More generally show that if a1 ellipsis an belongs to Rn are

More generally, show that if a_1, ellipsis, a_n belongs to R^n are mutually orthogonal, then | det(a_1, ellipsis, a_n)| = ||a_1|| ellipsis ||a_n||.

Solution

Let the matrix [a₁,a₂,…,a_n ] be denoted by A. We know that multiplying a row/column of a determinant by a constant, scales up the value of the determinant by that constant. Let b_i = a_i / ||a_i|| for 1 i n and let B = [b₁,b₂,…,b_n ]. Then B is an orthogonal matrix. We know that each element of an orthogonal matrix is equal to its cofactor. Since,B is a n x n matrix, we have det(B) = _j=1^{j =n} b_ijB_ij = _j=1^{j =n} (b_ij)²

We also know that multiplying a row/column of a determinant by a constant, scales up the value of the determinant by that constant. Hence det(A) = ||a₁||…||a_n||det(B) so that |det(A)| = ||a₁||…||a_n|||det(B)|= ||a₁||…||a_n|| * | _j=1^{j =n} (b_ij)²|. Now, since each of b₁,b₂,…,b_n is a unit vector, hence | _j=1^{j =n} (b_ij)²| =1. Therefore, |det(A)| =||a₁||…||a_n||.