Mathematics: Matrix Equations and Diagonalization

Running head: MATHEMATICS
MATHEMATICS
Name of the Student
Name of the University
Author Note

1MATHEMATICS
Question 1:
The population of rabbits and foxes are related by the following equations.
rn +1= 4 r n−2 f n
3
f n+1= rn + f n
3
a) The equations can be represented by a matrix form
[ rn +1
f n +1 ]=
[ 4
3
−2
3
1
3
1
3 ] [ rn
f n ]
Here, M =
[ 4
3
−2
3
1
3
1
3 ]
b) The inverse of the matrix M is calculated by the following formula.
M−1 = Adj(M)/Det(M)
Where, Adj(M) =
[ 1
3
−1
3
2
3
4
3 ]T
=
[ 1
3
2
3
−1
3
4
3 ]
Det(M) = (4/3)*(1/3) + (2/3)*(1/3) = 4/9 + 2/9 = 6/9 = 2/3.
Hence, M−1 = [ 1
3
2
3
−1
3
4
3 ](2/3)
=
[ 1
2 1
−1
2 2 ]
c) The eigenvalues of M are calculated from the following equation.

2MATHEMATICS
det(M – λI) = 0
Here, I is the identity matrix.
Hence, det(
[ 4
3
−2
3
1
3
1
3 ]− [ λ 0
0 λ ]) = 0
Now, the eigenvalues found from the above equation are λ 1 = 1 and λ 2 = 2/3.
d) Now, the eigenvectors for each of the eigenvalues are calculated by constructing the
augmented matrix for each Eigen value and then converting it to row echelon form and then
obtaining the eigenvectors from system of linear equation.
The augmented matrix is given by,
(M – λ 1∗I : 0) =
[ 4
3 −1 −2
3 0
1
3
1
3 −1 0 ] =
[ 1
3
−2
3 0
1
3
−2
3 0 ]
Now, converting to row echelon form
[ 1
3
−2
3 0
1
3
−2
3 0 ] = [ 1
3
−2
3 0
0 0 0 ] (R2 = R2 – R1)
Hence, x1*(1/3) – x2*(2/3) = 0 => x1 = 2*x2
Hence, the eigenvector x = [ 2 x 2
x 2 ] = [ 2
1 ] x 2
For any real x2 ≠ 0, [2
1 ] x 2 are the eigenvectors associated with the eigenvalue λ 1 = 1.
Now, similarly for λ 2 = 2/3

3MATHEMATICS
(M – λ 1∗I : 0) =
[ 4
3 − 2
3
−2
3 0
1
3
1
3 − 2
3 0 ] =
[ 2
3
−2
3 0
1
3
−1
3 0 ] = [ 2
3
−2
3 0
0 0 0 ](R2 [= R2 – (½)R1)
Hence, in this case eigenvector y = [ y 2
y 2 ] = [ 1
1 ]y2
For any real y2 ≠ 0, [1
1 ] y 2 are the eigenvectors associated with the eigenvalue λ 2 = 2/3.
e) Now, for diagonalization of matrix M the linearly independent eigenvectors are found.
From the above it can be seen that x = [ 2
1 ] and [ 1
1 ] are linearly independent.
Hence, the invertible matrix P = [x ; y] = [2 1
1 1 ]
Now, the diagonal matrix has the eigenvalues in the diagonal position.
Hence, D = [ 1 0
0 2
3 ]
Hence, the matrix M can be diagonalized by,
P∗D∗P−1 = M
It can be seen that by matrix multiplication and calculating P−1 gives
[ 4
3
−2
3
1
3
1
3 ]which is the
matrix M.
f) Now, the determinant of M = product of eigenvalues = (2/3)*1 = 2/3.
Also, the eigenvalues of diagonal matrix D is the diagonal elements.