Given a,b,c integer show that the system have unique solution and find solution?

(a+(1/3))x+by+cz=0

ax+(b+(1/3))y+cz=0

ax+by+(c+(1/3))z=0

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Given a,b,c are integers, to show that the system have unique solution we have to find the value of the following determinant:

`|[a+1/3,b,c],[a,b+1/3,c],[a,b,c+1/3]|`

The value of this determinant is:

`(a+1/3)(b+1/3)(c+1/3)+2abc-ac(b+1/3)-bc(a+1/3)-ab(c+1/3)`

`=3abc+(ab)/3+(bc)/3+(ac)/3+a/9+b/9+c/9+1/27-3abc -(ab)/3-(bc)/3-(ac)/3`

`=a/9+b/9+c/9+1/27`

`=1/9(a+b+c+1/3)`

Since, this determinant is non-zero, by Cramer's rule the given system of equations have **a unique solution**.

Hence, **x=0, y=0 and z=0** since each of the determinants `D_x` , `D_y` ,`D_z ` will contain a column of zeros (i.e the constant terms of the three equations).

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