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
George C.
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#

Related questions
Impact of this question
1150 views around the world
You can reuse this answer
Creative Commons License