Exercise 1.5.15. Given the matrices
find their factors L, D, and U and associated permutation matrix P such that PA = LDU. Confirm that the factors are correct.
Answer: In the first step in elimination for the first matrix we must exchange rows 1 and 2, by multiplying by permutation matrix . The intermediate matrices for U, L, and P are as shown:
There is already a zero in the (2,1) position, so the corresponding multiplier is zero. We next subtract 2 times row 1 from row 3:
and then subtract 3 times row 2 from row 3:
At this point elimination is complete and we have our final L and P:
We then need to determine D and U. We divide each row of the first matrix by diagonal entry in that row to produce U, with the divisor then going into D:
Finally, to confirm the factors are correct we compute LDU and PA:
We now turn to the second matrix:
As the first step of elimination we multiply row 1 by 2 and subtract it from row 2, producing the following intermediate matrices (which will become U and L respectively):
Since elimination produced a zero in the pivot position, we need to exchange rows 2 and 3 using the permutation matrix , and readjust the intermediate L to reflect the exchange:
We then subtract 1 times row 1 from (the new) row 2:
At this point elimination is complete and we have our final L and P respectively:
We then need to determine D and U. We divide each row of the first matrix by the diagonal entry in that row (if it’s nonzero) to produce U, with the divisor then going into D:
We then compute PA and LDU to confirm the factorization is correct:
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.