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MAE 107 – Computational Methods
Summer Session II 2009
Midterm sample – Solutions
Problem 1
function fact=factorial(n)
% Computes the factorial of n for an integer n positive
if(n<0)
error(’n must be a positive integer’)
end
fact=1;
for k=1:n,
fact=fact*k;
end
Note that the previous function, indeed computes
n
! for
n
≥
0. In particular for
n
= 0, the
for.
..end
loop is never started as
[1:n]
does not have any element.
Problem 2
Consider the matrix
A
=
2

1
3

2

2

2
2

7
4
.
1. The ﬁrst step of the Forward Elimination procedure aims at canceling the coeﬃcients below the
pivot coeﬃcient
a
11
= 2 in the ﬁrst column of
A
. To do so, we perform the following row operations:

(2)
←
(2)

λ
21
(1) with
λ
21
=

1,

(3)
←
(3)

λ
31
(1) with
λ
31
= 1.
The result of this operation can be written as a decomposition for
A
:
A
=
1
0 0

1 1 0
1
0 1
2

1 3
0

3 1
0

6 1

{z
}
A
(1)
The second step of the Forward Elimination procedure aims at canceling the coeﬃcients below the
second pivot coeﬃcient
a
22
=

3 in the second column of the modiﬁed
A
(1)
. The following row
1
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←
(3)

λ
32
(2) with
λ
32
= 2. The result of this operation can be written
as the following decomposition for
A
(1)
:
A
(1)
=
1 0 0
0 1 0
0 2 1
2

1
3
0

3
1
0
0

1
From these two steps we can write:
A
=
1
0 0

1 1 0
1
0 1
1 0 0
0 1 0
0 2 1
2

1
3
0

3
1
0
0

1
=
1
0 0

1 1 0
1
2 1
2

1
3
0

3
1
0
0

1
The LU decomposition of
A
is therefore:
A
=
L
·
U
,
with
L
=
1
0 0

1 1 0
1
2 1
and
U
=
2

1
3
0

3
1
0
0

1
.
2. Using the fundamental property of the determinant, the determinant of
A
is the product of the
determinants of
L
and of
U
. These two matrices are diagonal, therefore their determinants are
easily computed as the product of their diagonal terms.

L

= 1 and

U

= 6, therefore

A

= 6
3. To compute
Y
=
A

1
we need to solve 3 linear systems for the column vectors
y
(1)
,
y
(2)
and
y
(3)
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This note was uploaded on 10/11/2011 for the course MAE 107 taught by Professor Rottman during the Summer '08 term at UCSD.
 Summer '08
 Rottman

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