Let A be an m x n matrix with rank n, and let PA = LU and A = QR be the PA LU and A QR factorizations of A, respectively, where P is a permutation matrix, L is a lower triangular matrix with 1's on the diagonal, and U is an upper triangular matrix. Q is an orthogonal matrix, and R is an upper triangular matrix.
We will show that the last row of L-1 is equal to the last column of Q.
Since PA = LU and A = QR, we have:
A = P^-1LU
A = QR
Multiplying both sides by Q^-1, we get:
Q^-1A = R
Substituting A = P^-1LU, we get:
Q^-1P^-1LU = R
Multiplying both sides by L^-1, we get:
Q^-1P^-1U = RL^-1
Since L is lower triangular with 1's on the diagonal, L^-1 is also lower triangular. Therefore, the last row of L^-1 is of the form [0 0 ... 0 1], where the 1 is in the last column.
Similarly, since Q is orthogonal, Q^-1 is also orthogonal. Therefore, the last column of Q^-1 is of the form [0 0 ... 0 1], where the 1 is in the last row.
Thus, we have shown that the last row of L^-1 is equal to the last column of Q^-1.