Description
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
except the first column looks like `1 =
â 1
−l21
−l31
.
.
.
−`n1
ì
with lj1 =
aj1
a11
, j =
2, 3, · · · n. It is obvious that L1 = I +`1e
T
1
. Prove L
−1
1 = I −`1e
T
1
. 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.
2. Find the general solution to Ax = b with A =
Ñ
1 2 1 3
−3 2 1 0
3 2 1 1é
and b =
Ñ
2
−5
2
é
. You may use some of the information from the previous problem.
3. Given the matrix A =
Ñ
−1 2 1 0 2
2 0 0 3 −1
−1 6 3 3 5 é
, b =
Ñ
−1
0
c
é
, c ∈ R,
(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.
