Vectors in 2-Space and 3-Space : Properties of the determinant function
2. Verify that det(AB) = det(A) det(B) for
A= 2 1 0
3 4 0
0 0 2
and
B = 1 -1 3
7 1 2
5 0 1
Is det(A+B) = det(A) + det(B) ?
Proof. First, we compute AB and A+B.
1
9

AB   31 1
10 0

8 
3


17  and A  B  10
5
2 

0
5
0
3

2
3 
So, we expand the |AB| along the 3rd row and get
1
8
9
1
+2
det( AB)  10
 10*(-25)+2*40=-170
1
31
17
1
If we expand the |A+B| along the 2nd column, then we have
3
3
det( A  B)  5
 5(9  15)  30 .
5
3
Secondly, we calculate det(A) and det(B).
Expanding det(A) along the 3rd row, we have
2
1
=2(2*4-3*1)=10
det( A)  2
4
3
and expanding det(B) along the 3rd row, we get
1
3 1
1
+
det( B)  5
 5*(-5)+8=-17
1
2 7
1
It is clear that det(A)*det(B)=10*(-17)=-17=det(AB).
However, det(A+B)=-30 which is NOT equal to det(A)+det(B)=-7.
5. Let A = a b c
d e f
g h i
Assuming that det(A) = -7, find
b) det(A-1)
e) det a g d
b h e
c i f
Solution.
b) As AA-1 =I, where I is the identity matrix, we have
det(A)*det(A-1)=det(I)=1
So,
det(A-1)=1/det(A)=1/(-7)=  17
e) We know that
g
a d


AT   b e
h  and det( AT )  det( A)  7 .
c f
i 

By the properties of the determinant, if we exchange the 2nd and 3rd columns, then we
have
a
d
a
g
g
d
b
e  b
e
h   det( AT )  (7)  7
h
c
f
i
c
f
i
So,
a
g
d
b
h
e 7
c
i
f
9. Prove the identity without evaluation the determinants.
a1 + b1
a2 + b2
a3 + b3
a1 – b1
a2 – b2
a3 – b3
c1
c2
c3
=
a1 b1 c1
a2 b2 c2
a3 b3 c3
-2
Proof. By the properties of the determinant, if we add the 2nd column onto the 1st column,
then we have
a1  b1 a1  b1 c1
2a1 a1  b1 c1
a 2  b2
a 2  b2
c 2  2a 2
a 2  b2
c2
a 3  b3
2a 3 a 3  b3 c3
a 3  b3 c3
Then we take out a factor 2 from the 1st column, we have
a1  b1 a1  b1 c1
a1 a1  b1 c1
a 2  b2
a 2  b2
c2  2 a2
a 2  b2
c2
a 3  b3
a 3 a 3  b3 c3
a 3  b3 c3
st
Now we multiply the 1 column by -1, then we add the 1st column onto the 2nd column,
so we have
a1  b1 a1  b1 c1
a1  b1 c1
a 2  b2
a 3  b3
a 2  b2
c2  2 a2
 b2
c2
a 3  b3 c3
a 3  b3 c3
Finally, we take out a factor (-1) from the 2nd column, we have
a1  b1
a1  b1
c1
a1
b1
c1
a 2  b2
a 2  b2
b2
c2
a 3  b3
a 3  b3
c 2  2 a 2
c3
a3
b3
c3
16. Let A and B be n x n matrices. Show that if A is invertible,
then det(B) = det(A-1BA)
Proof. We use a fact: For the square matrices A and B, det(AB)=det(A)*det(B).
Hence,
det(A-1BA)= det(A-1)*det(B)*det(A). …………………..(1)
Note that det(A-1)* det(A)= det(A-1A)=det(I)=1 …………………..(2)
Note: I is the identity matrix whose determinant is 1.
So, by (1)(2), we have
det(A-1BA)= det(B)
as desired.
18. Prove that a square matrix A is invertible if and only if ATA is invertible.
Proof. We use the following theorem (which can be found on any book of Linear
Algebra): Theorem: A square matrix A is invertible if and only if |A| is non-zero.
First, we prove that, if a square matrix A is invertible, then ATA is invertible.
As A is invertible, by the above theorem, we know that | A | 0 . As
| AT A || AT || A || A | 2  0 , we conclude by the above theorem that ATA is invertible.
Secondly, we prove that, if ATA is invertible, then A is invertible.
As ATA is invertible, by the above theorem, we know that | AT A | 0 . As
| AT A || AT || A || A | 2 , we conclude that | A | 0 . So, by the above theorem, we know
that A is invertible.
10. a) In the accompanying figure, the area of the triangle ABC can be expressed as
area ABC = ½ x1 y1 1
x2 y2 1
x3 y3 1
Note: In the derivation of this formula, the vertices are labeled such that the
triangle is traced counterclockwise proceeding from (x1, y1) to (x2, y2) to (x3, y3). For a
clockwise orientation, the determinant above yields the negative of the area.
Proof. We consider two vectors AB and AC. As
AB=(x2 -x1, y2 -y1 ) and AC=(x3 -x1, y3 -y1 ), the cross product of these two
vectors is
AB  AC  (x2 -x1, y2 -y1 )  (x3 -x1, y3 -y1 )

=[( x2 -x1 )(y3 -y1 )- (x3 -x1 )(y2 -y1 )] k
So, the area of the triangle ABC is
1
1
( x2 -x1 )(y3 -y1 )- (x3 -x1 )(y2 -y1 )] ……………….(1)
2 | AB  AC | 2 [
It is easy to prove the right hand side of (1) is equal to
½ | x1
| x2
| x3
We are done.
y1
y2
y3
1|
1|
1|
Note: Though the vectors AB and AC are 2-dimensional, we can think of them as three
dimensional, namely,
AB=(x2 -x1, y2 -y1, 0) and AC=(x3 -x1, y3 -y1, 0)
b) Use the result in (a) to find the area of the triangle
with vertices (3,3), (4,0), (-2, -1).
C(x3, y3) 
·
B(x2, y2)

A(x1, y1)
D
E
F
Figure Ex-10
Solution. Use the formula in (a), we can get the area of this triangle as follows.
3
3
1
Area   12 4
0
1
1 1 1
We evaluate it by the properties of the determinant. Adding the 3rd column onto 1st and
2nd columns, we get
3
1
4 4 1
3
1
1
Area   2 4
1 2 5 1 1
0
1 1 1
0 0 1
Then we expand the last determinant along the 3rd row. We have
3
3
1
Area   12 4
0
1
  12
1 1 1
4
4
5
1
8
Euclidean Vector Spaces: Euclidean n-Space
6. Let u = (4, 1, 2, 3), v = (0, 3, 8, -2), and w = (3, 1, 2, 2). Evaluate each expression.
a) u + v
b) u + v
c) -2u + 2 u
d) 3u – 5v + w
e) 1 w
w
f) 1 w
w
Solution. a) As u+v=(4, 4, 10, 1), we have
|| u  v || 4 2  4 2  10 2  12  133 ;
b) || u ||  || v || 4 2  12  2 2  32 
0 2  3 2  8 2  (2) 2  30  77 ;
c) || 2u || 2 || u || 4 || u || 4 4 2  12  2 2  32  4 30
d) As 3u-5v+w=3(4, 1, 2, 3)-5(0, 3, 8, -2)+ (3, 1, 2, 2)=(15, -11, -32, 21), we have
|| 3u  5v  w || 15 2  (11) 2  (32) 2  212  1811
e) As || w || 32  12  2 2  2 2  3 2 , we have
1
||w||
w  3 1 2 (3,1,2,2)  (
f) || ||w1 || w ||
1
||w||
1
2
, 3 12 ,
2
3
,
2
3
)
|| w || 1
16. Find two vectors of norm 1 that are orthogonal to the tree vectors u = (2, 1, -4),
v = (-1, -1, 2, 2), and w = (3, 2, 5, 4).
The question is NOT well-defined, as u is 3-dimensional, but v and w are 4-dimensional
vectors. Please check and give me the correct one so that I can help with it. Thanks.
20. Find u  v given that u + v = 1 and u – v = 5
Solution. Note that || u  v || 2 || u || 2  || v || 2 2u  v and || u  v || 2 || u || 2  || v || 2 2u  v .
Hence,
|| u  v || 2  || u  v || 2  4u  v
i.e.,
12  5 2  4u  v
So,
i.e.,
 24  4u  v
u  v  24 / 4  6
24. Prove the following generalization of Theorem 4.1.7. If v1, v2, …, vr are pairwise
orthogonal vectors in Rn, then
v1 + v2 + … + vr
2
= v1
2
+ v2
2
+ … + vr
2
Proof. Note that || v1  v2  ...  vr || 2   v1  v2  ...  vr , v1  v2  ...  vr  . According to
the properties of the inner product, we have
 v1  v2  ...  vr , v1  v2  ...  vr 
=  v1 , v1  v2  ...  vr  +  v2 , v1  v2  ...  vr  +…+  vr , v1  v2  ...  vr 
…………………..(1)
Now we evaluate the first term on the right hand side(RHS):  v1 , v1  v2  ...  vr 
 v1 , v1  v2  ...  vr  =  v1 , v1  +  v1 , v2  +…+  v1 , vr 
As v1, v2, …, vr are pairwise orthogonal vectors, we have
 v1 , v2  =…=  v1 , vr  =0
Hence,  v1 , v1  v2  ...  vr  =  v1 , v1 || v1 || 2 . Similarly, the second term on RHS is
 v2 , v1  v2  ...  vr  = || v2 || 2
…
and the last term equals  vr , v1  v2  ...  vr  = || v r || 2
So, together with (1), we get
 v1  v2  ...  vr , v1  v2  ...  vr  = ||v1 || 2 + ||v2 ||2 + … + ||vr||2
As || v1  v2  ...  vr || 2   v1  v2  ...  vr , v1  v2  ...  vr  , we complete our proof.
26. Use the Cauchy-Schwarz inequality to prove that for all real values of a, b, and ,
(a cos + b sin)2  a2 + b2
Proof.
The Cauchy-Schwarz inequality is
| u, v ||| u ||  || v ||
……………………………….(1)
for any vectors u and v. Now letting u=(a,b) and v=(cos, sin), we have
 u, v  a cos   b sin  ,
|| u || a 2  b 2 ,
and
|| v || (cos  ) 2  (sin  ) 2  1  1.
Hence, together with (1), we get
| a cos   b sin  | a 2  b 2
Then squaring both sides, we get
(a cos   b sin  ) 2  a 2  b 2