Linear Algebra Homework: Matrix Determinants & Eigenvalues

This assignment solution covers a range of linear algebra concepts, including matrix determinants, non-singularity, and trace calculations. It delves into matrix operations, such as finding cofactors, adjoints, and inverses. The solution also explores eigenvalues and eigenvectors, demonstrating ho...Read More

Solution: Given matrix is
a)
Note that, if A is any 3 by 3 matrix such that , then by using Laplace-
expansion by column 3, the determinant of the matrix A is
So for the given matrix,
Hence, the determinant of given matrix is 0.
b (i): we know that a matrix is non-singular if its determinant is non zero. Since the
determinant of the given matrix is 0. Hence, matrix A is not singular, that is matrix is
singular.
b (ii): The trace of A10 is the sum of diagonal elements of matrix A10.
c): Given matrix is
(i): The determinant of matrix B is
(ii): Cofactors
(iii): the cofactor matrix is
So, the adjoint of the matrix B is the transpose of matrix P. that is
(iv): The inverse of matrix B is

Solution 3:
Given matrix is
(a):
To find eigenvalues of A, solve this implies that
This implies that the eigenvalues are
(b): Since, the largest eigenvalue is . So solve we get
Now, apply gauss elimination method to find value of eigenvector u.
We get
Suppose that
This implies that
Hence, for , eigenvector corresponding to the eigenvalue is
Therefore, the eigenvector of unit length corresponding to the largest eigenvalue of A is
(c): Since, the 2nd largest eigenvalue is . So solve we get
Now, apply gauss elimination method to find value of eigenvector u.
We get
Suppose that
This implies that

Hence, for , eigenvector corresponding to the eigenvalue is
Therefore, the eigenvector of unit length corresponding to the 2nd largest eigenvalue of A
is
(d): Since and are unit vectors because norm of both vectors is 1. And the inner
product
Hence, eigenvectors are orthonormal.
(e): We know that if two vectors are orthonormal, then they are linearly independents.
From part (d), the vectors are linearly independents.
(f): We know that the determinant of the matrix is the product of eigenvalues. Since, the
eigenvalues of A are 0, 1 and 3. This implies that
. Since, determinant is 0, so the rank of matrix must be less that 3 and
hence rank of matrix is 2.
Solution 4a: To prove that if A and B are similar matrices, they have the same
eigenvalues.
Since, A and B are similar matrices, there exists a nonsingular matrix C such that C−1AC
 B.
Suppose that PA( ) and PB( ) be the characteristic polynomials of matrices A and B
respectively. This implies that
Since this implies that
… (1)
We know that if the characteristics equations of two matrices are same then their
eigenvalues are also same. From equation (1) , hence, eigenvalues of two matrices
A and B are same. This completes the proof.
Solution 4b: To prove that if A is an idempotent matrix, the eigenvalues of A are either 0
or 1.
Suppose that be an eigenvalues of idempotent matrix A and x be the corresponding
non zero Eigen vector. This implies that
.
Since, matrix A is idempotent matrix this implies that .
Now,
And,

Form equation (1) and equation (2) we get
Since, x is non zero,
Solution 4c: Given that
This implies that
We know that and also we know that trace is the sum of
diagonal elements of the given matrix. So,