2 and Add it to Row 3

`[[1,2,-3,-28],[0,4,2,0],[0,11,0,-33]]`

Now the equations can be written as,

`x+2y-3z=-28`     ----- equation 1

`4y+2z=0`            ------ equation 2

`11y=-33`               ----- equation 3

From equation 3,

`y=-33/11=-3`

Substitute back y in equation 2,

`4(-3)+2z=0`

`-12+2z=0`

`2z=12`

`z=12/2=6`

substitute back y and z in equation 1,

`x+2(-3)-3(6)=-28`

`x-6-18=-28`

`x=-28+18+6`

`x=-4`

So the solutions are x=-4, y=-3, z=6

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