1. Let A be a n×n matrix with a11 6=0, the first step in LU decomposition is to introduce zeros below the first diagonal a11. This can be done by multiplying A by a lower triangular matrix L1 that is equal to the n × n diagonal matrix
â
1 ì
−l21
except the first column looks like `1 = −l31 with
...
−`n1
2,3,···n. It is obvious that . Prove . This is the first stroke of luck in LU decomposition: find the inverse of L1 can be done by simply negativing the entries below the first diagonal.
Ñ 1 2 1 3é
2. Find the general solution to Ax = b with A = −3 2 1 0 and b =
3 2 1 1
Ñ 2 é
−5 . You may use some of the information from the previous problem.
2
Ñ−1 2 1 0 2 é Ñ−1é
3. Given the matrix A = 2 0 0 3 −1 ,b = 0 ,c ∈ R,
−1 6 3 3 5 c
(a) For which value of c does the equation Ax = b have a solution?
(b) After choosing c so that the system has a solution, find a particular solution to Ax = b.
