Tutorial week 15
1. Row Vectors and Column Vectors. List the row vectors and the column vectors of the Matrix
2
3
1
Solutions: the row vector are simply
1 0
5 7
4 2
1
1
7
2 1 0 1 3 5 7 1 1 4 2 7 the column vectors are
2
3
1
1
5
4
0
7
2
1
1
7
2. Express the product Ax as a linear combination of the column vectors of A
4
3
0
0
6
1
1
2
4
2
3
5
Solution: the product Ax is
4
3
0
0
6
1
1
2
4
2
3
5
13
22
17
we want to show that
a
4
3
0
b
This is clearly true for the coeficients a
2
4
3
0
0
6
1
c
2, b
3
0
6
1
1
2
4
3 and c
5
13
22
17
5
1
2
4
13
22
17
3. Determine if (a) b is in the column space of A and if so (b) express b as a linear combination of the
column vectors of A.
1 3
2
A
b
4 6 10 1
Solution:
(a) To determine if b is in the column space of A we need to show that Ax
the system of equations
x1
3x2
4x1 6x2
which gives the results
b is consistent. We solve
2
10
x1
1
x2
1
(b) because the column space of A is define by all linear combinations of the column of A. So if b is a
vector of this space then
2
1
3
k
l
10 4 6 where k l are some reals to be determine. This equation
implies that
2
k
3l
4k 6l
10
has to be well defined. Indeed we can solve this system
and find that
k
and
4. Suppose that x1
linear system Ax
1
l
1
1, x2 2, x3 4, x4 3, constitute a particular solution of a inhomogeneous
b and the homogeneous system
Ax 0 is given by the formulas
x1
x2
4s
r s
x3
3r
r
x4
s
Find the vector form of the general solution of the inhomogenous of Ax 0. Solution: According to
our notes the solution of the general solution of the inhomogeneous equation, x, and the homogenous
one, xh are related via the particular solution of the inhomogenous
equation, x 0 , via the formula
x
x
3r 4s
r s
r
s
x0
xh
r
3
1
1
0
s
4
1
0
1
5. Find a basis for (a) the nullspace, (b) the row space and (c) the column space of A
1 2 0
3 6 1
A
2
(a) the nullspace is defined by the solution of the equation
Ax
x1
2x2
3x1
6x2
0
x3
the solution is find by assumig a value of lets say, x 1
the solution is
x
0
0
s, which fixes x2
s
2
and therfore x3
0 so
1
1
2
s
0 then
0
and therefor the basis S for the nullspace is made of only one vector
S
1
1
2
0
(b) the row-echelon matrix R is found simply by multiplying the first row by 3 and subtracted to the
second one
1 2 0
R
0 0 1 with leading
The basis for the row space consist of the rows
1’s so
1 2 0 0 0 1 S
(c) The basis of the column space of A will correspond to the equivalent columns that constitute the
basis of the column space of R. So if the basis of the column space of R is
1
0
W
0
1
that is the columns with leading 1’s. The basis of the column space of A will correspond to the equivalent
columns in A:
1
0
S
3 1 3