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Lecture 8 – Appendix B: Some sample problems from Boas
Here are some solutions to the sample problems assigned for Chapter 3.
Solution: We want think about the following set of equations in terms of matrices
and their properties.
x y z x
x y z y
x y z z
Thus we have M = N = 3 and can proceed to row reduce the relevant matrices. We
find
3 3 2 1 1 1 1 5 2
2 2 2 1
R R R R R R
R R R
A
M
Aug 3 3 2 1
2 2 2 1
1 1 1 5 2
Aug 1 1 19 20 3
2 2 3 2 3
R R R
R R R
R R R
R R R
R R R
A
M
Since the ranks of the 2 matrices satisfy
Aug
M M
, we conclude that the equations
are inconsistent (over constrained) and that there is no (consistent) solution.
Solution: We want to determine the rank of the following matrix using row reduction
2 2 1 3 3 2
3 3 2 1 4 4 2
R R R R R R
R R R R R
We cannot increase the number of zeros by combining these 2 rows and conclude that
the rank of this matrix is 2. In Mathematica this looks like
Ex 3.2:
A={{1,1,4,3},{3,1,10,7},{4,2,14,10},{2,0,6,4}}
{{1,1,4,3},{3,1,10,7},{4,2,14,10},{2,0,6,4}}
MatrixForm[A]
MatrixRank[A]
2
Solution: Now we want to evaluate a determinant. First simplify by combining the
second and fourth rows and then proceed to find the determinant
2 3
2 2 4
1 3
R R R
1 1 4 3
3 1 10 7
4 2 14 10
2 0 6 4
x y z
x y
x z
x
y
z
In order for there to be a nontrivial solution the determinant of the coefficient matrix
must vanish so that it has no inverse. This leads to
2
So now for each (eigen)value we want to solve the equations (find the eigenvectors).
We have
3 1 3 0
0
2 : 1 0 0 0 ,
3
3 0 0 0
2 1 3 0 5
3 : 1 5 0 0 3 5 ,
3 0 5 0 3
5 1 3 0 2
4 : 1 2 0 0 3 2.
3 0 2 0 3
x
x
y
y z
z
x x y
y x z
z y z
x x y
y x z
z y z
Solution: Now think about a 2-D matrix as a rotation. We make the identification
cos sin
5 5
2 1 sin cos
C
So the matrix C can be identified as a (passive) rotation through
1
sin 2 5 1.
radians 63.4degrees.
Solution: Here we practice finding eigenvalues and eigenvectors. We have for the
eigenvalues
1 2 1
1 2 2
1 3 Tr 3 1
2 2 det 4 4
A
A
A
So we can determine the eigenvectors (up to an arbitrary normalization and phase;
here we choose unit vectors) to be
1 1
2 2
x x x y x
v
y y x y y
x x x y x
v
y y x y y
Note that, since the original matrix is not Hermitian, the eigenvectors are not
orthogonal. Using Mathematica we find
Ex 3.11:
A={{1,3},{2,2}}
{{1,3},{2,2}}
MatrixForm[A]
1 1
2 2
x x x y x x y
y y x y y x y v
z z z z
x x x y x x y
y y x y y x y v
z z z z z
3 3
x x x y x x y
y y x y y x y v
z z z z z
Note that in this case the original matrix is Hermitian (symmetric and real) and the
eigenvectors are orthogonal. Using Mathematica we find
Ex 3.11:
A={{2,3,0},{3,2,0},{0,0,1}}
{{2,3,0},{3,2,0},{0,0,1}}
MatrixForm[A]
Eigenvalues[A]
{5,-1,1}
Eigenvectors[A]
{{1,1,0},{-1,1,0},{0,0,1}}
Solution: Finally consider a more complicated (but still Hermitian) 3x3 matrix. We
have
2 3 0
3 2 0
0 0 1
2 2 2 1 2 3
3 2 2 3 2 2
2 1 3 2 1 3
2 3 1 2 3 1
3 1 9 2 2 1 6 2 6 2 1
3 2 8 8 2 3 4 2 8 2
2 20 2 5 4
2
0 4.
5
A
Now we can proceed to find the corresponding eigenvectors by solving the equations
(and checking that we get orthogonal vectors, but specified only up to a normalization
and phase, here we choose unit vectors). We find
1
1
2
3 2 2 3 2 2 2 2
2 : 2 1 3 2 2 3 2 3 2
2 3 1 2 3 2 3 2
0
0
1
1 ,
2
1
3 2 2
4 : 2 1 3 4
2 3 1
x x x y z x y z x
y y x y z y y z x
z z x y z z y z x
x
v
y z
x x
y y
z z
2
3 2 2 4 4 2
2 3 4 5 3 2
2 3 4 3 5 2
4
1
1 ,
4 3 2
1
x y z x y z x
x y z y y z x
x y z z y z x
y z
v
x y
2 2 2
2
2 3 2 2
1
2
3
x
x y z y x y z xy xz yz
z
So we know that, expressed in terms of the principal axes, the coefficient matrix for
the equation of the surface will be diagonal with these eigenvalues down the diagonal
(although the choices of which axis corresponds to the specific axes, x ’, y ’ or z ’ is
arbitrary, as we can interchange any pair with a further rotation). We have one choice
as (note that, as expected, the trace and determinant of the matrix are unchanged)
2 2 2
x
x y z y x y z
z
