# what is a if ideterminate with infinitelly solutions?ax+y+z=1 x+ay+z=2-a x+y+az=3a+1

*print*Print*list*Cite

You should notice that the number of equations equals the number of variables, hence only determinant of matrix of system may tell if the system is indeterminate.

Forming the matrix of the system yields:

`A = ((a,1,1),(1,a,1),(1,1,a))`

You need to calculate determinant of matrix A such that:

`Delta = [[a,1,1],[1,a,1],[1,1,a]]` = `a^3 + 1 + 1 - a - a - a`

You need to remember that if `Delta = 0` , then the system is not determinate, hence `a^3 - 3a + 2 = 0` .

You need to solve the equation by factoring such that:

`a^3 -3a +3 - 1= 0 =gt (a^3 - 1) - 3(a - 1) = 0`

`` You need to use a special product instead of difference of cubes such that:

`a^3 - 1 = (a-1)(a^2 + a + 1)`

`` `(a-1)(a^2 + a + 1) - 3(a - 1) = 0`

Factoring out a-1 yields:

`(a-1)(a^2 + a + 1 - 3) = 0 =gt a - 1 = 0 =gt a = 1`

`a^2 + a - 2 = 0`

`a_(1,2) = (-2+-sqrt(1 + 8))/2 =gt a_(1,2) = (-2+-sqrt9)/2`

`a_(1,2) = (-2+-3)/2 =gt a_1 = 1/2 ; a_2 = -5/2`

Since the determinant of matrix of system is zero, then at least a minor is not zero, hence `delta = [[a,1],[1,a]] != 0 =gt a^2 - 1 != 0 =gt a_(1,2) != +- 1` for the number of solutions to the indeterminate system to be infinitely.

**Hence, evaluating the values of a under specified conditions yields: `a != +- 1; a in {-5/2 ; 1/2}.` **