Scientific Computing Using Python - PHYS:4905 - Fall 2018
Practice Exam #1 - 10/4/2018
All work must be shown to receive full credit for each problem.
1. Find the augmented matrix for the system of linear equations below. Then apply Gaussian elimination and write down your steps as you go. What is the solution?
(
3
1
1
3
0
6
3
2
2
)
(
x
y
z
)
=
(
1
2
2
)
\begin{pmatrix} 3 & 1 & 1 \\ 3 & 0 & 6 \\ 3 & 2 & 2 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 2 \\ \end{pmatrix}
2. L is a linear operator and
L
(
2
2
)
=
(
5
3
)
L \,\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}
and
L
(
1
-1
)
=
(
3
7
)
L \, \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}
. Find
L
(
1
3
)
L \, \begin{pmatrix} 1 \\ 1 \end{pmatrix}
.
3. Are the following vectors linearly independent and what vector space do they span?
(
1
1
1
)
(
2
2
1
)
(
1
1
0
)
\begin{pmatrix} 1 \\ 1 \\1 \end{pmatrix} \;\;\; \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \;\;\; \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}
4. What is the determinant of the following matrix?
(
5
1
3
7
4
2
0
2
6
)
\begin{pmatrix} 5 & 1 & 3 \\ 7 & 4 & 2 \\ 0 & 2 & 6 \end{pmatrix}
Equations and definitions
det
(
M
)
=
∑
σ
s
g
n
(
σ
)
m
σ
(
1
)
1
m
σ
(
2
)
2
⋯
m
σ
(
n
)
n
\det(M) = \underset{σ}{∑} sgn(σ) \, m_{σ(1)}^1 \, m_{σ(2)}^2 \, ⋯ \, m_{σ(n)}^n
the sum is over all permutations of the set (1, 2, ...
n
) where
n
is the dimension of the matrix and
sgn
(σ) is +1 if the permutation is even and -1 if odd.
adj
(M) = (cofactor
(
m
j
i
)
(m_j^i)
)
T
where
i
and
j
run from 1 to
n
and the cofactor is the product of the minor associated to
m
j
i
m_j^i
and (-1)
i
+
j
.
A
minor
of an
n
×
n
matrix
M
is the determinant of any square matrix obtained from
M
by deleting one row and one column. If we pick entry
m
j
i
m_j^i
, you then delete row i and column j of M and the resulting matrix is the minor associated with that entry.
Inverse:
M
-
1
=
1
det
(
M
)
a
d
j
(
M
)
M^{-1} = \frac{1}{\det(M)} adj(M)
If the inverse of M exists, then
det
(
M
-
1
)
=
1
det
(
M
)
\det(M^{-1}) = \frac{1}{\det(M)}