Question 1044577
{{{abs(matrix(4,4,a,1,1,1,1,a,1,1,1,1,a,1,1,1,1,a))}}}
Perform ERO3's (which do not change the value of the determinant)
{{{-a*R[2] + R[1]}}}, {{{-1*R[2] + R[3]}}} and {{{-1*R[2] + R[4]}}}, giving


{{{abs(matrix(4,4,0,1-a^2,1-a,1-a,1,a,1,1,0,1-a,a-1,0,0,1-a,0,a-1))}}}.

= {{{-abs(matrix(3,3,1-a^2,1-a,1-a,1-a,a-1,0,1-a,0,a-1))}}}, 

by Laplace expansion with the the first column as pivot.

Factoring 1-a from all three rows give

= {{{-(1-a)^3*abs(matrix(3,3,1+a,1,1,1,-1,0,1,0,-1))}}}.

Now {{{-(1-a)^3 = (a-1)^3}}}, and so the last determinant is equal to

{{{(a-1)^3*abs(matrix(3,3,1+a,1,1,1,-1,0,1,0,-1))}}}.

Now perform the following ERO3 {{{1*R[2] + R[1]}}}, and ERO3 {{{1*R[3] + R[1]}}} IN SUCCESSION, to get

={{{(a-1)^3*abs(matrix(3,3,3+a,0,0,1,-1,0,1,0,-1))}}}.

={{{(a+3)(a-1)^3*abs(matrix(2,2,-1,0,0,-1))}}}

By performing Laplace expansion with the first row as pivot.

={{{(a+3)(a-1)^3*(-1*-1-0*0) = highlight((a+3)(a-1)^3)}}}.