# Solve the next system 2x + 3y + 4z = 0 4x + 9y + 16z =0 8x + 27y + 64z = 0

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2x+3y+4z=0.....(1)

4x+9y+16z=0....(2)

8x+27y+64z=0...(3)

What we have is three equation with three variables, we can solve it the same way we solve equations with 2 variables:

Let us multiply (1) with -2 and add to (2)

==> -4x-4y-8z=0

4x+9y+16z=0

==> 5y+8z=0 ....(4)

Now multiply (2) with -2 and add to (3)

-8x-18y-32z=0

8x+27y+64z=0

==> 9y+32z=0 ...(5)

Now lultiply (4) with -4 and add to (5)

-20y-32z=0

9y+32z+0

==> -11y=0

==> y=0

==> z=0

==> x=0

To solve

2x+3y+4z = 0..........(1)

4x+9y+16z = 0.........(2)

8z+27y+64z = 0.......(3)

Solution:

4*(1)- (2) gives: 4x+3y = 0........(4)

16(1)-(3) give: 24x+ 21y = 0.......(5)

7*(4) - (5) gives: 4x = 0. So y= 0, So from (4)4x+3*0 = 0. Or x = 0. Substititing x=0 and y =0 in eq (1), we get: z= 0.

So x =0

y=0 and

z= 0.

We notice that the system is homogeneous type (all the free terms are equal to 0).

If the system is determined, the only solution is the null one, where x = 0, y = 0 and z = 0.

If we create the matrix of the system and after that we calculate it's determinant, and if this one is different from zero, then the only solution of the system is x = 0 , y = 0 and z = 0.