`3x - 2y + z = -15, -x + y + 2z = -10, x - y - 4z = 14` Use matricies to solve the system of equations. Use Gaussian elimination with back-substitution.

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`3x-2y+z=-15`

`-x+y+2z=-10`

`x-y-4z=14` 

`A=[[3,-2,1],[-1,1,2],[1,-1,-4]]`

`b=[[-15],[-10],[14]]`

`[A|b]=[[3,-2,1,-15],[-1,1,2,-10],[1,-1,-4,14]]`

Multiply 2nd Row by 3 and add Row 1

`[[3,-2,1,-15],[0,1,7,-45],[1,-1,-4,14]]`

Multiply 3rd Row by 3 and subtract it from Row 1

`[[3,-2,1,-15],[0,1,7,-45],[0,1,13,-57]]`

Subtract Row 2 from Row 3

`[[3,-2,1,-15],[0,1,7,-45],[0,0,6,-12]]`

Now the equations can be written as,

`3x-2y+z=-15`     ----equation 1

`y+7z=-45`       ...

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`3x-2y+z=-15`

`-x+y+2z=-10`

`x-y-4z=14` 

`A=[[3,-2,1],[-1,1,2],[1,-1,-4]]`

`b=[[-15],[-10],[14]]`

`[A|b]=[[3,-2,1,-15],[-1,1,2,-10],[1,-1,-4,14]]`

Multiply 2nd Row by 3 and add Row 1

`[[3,-2,1,-15],[0,1,7,-45],[1,-1,-4,14]]`

Multiply 3rd Row by 3 and subtract it from Row 1

`[[3,-2,1,-15],[0,1,7,-45],[0,1,13,-57]]`

Subtract Row 2 from Row 3

`[[3,-2,1,-15],[0,1,7,-45],[0,0,6,-12]]`

Now the equations can be written as,

`3x-2y+z=-15`     ----equation 1

`y+7z=-45`          ------ equation 2

`6z=-12`                 ------ equation 3

From equation 3,

`z=-12/6=-2`

Now substitute back z in equation 2,

`y+7(-2)=-45`

`y-14=-45`

`y=-45+14`

`y=-31`

Substitute back the value of y and z in equation 1,

`3x-2(-31)+(-2)=-15`

`3x+62-2=-15`

`3x=-15-60`

`3x=-75`

`x=-75/3`

`x=-25`

So the solution is x=-25, y=-31 and z=-2``

 

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