0.8 Number 7
We'll show that uu^{T} is singular by showing that the system (uu^{T})X has
a non-trivial solution. Let's consider two cases. First, if u_{1}=0, then
is a solution, since
On the other hand, if , then
is a non-trivial solution to (uu^{T})X=0, since
In either case (u_{1}=0 or ), we've exhibited a non-trivial
solution, so the coefficient matrix uu^{T} is singular.
0.9 Number 1
so the matrix is invertible, and its inverse is
0.9 Number 2
Since the first row is twice the second, the matrix is not invertible.
0.9 Number 7
Since the left block of the reduced matrix has a row of zeroes, the given
matrix is singular.
0.9 Number 8
so the the matrix is invertible, and its inverse is
0.10.1 Number 3
0.10.1 Number 7
0.10.2 Number 3
Subtracting twice the last row from the first gives
A cofactor expansion along the last column gives
Expanding in cofactors along the last row gives
0.11 Number 1
a
Try to solve the system
The coefficient matrix row reduces to
which is the coefficient matrix of an equivalent homogeneous system, which
can be easily solved for x_{1} and x_{2} in terms of x_{3}:
Taking x_{3}=1 gives the solution
and the dependency relation
b
The vectors are linearly independent since
c
The matrix
row reduces to the identity matrix, so the columns are independent.
d
The matrix
row reduces to
As in part a, the vectors are linearly dependent, with dependence relation
e
The matrix
row reduces to
so the system AX=0 has the solution
giving the dependence relation
f
The matrix
row reduces to
giving the dependence relation
g
The matrix with the given columns row reduces to a identity
matrix, so the columns are linearly independent.
h
The matrix with the given columns row reduces to
which is the coefficient matrix of a system with the non-trivial solution
This gives the dependency relation