How do I solve matrix for a in CC?

det A=|[1,1,1,-a],[1,1,-a,1],[1,-a,1,1],[-a,1,1,1]|!=0

1 Answer
Feb 24, 2018

det A = (x+1)^3(x-3), so det A != 0 if and only if a != -1 and a != 3

Explanation:

abs((1,1,1,-a),(1,1,-a,1),(1,-a,1,1),(-a,1,1,1))

= abs((-a,1,1,1),(1,-a,1,1),(1,1,-a,1),(1,1,1,-a))

= 1/(a-2) abs((-a,1,1,1),(1,-a,1,1),(1,1,-a,1),(a-2,a-2,a-2,2a-a^2(-a)))

= 1/(a-2) abs((-a,1,1,1),(1,-a,1,1),(1,1,-a,1),(0,0,0,3+2a-a^2))

= (3+2a-a^2)/(a-2) abs((-a,1,1),(1,-a,1),(1,1,-a))

= (3+2a-a^2)/((a-1)(a-2)) abs((-a,1,1),(1,-a,1),(a-1,a-1,a-a^2))

= (3+2a-a^2)/((a-1)(a-2)) abs((-a,1,1),(1,-a,1),(0,0,2+a-a^2))

= ((3+2a-a^2)(2+a-a^2))/((a-1)(a-2)) abs((-a,1),(1,-a))

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

= ((a^2-2a-3)(a^2-a-2)color(red)(cancel(color(black)((a-1))))(a+1))/(color(red)(cancel(color(black)((a-1))))(a-2))

= ((a-3)(a+1)color(red)(cancel(color(black)((a-2))))(a+1)(a+1))/color(red)(cancel(color(black)((a-2))))

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

So:

det A != 0 effectively means a != -1 and a != 3