Universidad San Pablo - CEU
Mathematical Fundaments of Biomedical
Engineering 1
Problems
Author:
First Year Biomedical
Engineering
Supervisor:
Carlos Oscar S. Sorzano
September 15, 2013
1
Chapter 3
Lay, 3.1.42
Let
a
u=
b
c
v=
,
0
and
where
a, b, c
are positive (for simplicity). Com-
u, v, u+v and 0, and compute
v u . Draw a picture and ex-
pute the area of the parallelogram determined by
the determinants of the matrices
u
v
and
plain what you nd.
Solution: The area of the parallelogram is base times height. In this case:
2
1.5
u
y
1
u+v
0.5
0
v
−0.5
−1
−1
0
1
2
3
4
x
A = cb
v is
a c b 0 = a · 0 − bc = −bc
u is
The determinant of v
c a 0 b = cb − a · 0 = cb
v = abs v u We see that A = abs u
The determinant of
u
Lay, 3.2.14
Combine the methods of row reduction
and cofactor expansion to compute
the determinant
−3
1
−3
3
−2
1 −4 3
0 −3 4 −2
8 −4
0
4 Solution:
1
−3 −2
1 −4 1
3
0 −3 = 4 −2
8 −3
3 −4
0
4 −3 −2 1 −4 1
3 −3
1
3 0 −3 1+3 −9
0
0
= (−1) 1 =
0 0
0 3 −4
−9
4
3 −4 0
4
1
3 −3 3 −3 2+1
0
0 = (−1) (−9) = −9
−4
4 3 −4
4
= 9(3 · 4 − (−3) · (−4)) = 0
D
r3 ← r3 + 2r1
D
D
D
Lay, 3.2.15
Assume
a
d
g
b
e
h
c
f
i
= 7.
Calculate
a
b c
d
e
f
5g 5h 5i
.
Solution:
a
b c
d
e
f
5g 5h 5i
=
a
5 d
g
b
e
h
c
f
i
= 5 · 7 = 35
Lay, 3.2.18
Assume
a
d
g
b
e
h
c
f
i
= 7.
Calculate
g
a
d
h
b
e
i
c
f
.
Solution:
g
a
d
h
b
e
i
c
f
a
= − g
d
b
h
e
c
i
f
a
= d
g
b
e
h
c
f
i
=7
Lay, 3.2.19
Assume
a
d
g
b
e
h
c
f
i
= 7.
Calculate
a
2d + a
g
b
c
2e + b 2f + c
h
i
.
Solution:
a
b
c
2d + a 2e + b 2f + c
g
h
i
=
a
a
b
c 2d 2e 2f = 2 d
g h
g
i b
e
h
c
f
i
= 2 · 7 = 14
Lay, 3.2.24

Use determinants to decide if the set of vectors
is linearly independent.
Solution:
2
  
4
−7
 6 ,  0 ,
−7
2

and

−3
−5
6
−7 −3 0 −5 = 11
2
6 4
6
−7
The three vectors are linearly independent because their determinant is dierent
from 0.
Lay, 3.2.31
A
Show that if
Solution: If
A
is invertible, then
det{A−1 } =
1
det{A}
is invertible, then
AA−1 = I
Taking determinants on both sides
det{AA−1 } = det{I}
det{A} det{A−1 } = 1
1
det{A−1 } = det{A}
Lay, 3.2.32
Find a formula for
det{rA}
when
A
is an
n × n matrix.
A
Solution: Consider the column decomposition of
A = a1
a2
... an
Then
rA = ra1 ra2 ...
det{rA} = ra1 ra2
= r a1 ra2
= r2 a1 a2
= rn a1 a2
= rn det{A}
ran
...
...
...
...
ran ran ran an Lay, 3.2.33
Let
A
and
B
square matrices. Show that even though
be equal, it is always true that
AB
det{AB} = det{BA}
Solution: By applying properties of the determinants
det{AB} = det{BA}
det{A} det{B} = det{B} det{A}
Lay, 3.3.1
Use Cramer's rule to solve the following equation system
5x1 + 7x2 = 3
2x1 + 4x2 = 1
3
and
BA
may not
Solution:
3
1
x1 =
5
2
5
2
x2 = 5
2
7 4 = 3·4−7·1 = 5
5·4−7·2
6
7 4 3 1 = 5·1−3·2 = − 1
5·4−7·2
6
7 4 Lay, 3.3.7
Determine the values of the parameter
s
for which the system below has a
unique solution.
6sx1 + 4x2 = 5
9x1 + 2sx2 = −2
Solution: Applying Cramer's rule
5 4 8
8
−2 2s )
5(s+ 10
)
10(s+ 10
= 5·2s−4·(−2) = 10s+8
√
√
√
√
x1 =
6s·2s−4·9
12s2 −36 = 12(s+ 3)(s− 3) = 6(s+ 3)(s− 3)
6s 4 9 2s 6s
5 9 −2 )
12(s+ 45
= 6s·(−2)−5·9 = − 12s+45
√ 12 √
x2 = 6s·2s−4·9
12s2 −36 = − 12(s+ 3)(s− 3) =
6s
4
9 2s s+ 45
12 √
− (s+√3)(s−
3)
This equation system has a unique solution if the denominator of the fractions above do not vanish, that is,
√
s 6= ± 3.
Lay, 3.3.11

Calculate the adjugate of the matrix
calculate
0
A= 3
−1
−2
0
1

−1
0 .
1
Then, use it to
A−1 .
Solution: For calculating the adjugate of the matrix
its cofactors
4
A we need to calculate all
C11
=
C12
=
C13
=
C21
=
C22
=
C23
=
C31
=
C32
=
C33
=
0 0 (−1)
1 1 3 0 (−1)1+2 −1 1 3 0 (−1)1+3 −1
1 −2 −1 (−1)2+1 1 1
0 −1 (−1)2+2 1 −1
0 −2 (−1)2+3 1 −1
−2 −1 (−1)3+1 0 0
0 −1 (−1)3+2 0 3
0 −2 (−1)3+3 3
0 1+1
=
0
=
−3
=
3
=
1
=
−1
=
2
=
0
=
−3
=
6
The adjoint is


−3 3
−1 2 
−3 6
0
A∗ =  1
0
For calculating
A−1
we need the determinant of
A.
We use the cofactor expan-
sion along the second row
|A| = a21 C21 + a22 C22 + a23 C23 = 3 · 1 = 3
Now

A−1
0
1
= |A|
(A∗ )T = 13  −3
3
 
0
1
0
−1 −3  =  −1
2
6
1
1
3
− 31
2
3

0
−1 
2
Lay, 3.3.21
Find the area of the parallelogram whose vertices are (-1,0), (0,5), (1,-4),
(2,1).
Solution: Let us draw the parallelogram:
5
6
4
x
2
2
0
−2
−4
−6
−3
Calling
−2
−1
0
x1
1
2
3
xA = (−1, 0), xB = (0, 5), xC = (1, −4), the sought area is the
xB − xA and xC − xA .
xB − xA = 0 5 − −1 0 = 1 5 xC − xA = 1 −4 − −1 0= 2 −4
1 2 = 14
abs xB − xA xC − xA = abs 5 −4 absolute
value of the determinant of the vectors
Lay, 3.3.25
Use the concept of volume to explain why the determinant of a
is zero i
A
3×3
matrix
is not invertible.
Solution: From the invertible matrix theorem, we know that a matrix is invert-
ible i its columns are linearly independent. So the statement of this problem
can be restated as the determinant of a
of
A
3 × 3 matrix is zero i the three columns
are linearly dependent. On the other side interpreting the determinant of
A as the volume of the parallelepiped formed by the three columns, the problem
is the volume of the parallelepiped formed by three vectors is zero i the three
columns of
A
are linearly dependent.
If the three vectors are linearly dependent, they span a subspace of dimension
2 or 1. In both cases, there is no real parallelepiped but a parallelogram or a
segment and the volume of the parallelepiped is 0.
Let us show that if the volume of the parallepiped is zero, then three columns
are linearly dependent. Let's assume they are linearly independent. Then, they
would actually span a three-dimensiional space, and the volume of the parallelepiped formed by the three would not be zero.
But this is a contradiction
with our hypothesis. So the three vectors have to be linearly dependent.
Lay, 3.3.26
T : Rn → Rm be a linear transformation, and let p be a vector and
S a set in Rn . Show that the image of p + S under T is the translated set
T (p) + T (S) in Rm .
Solution: Any vector of the set p + S is of the form
Let
x=p+s
6
where
s ∈ S.
If we apply
T
to
x
and exploiting the fact that
T
is a linear
transformation, we get
T (x) = T (p + s) = T (p) + T (s)
The set of all vectors of the form
T (s)
is actually
T (S),
so we have that, as
stated by the problem,
T (x) ∈ T (p) + T (S)
Lay, 3.3.32
S be the tetrahedron in R3 with vertices at the vectors 0, e1 , e2 and e3
0
let S be the tetrahedron with vertices at vectors 0, v1 , v2 and v3 . See the
Let
and
gure.
a. Describe a linear transformation that maps
S
into
b. Find a formula for the volume of the tetrahedron
Volume of
S = 13
Area of the base
S0.
S0
·
using the fact
Height.
Solution:
a. Consider the matrix
A = v1
The tetrahedron
S
v2
v3
is formed by all those points that can be written in the
form
x = λ0 0 + λ1 e1 + λ2 e2 + λ3 e3
with
λ0 + λ1 + λ2 + λ3 ≤ 1
If we consider now
Ax,
we have
7
Ax
= A(λ0 0 + λ1 e1 + λ2 e2 + λ3 e3 )
= λ0 A0 + λ1 Ae1 + λ2 Ae2 + λ3 Ae3
= λ0 0 + λ1 v1 + λ2 v2 + λ3 v3
So this is a point in the tetrahedron
b. The base of the tetrahedron
S
S0
as required by the problem.
is a triangle with vertices
0, e1
and
e2 ,
area is
Area triangular base =
1
2 Base
·
Height =
The height of the tetrahedron is the length of
Volume of
S = 31
Area of the base
·
e3 ,
1
21
S 0 =| det{A}|
Height =
8
Volume of
1
2.
that is, 1. Finally
According to Theorem 5.2 in Chapter 4, the volume of
Volume of
·1=
11
3 21
S0
=
is
S = 16 | det{A}|
1
6
whose