Notes on Homework 2
1.
(a)
2 −3 1 8
1 −2 −2 4
(b)
 
x
2 3
1  
8
y =
1 −2 −2
4
z
2. Suppose ~v ∈ Rn is a solution to the homogeneous matrix equation
Ax = 0.
Suppose λ ∈ R. Show that λ~v is also a solution to this equation.
We need to show that A · (λv) = 0, i.e., that each component of this vector is zero. But the
ith component is
n
( A · λv)i =
∑ ai j · (λv) j
j=1
n
=
∑ ai j · λv j definition of λv
j=1
n
=λ
∑ ai j v j
j=1
= λ · ( A · v)i def of Av
= λ · 0 since Av = 0
= 0.
3. We need to show that for 1 ≤ i ≤ m and 1 ≤ j ≤ p, the (i, j) entries of BD + CD and of
( B + C ) D are the same. On one hand, we have
( BD + CD )i j = ( BD )i j + (CD )i j def matrix addition
n
n
k =1
n
k=1
= ( ∑ Bik Dk j ) + ( ∑ Cik Dk j ) def matrix mult
=
∑ (Bik Dk j + Cik Dk j )
k =1
since the indexing sets for the summations are the same; k goes from1 to n
n
=
∑ (Bik + Cik )Dk j
k =1
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008
because multiplication distributes.
On the other hand, the i j entry of ( B + C ) D is
n
(( B + C ) D )i j =
∑ (B + C)ik Dk j def matrix mult
k =1
n
=
∑ (Bik + Cik )Dk j def matrix addition
k =1
which is the same as before. Therefore, ( B + C ) D = BD + CD.
4. You should, of course, check that these work.


1 0 0
(a) 0 α 0.
0 0 1


1 0 β
(b) 0 1 0 .
0 0 1
5.
(a) Multiply out;
AB =
2 4
3 1
1
− 10
3
10
2
5
− 15
3
1
) + 4 10
2( 25 ) + 4(− 51 )
2(− 10
=
1
3
3(− 10
) + 1 10
7 3( 52 ) + 1(− 51 )
1 0
=
0 1
(b) We need to solve
x
3
A
=
.
y
1
So, left-multiply by B:
x
3
=
y
1
x
3
BA
=B
y
1
1
x
= 10
7
y
10
A
Check that this solution works.
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008
6.
(a) Start row-reducing:

1 0 3 1 0
 2 1 8 0 1
1 2 7 0 0

1 0 3 1 0
 0 1 2 −2 1
0 2 4 −1 0

0
1 0 3 1
 0 1 2 −2 1
0 0 0 3 −2

0
0 
1

0
0 
1

0
0 
1
The zeros in the last row signify
  that there’s no inverse. (Somewhat more precisely, the
0
problem of solving Ax = 0 is inconsistent.)
1
(b) Apply the same method to


2 0 4 1 0 0
 0 6 0 0 1 0 
6 0 8 0 0 1
and obtain
so that the desired inverse is
Professor Jeff Achter
Colorado State University

1 0 0 −1 0 12
 0 1 0 0 1 0 
6
0 0 1 34 0 − 14


−1 0 21
 0 1 0 
6
3
0 − 14
4

M369 Linear Algebra
Fall 2008