Notes on Homework 5
Some of the computational aspects of the problems have been omitted here.
1.
(a)
1 2 −1 6
2 0 3 4
(b) Start row-reducing:
1 2 −1 6
2 0 3 4
1 2 −1
6
0 −4 5 −10
6
1 2 −1
0 1 −5/4 5/2
1 0 3/2
2
0 1 −5/4 5/2
(c) Then z is a free variable, and the most general solution to the system of equations is


2 − 23 z
5/2 − 5 z
4
z
or
  3 

−2z
2
5/2 + − 5 z .
4
0
z
2.
(a) We calculae A−1 by row-reducing ( A| I ):

1 4 0 1
 0 0 1 0
2 2 6 0

1 4 0 1
 0 0 1 0
0 −6 0 −2

1 4 0 1 0
 0 1 0 1/3 0
0 0 1 0 1

1 0 0 −1/3 0
 0 1 0 1/3 0
0 0 1
0
1
Professor Jeff Achter
Colorado State University

0 0
1 0 
0 1

0 0
1 0 
0 1

0
−1/6 
0

2/3
−1/6 
0
M369 Linear Algebra
Fall 2008


−1/3 0 2/3
Therefore, A−1 =  1/3 0 −1/6.
0
1
0
(b) May as well compute by expanding along the second row:
4 0
1 0
1 4
2+2
2+3
2+1
+ (−1) 0 · det
+ (−1) 1 · det
det( A) = (−1) 0 · det
2 6
2 6
2 2
= 6.
(c) We know that det( A) det( A−1 ) = det( AA−1 ) = det( I3 ) = 1, so det( A−1 ) = det( A)−1 =
1/6.
 
 
2
3
3. The vectors v1 = 2 and v2 = 6 are linearly independent if the only solution to
9
9
 
0
x1 v1 + x2 v2 = 0 is x1 = x2 = 0. So, try to solve the system
0
 


0
2 3 2 6 x1 = 0 .
x2
0
9 9
The augmented matrix is


2 3 0
 2 6 0 ;
9 9 0
its reduced row-echelon form is


1 0 0
 0 1 0 .
0 0 0
From this, we see that the only solution is x1 = x2 = 0, and thus the vectors are linearly
independent.
For similar reasons, we compute the RREF of the augmented matrix


1 2 3 0
 3 1 −1 0  ,
−1 4 9 0
which is
Professor Jeff Achter
Colorado State University


1 0 −1 0
 0 1 2 0 .
0 0 0 0
M369 Linear Algebra
Fall 2008
The third variable is free, which indicates that there is a nontrivial solution to the problem of representing 0. In particular (use z = 1, y = 0 − 2z = −2, x = z = 1, we
have
 
     
1
2
3
0
 3  − 2 1 + −1 = 0
−1
4
9
0
and the vectors are linearly dependent.
(b)
(a) Could do the same calculation, or simply observe that any set of vectors containing the
zero vector is always linearly dependent; in particular,
0
1
0
0
1
+0
+0
=
0
0
1
0
and the vectors are linearly dependent.
4. We need to show that for each 1 ≤ i, j ≤ n, the i j entry of α ( A + B) is equal to the i j entry of
αA + αB. We compute each of these separately, and hope they agree. On one hand,
(α ( A + B))i j = α ( A + B)i j def scalar-matrix mult
= α ( Ai j + Bi j ) def matrix add
= αAi j + αBi j distributive law for numbers.
On the other hand,
(αA + αB)i j = (αA)i j + (αB)i j def matrix add
= αAi j + αBi j def scalar-matrix mult – twice!.
Therefore, α ( A + B) = αA + αB.
5. Suppose {v1 , · · · , vs } are linearly independent; we want to show that any subset of them
is also linearly independent. After relabeling or reordering, we assume that the subset is
{v1 , · · · , vr } ⊆ {v1 , · · · , vr , vr+1 , · · · , vs }, with r ≤ s. Suppose there are numbers { a1 , · · · , ar }
with
r
∑ ai vi = 0;
i =1
we need to show that each ai = 0. But we can extend this linear relation to one involving all
the v j ’s:
a1 v1 + · · · + ar vr + 0 · vr+1 + · · · + 0 · vs = 0.
Since {v1 , · · · , vs } is linearly independent, each of these coefficients must be zero. In particular, each of a1 , · · · , ar is equal to zero.
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008
6.
(a) Compute:
λa λb
det(λA) = det
λc λd
= λaλd − λbλc
= λ 2 ( ad − bc)
a b
2
= λ det
.
c d
(b) We proceed by induction on n; the base case n = 2 was provided by (a). As an inductive
hypothesis, suppose it’s known that for any B ∈ Matn,n (R), det(λB) = λ n B. We want
to show that if A ∈ Matn+1,n+1 (R), then det(λA) = λ n+1 det( A).
So, compute by expanding along the first row:
n+1
det(λA) =
∑ (−1)1+ j (λA)1 j det((λA)(1| j))
j=1
But (λA)1 j = λ · A1 j , while (λA)(1| j) = λ · ( A(1| j)). (Check this!) So
n+1
=
∑ (−1)1+ j λ · A1 j · det(λA(1| j))
j=1
Now apply the inductive hypothesis to each matrix λA(1| j), since each of them is a
n × n matrix:
n+1
=
∑ (−1)1+ j λA1 j λn det A(1| j)
j=1
= λ n+1
n+1
∑ (−1)1+ j A1 j det A(1| j)
j=1
= λ n+1 det( A).
7. Since the determinant of A isn’t zero, A has an inverse, A−1 . Therefore, for any b ∈ R4 , we
can solve the equation Ax = b by letting x = A−1 b. But solving the equation Ax = b is
exactly the same as demonstrating that b is in the column space of A. So, yes, every element
of R4 is in the column space of A.
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008