Page 355 - 35Linear Algebra

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Now zero out the top row by subtracting x 1 times the first column from the
second column, x 1 times the second column from the third column et cetra.
Again these column operations do not change the determinant. Now factor
out x 2 − x 1 from the second row, x 3 − x 1 from the third row, etc. This
does change the determinant so we write these factors outside the remaining
determinant, which is just the same problem but for the (n − 1) × (n − 1)
case. Iterating the same procedure gives the result

 2 n−1 
1 x 1 (x 1 ) · · · (x 1 )
1 x 2
 (x 2 ) 2 · · · (x 2 ) n−1
 2  Y
1 x 3  = (x i − x j ) .
 (x 3 ) · · · (x 3 ) n−1
det 
 . . . . .
 . . . . . . . . . .  i>j

1 x n (x n ) 2 · · · (x n ) n−1
Q
(Here stands for a multiple product, just like Σ stands for a multiple
sum.)
3
8. (a) No, a basis for R must have exactly three vectors.
4
(b) We first extend the original vectors by the standard basis for R and
then try to eliminate two of them by considering

           
1 4 1 0 0 0
2 3 0 1 0 0
           
α   + β   + γ   + δ   + ε   + η   = 0 .
3 2 0 0 1 0
           
4 1 0 0 0 1
So we study

   
1 4 1 0 0 0 1 4 1 0 0 0
2 3 0 1 0 0 0 −5 −2 1 0 0
   
  ∼  
3 2 0 0 1 0 0 −10 −3 0 1 0
   
4 1 0 0 0 1 0 −15 −4 0 0 1
1 0 − −4 0 0 1 0 0 2 0
 3   3 
5 5
0 1 2 1 0 0 0 1 0 − 19 − 2 0
∼  5 5  ∼  5 5 
0 0 1 10 1 0 0 0 1 10 1 0
   
0 0 2 15 0 1 0 0 0 − 5 −10 1
2 2
From here we can keep row reducing to achieve RREF, but we can
already see that the non-pivot variables will be ε and η. Hence we can

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