Proof Let y be a solution to By=d. Then y is the linear combination of the basis that
represents d. This can be solved by Cramer's rule as Yk = det(Bk ) where Bk is the matrix
det(B)
obtained by replacing the kth column of B by d. Since B is triangular, it may be put into
lower triangular form with 1 's on the diagonal by a combination ofrow and column
interchanges. Therefore det(B)= + 1 or -1. Because any square submatrix of A will only
contain entries of 0 or 1 with a maximum of two 1 's in each column by the design ofthe
matrix A, every determinant of any submatrix of A will have a value of + 1, -1, or 0, so
det(Bk
)= 0, +1, or-I. ThereforeYk=O, +1, or-1.+
Proof Let y be a solution to By=d. Then y is the linear combination of the basis thatrepresents d. This can be solved by Cramer's rule as Yk = det(Bk ) where Bk is the matrixdet(B)obtained by replacing the kth column of B by d. Since B is triangular, it may be put intolower triangular form with 1 's on the diagonal by a combination ofrow and columninterchanges. Therefore det(B)= + 1 or -1. Because any square submatrix of A will onlycontain entries of 0 or 1 with a maximum of two 1 's in each column by the design ofthematrix A, every determinant of any submatrix of A will have a value of + 1, -1, or 0, sodet(Bk)= 0, +1, or-I. ThereforeYk=O, +1, or-1.+
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