Question 22721
 Note: You should say that you mean to find the determinant of C and forget 
 about those uncommon terms as rank-one updated matrices, or alpha variables, since you should not suppose other people sitting in the same class and use the same textbook. In fact, because of that ,I replied to you a little bit tardy.
 

matrix C: (I change it to Cn)
[(1+A1) A2 ......An]
[A1 (1+A2) ......An]
[..................]
[A1 A2.......(1+An)]


to show det(Cn) =
|(1+A1) A2 ......An|
|A1 (1+A2) ......An|
|..................| 
|A1 A2.......(1+An)|
=  1+ {{{SIGMA}}}Ai (i from 1 to n) ...(*)

 Usually, we prove it by math. induction.

 Basic, when n = 1, C1 is a 1x1 matrix, we have C1 = [1+A1]
  so, det(C1) = 1 + A1, so (*) is true when n = 1.
 
 Induction Hypothesis: Assume that (*) is true, when n = k. i.e. det(Ck)  =   1+ {{{SIGMA}}}Ai (i from 1 to k). To show det(C(k+1))  =  
1+ {{{SIGMA}}}Ai (i from 1 to k+1). 
Since det(C(k+1))
|(1+A1) A2 ......A(k+1)|
|A1 (1+A2) ......A(k+1)|
|..................| 
|A1 A2.......(1+A(k+1))|
 (multiply -1 to the 1st row and add it to the n+1 th row )
=
|(1+A1) A2 ......A(k+1)|
|A1 (1+A2) ......A(k+1)|
|..................| 
|-1 0.......1|
 (expand along the last row)
= {{{(-1)*(-1)^(k+1) }}}
|A2 A3 ......  A(k+1)|
|(1+A2) A3......A(k+1)|
 |..................|
 |A2 A3 ......1+Ak A(k+1)|
 +  
|(1+A1) A2 ......Ak|
|A1 (1+A2) ......Ak|
|..................| 
|A1 A2.......(1+Ak)|
 (set the first determinant to be Bk and use Induction Hyp.)
= {{{(-1)^(k+2)}}} Bk + (1+ {{{SIGMA}}}Ai ) (i from 1 to k)) 

To compute Bk, by adding 1st row *-1 to each of the other k-1 rows, we have  
  |A2 A3 ......  A(k+1)|
  |1 0 0..0|
  |.............|
  |0 0.......1 0|
, expanding along the last column, we can see that Bk = {{{(-1)^k}}} * A(k+1) .
Hence, det(C(k+1)) = {{{(-1)^(2k+2)}}} A(k+1)+ 1 + {{{SIGMA}}}Ai (i from 1 to k)
=  1+ {{{SIGMA}}}Ai (i from 1 to k+1). 
This proves that (*) is true, when n = k+1 and the induction is complete.

The idea is the same as , the case when n = 3.
Det(C3)=
|1+A1 A2 A3|
|A1 1+A2 A3| 
|A1 A2 1+A3|
=  |1+A1 A2 A3|
   |A1 1+A2 A3|
    | -1  0 1|
=  (-1) | A2 A3|
        |1+A2 A3|
  + det(C2)
 =  (-1)| A2 A3|
        |1 0|
  + det(C2)

=  -(-1)*A3*1 + (1+A1+A2)
= 1+A1+A2+ A3.

If you have learned eigenvalues and character polynomials,then we can have a direct proof as below.
 Let Rn be the (1xn) row = [A1 A2.. An] , and the nxn matrix Un = [Rn, Rn,..Rn] (i.e. each row equals to Rn). By solving the possible eignvalues & eigenvectors of Un, we see that the char. poly. of Un = det(xI-Un) 
= {{{x^(n-1)}}}(x- {{{SIGMA}}}Ai (i from 1 to n)).
 (Note I+Un = Cn and det(-I-Un) = {{{(-1)^n}}} det(I + Un))
Set x =-1, we obtain det(-I-Un) = {{{(-1)^(n)}}}*det(Cn) = 
{{{(-1)^n}}} * (1+ {{{SIGMA}}}Ai (i from 1 to n)) . 
Hence, det(Cn) = 1+ {{{SIGMA}}}Ai (i from 1 to n).


 Try to read carefully and understand the details.
 Good luck !

 Kenny