Exercise 2.1.1
a)

u=Table[i!,{i,5}]

b)

v=Array[#^#&,5]

c)

w={-1,1,-1,1,-1}

Exercise 2.1.2
a)

u.v

b)

(u.u)^(1/2)

%//N

c)

norm[a_]:=(a.a)^(1/2)

norm[u-v]+norm[v-w]-norm[u-w]//N

d)

norm[u]norm[v]-(u.v)//N

Exercise 2.1.3

$Post:=If[MatrixQ[#],MatrixForm[#],#]&
A=Array[Max,{3,5}]

u=Array[#!&,{3}]

v=Array[(-1)^#&,{5}]

a)

u.A

b)

A.u : not defined

c)

v.A : not defined

d)

A.v

Exercise 2.1.4

M=Table[Append[A[[i]],u[[i]]],{i,3}]

R=RowReduce[M]

Clear[x]
Do[x[i]=R[[i,6]],{i,1,3}]
x[3]+=-5x[5]/3-4x[4]/3;
{x[1],x[2],x[3],x[4],x[5]}

Exercise 2.1.5

Transpose[Append[Transpose[A],u]]

Exercise 2.1.6

xA=u <==>Transpose[A].x=u

u={1,2,3,4,5};
LinearSolve[Transpose[A],u]

Since the rank of A is less or equal 3, we check
that this is indeed the most general solution:

RowReduce[Transpose[Append[A,u]]]

Exercise 2.1.7

There is no such example, i.e. associativity holds,
and one can check it with symbolic matrices/vectors
(valid for other dimensions than 5, too):

Clear[a,b,u];
A=Array[a,{5,5}];
B=Array[b,{5,5}];
U=Array[u,5];
Together[(A.U).B-A.(U.B)]

Exercise 2.1.8

A=Array[Max,{5,5}];
Id=IdentityMatrix[5];
Inv=Array[0&,{5,5}];

We solve by appending the identity matrix one column
at a time and solving by row reduction, for each one
column of the inverse. No check for the row reduction
is made here, since the inverse is verified in the end
by multiplication.
The symmetry of both A and Id is used to skip trans-
posing. Same for the inverse of A.

Do[Inv[[j]]=Transpose[RowReduce[
            Transpose[Append[A,Id[[j]]]]]][[6]],{j,1,5}];
Inv

A.Inv