generate a random matrix:
>>A=rand(4)

A = 0.8147    0.6324    0.9575    0.9572
    0.9058    0.0975    0.9649    0.4854
    0.1270    0.2785    0.1576    0.8003
    0.9134    0.5469    0.9706    0.1419

 type help lu to see documentation of LU decomposition.

Perform an LU decomposition of A: USE THE COMMAND

>> [LL,UU,PP] = lu(A)

to get

LL =
    1.0000         0         0         0
    0.9917    1.0000         0         0
    0.8920   -0.3250    1.0000         0
    0.1390   -0.4552    0.2567    1.0000
UU =
    0.9134    0.5469    0.9706    0.1419
         0   -0.4448    0.0024    0.3447
         0         0    0.0925    0.9426
         0         0         0    0.6955

As you can see LL and UU have the structure we expect. 
The matrix PP is called the Permutation Matrix: this performs ROW EXCHANGES.
matlab “performed” row exchanges before applying elementary row operations (save for
the row exchanges) to get an LU factorization. More details on this? take a numerical
linear algebra course.
 
PP =
     0     0     0     1
     0     1     0     0
     1     0     0     0
     0     0     1     0

===================================================

the lu command in matlab can be used differently:
Take another matrix,

A =
    0.8147    0.6324    0.9575    0.9572
    0.9058    0.0975    0.9649    0.4854
    0.1270    0.2785    0.1576    0.8003
    0.9134    0.5469    0.9706    0.1419
Perform the decomposition now this way:

>> [LL,UU]=lu(A)
LL =
    0.8920   -0.3250    1.0000         0
    0.9917    1.0000         0         0
    0.1390   -0.4552    0.2567    1.0000
    1.0000         0         0         0
UU =
    0.9134    0.5469    0.9706    0.1419
         0   -0.4448    0.0024    0.3447
         0         0    0.0925    0.9426
         0         0         0    0.6955
note: see help on LU to understand why LL looks "strange" 

Check that A=LL*UU
>>LL*UU
ans =
    0.8147    0.6324    0.9575    0.9572
    0.9058    0.0975    0.9649    0.4854
    0.1270    0.2785    0.1576    0.8003
    0.9134    0.5469    0.9706    0.1419

So if you want L to be explicit, use [LL,UU,PP] = lu(A), rather than
[LL,UU] = lu(A)


=======================================
One way to invert a matrix (but not the recommended way) is by
>>inv(A)
ans =
  -15.2997    3.0761   14.7235    9.6445
   -0.2088   -1.8442    1.0366    1.8711
   14.5694   -1.9337  -14.6497   -9.0413
   -0.3690    0.5345    1.4378   -0.4008

The more general way is that Ainverse = A\I, where I is the identity
matrix of the same size as A.

Generate an identity matrix of size 4X4:
>>B = eye(4);

Perform the inversion:
>>Ainv=A\B
Ainv =
  -15.2997    3.0761   14.7235    9.6445
   -0.2088   -1.8442    1.0366    1.8711
   14.5694   -1.9337  -14.6497   -9.0413
   -0.3690    0.5345    1.4378   -0.4008

======================================
Computing determinants via matlab is a bad idea...it is very unstable
numerically. Take a numerical analysis class to learn more. But for
small balanced matrices it's ok.

Generate a special matrix called a Hilbert matrix (type help hilb)

>>A=hilb(4)
A =
    1.0000    0.5000    0.3333    0.2500
    0.5000    0.3333    0.2500    0.2000
    0.3333    0.2500    0.2000    0.1667
    0.2500    0.2000    0.1667    0.1429

compute the determinant:
>>det(A)
ans =
   1.6534e-07
===================================
Generate another matrix
>>AA=hilb(3)
AA =
    1.0000    0.5000    0.3333
    0.5000    0.3333    0.2500
    0.3333    0.2500    0.2000

compute the rank of AA:
>> rank(AA)
ans = 3
compute the null space of AA
>>null(AA)
ans =
   Empty matrix: 3-by-0

Make a copy of AA call it C:
>>C=AA
replace the 3rd row of C by zeros:
>>C(3,:)=zeros(1,3)
C =
    1.0000    0.5000    0.3333
    0.5000    0.3333    0.2500
         0         0         0
compute rank of C
>>rank(C)
ans= 2
compute null space of C:
>>null(C)
ans =
    0.1170
   -0.7022
    0.7022
compute the det of C
det(C)
ans = 0

==============================


>> A = hilb(4)
Compute eigenvalues and eigenvectors of matrix:
>>[VV,DD]=eig(A)

VV =
   -0.6621   -0.7149    0.1745    0.1821
   -0.4819   -0.0292   -0.6291   -0.9288
   -0.2766    0.6972    0.5995    0.3178
   -0.5029   -0.0438   -0.4630    0.0570
DD =
    2.4021         0         0         0
         0   -0.0346         0         0
         0         0   -0.7158         0
         0         0         0   -0.4400
Compute eigenvalues/vectors of A transpose = A'	 
>>[VVp,DDp]=eig(A')
VVp =
   -0.5308   -0.6543   -0.3872   -0.0825
   -0.3371    0.1107   -0.1442   -0.6424
   -0.5902    0.6246   -0.3503   -0.0637
   -0.5062    0.4117    0.8406    0.7593
DDp =
    2.4021         0         0         0
         0   -0.0346         0         0
         0         0   -0.7158         0
         0         0         0   -0.4400
>>rank(A)
ans =
ans =    4
>>rank(A')
ans = 4