$Post:=If[MatrixQ[#],MatrixForm[#],#]&

Exercise 1.4.1

A=Array[(1/(#1+#2-1))&,{5,5}];
B=Array[(#1+#2-1)&,{5,6}];
Unprotect[C]; C=Table[1/(j+1),{i,5},{j,6}];

a)

A.B

b)

A.(B+C)

c)

A.B+A.C

d)

A.(B-C)

e)

A.B-A.C

f)

A.(B.C) : not defined

g)

(A.B).C : not defined

h)

MatrixPower[A,5]

i)

Inverse[A].MatrixPower[A,6]

j)

Id[n_]:=Array[If[#1==#2,1,0]&,{n,n}]

A+3Id[5]

k)

Id[5].A

l)

MatrixPower[A,3]-3A+Id[5]

Exercise 1.4.2

L=Array[Min,{2,2}];
M=Array[Max,{2,2}];
M.L-L.M

L=Array[Min,{3,3}];
M=Array[Max,{3,3}];
M.L-L.M

L=Array[Min,{10,10}];
M=Array[Max,{10,10}];
M.L-L.M

L=Array[Min,{11,11}];
M=Array[Max,{11,11}];
M.L-L.M

Exercise 1.4.3

g[i_,j_]=i^2-j^2;
A=Array[g,{3,3}]

p[A]=MatrixPower[A,3]+98A

If A were invertible, then MatrixPower[A,3]+98A = 0 ==>
A.(MatrixPower[A,2]+98Id) = 0 ==>
Inverse[A].A.(MatrixPower[A,2]+98Id) = Inverse[A].0 ==>
MatrixPower[A,2]+98Id = 0, which is not true:

MatrixPower[A,2]+98Id[3]

Now, see what error message Mathematica gives:

Inverse[A]

Exercise 1.4.4

A=Array[Max,{5,5}]

p[x]=Det[Id[5]x-A]

-5Id[5]-39A-98A.A-85A.A.A-15A.A.A.A+A.A.A.A.A

From p[A]=0 we have A.(-39Id-98A-85A.A-15A.A.A+A.A.A.A)=5Id
    ==> (-39Id-98A-85A.A-15A.A.A+A.A.A.A)/5 = Inverse[A]

(-39Id[5]-98A-85A.A-15A.A.A+A.A.A.A)/5

Inverse[A]           

Exercise 1.4.5

A:=Array[Random[Integer,{-10,10}]&,{4,4}];

Do[Print[Inverse[A]//MatrixForm],{i,1,10}]

The probability for random integer matrices (in a given
range) to be invertble is quite high, as our example
shows. For matrices of real numbers ("true reals", as 
opposed to their computer representations), that proba-
bility actually equals 1.

Exercise 1.4.6

A=Array[a,{3,3}];
B=Array[b,{3,3}];
C=Array[c,{3,3}];
lhs=(A+B).C;
rhs=A.C+B.C;
Simplify[lhs-rhs]

Exercise 1.4.7

A=Array[a,{5,5}];
lhs=(k*Id[5]).A;
mhs=k*A;
rhs=A.(k*Id[5]);
lhs-mhs

rhs-mhs

Exercise 1.4.8

Myrand:=Random[Integer,{-100,-1}];
Table[Myrand/Myrand,{i,5},{j,5}]

Exercise 1.4.9
a)

f:=Function[{i,j},If[i<=j,Myrand,0]]

b)

L=Array[f,{6,6}]

M=Array[f,{6,6}]

c)

L.M

Exercise 1.4.10

a)

g:=Function[{i,j},If[i<j,Myrand,0]]

b)

A=Array[g,{5,5}]

c)

MatrixPower[A,5]

d)

A.A.A

MatrixPower[A,4]

A=Array[g,{6,6}]

MatrixPower[A,3]

MatrixPower[A,6]

We observe that, if A is a strictly upper triangular 
n*n matrix, MatrixPower[n,i] will have exactly n-i
nonzero lines/columns.

Exercise 1.4.11

From our experiments in exercise 10, we can conjecture
that a strictly upper triangular n*n matrix will be
nilpotent of index <= n, since MatrixPower[A,n-1] can
have one nonzero element in the upper right corner,
while MatrixPower[A,n] = 0.

Exercise 1.4.12

ut:=Function[{i,j},If[i<j,x[10i+j],0]]

T=Array[ut,{5,5}]

a)

Clear[$Post]
T.T

T.T.T

MatrixPower[T,4]

The powers 2,3,4 of T are not the 0 matrix. Moreover,
they follow the pattern that we observed in our random
matrices.

b)

MatrixPower[T,5]//MatrixForm

Exercise 1.4.13

By induction, we'll show that T^i is stricly upper
triangular and its first i-1 superdiagonals are 0,
if T is strictly upper triangular, which obviously
suffices for our purpose.
i=1 statement follows immediately from T=T^1 strictly
upper triangular.
Suppose T^i strictly upper triangular and having i-1
superdiagonals equal 0. Then the j-th row of T^i will
have (at least) Min[j+i-1,n] leading 0's, whereis the
k-th column of T will always have (at most) k-1 nonzero
entries, from top. That is, in T^(i+1)=(T^i).T, the
[[j,k]]-th entry will be 0 for all j+i >= k, which
proves our statement.