format long e
x

x =

    9.500000000000000e-001

diary off
z

z =

    1.140000000000000e+000

diary off
% to suspend the diary or just to flush the
% buffers we do diary off and then diary on
% The file created by diary is just a text file,
% only good for us, users.  To save the variables
% in the workspace, we do save:
save 'matdata1'
clear
load 'matdata1'
uiimport('C:\Documents and Settings\janv\My Documents\matdata1.mat')
x

x =

    9.500000000000000e-001

% by default, all variables are of type double
format short
x

x =

    0.9500

pi

ans =

    3.1416

format long
ans

ans =

   3.141592653589793

% There is only one working precision:
% hardware machine precision.
% It is very easy and convenient to work
% with matrices in MATLAB
a = rand(3,4)

a =

  Columns 1 through 2

   0.814723686393179   0.913375856139019
   0.905791937075619   0.632359246225410
   0.126986816293506   0.097540404999410

  Columns 3 through 4

   0.278498218867048   0.964888535199277
   0.546881519204984   0.157613081677548
   0.957506835434298   0.970592781760616

format short
a

a =

    0.8147    0.9134    0.2785    0.9649
    0.9058    0.6324    0.5469    0.1576
    0.1270    0.0975    0.9575    0.9706

help rand

rand('seed',123)
rand(2,2)

ans =

    0.0878    0.0986
    0.6395    0.6906

rand(2,2)

ans =

    0.3415    0.2641
    0.2359    0.6044

rand('seed',123)
rand(2,2)

ans =

    0.0878    0.0986
    0.6395    0.6906

a

a =

    0.8147    0.9134    0.2785    0.9649
    0.9058    0.6324    0.5469    0.1576
    0.1270    0.0975    0.9575    0.9706

% Suppose that a contains the coefficient matrix
% and righthandside vector for a linear system.
A = a(:,1:3)

A =

    0.8147    0.9134    0.2785
    0.9058    0.6324    0.5469
    0.1270    0.0975    0.9575

b = a(:,4)

b =

    0.9649
    0.1576
    0.9706

x = A\b

x =

   -2.5775
    3.0365
    1.0462

r = b - A*x

r =

  1.0e-015 *

    0.4441
    0.1110
   -0.2220

norm(r)

ans =

  5.0877e-016

% The norm of the residual vector
% showed that we solved a linear system
% and obtained a solution with accuracy
% close to the machine precision.
[l,u] = lu(A)

l =

    0.8995    1.0000         0
    1.0000         0         0
    0.1402    0.0258    1.0000


u =

    0.9058    0.6324    0.5469
         0    0.3446   -0.2134
         0         0    0.8863

[l,u,p] = lu(A)

l =

    1.0000         0         0
    0.8995    1.0000         0
    0.1402    0.0258    1.0000


u =

    0.9058    0.6324    0.5469
         0    0.3446   -0.2134
         0         0    0.8863


p =

     0     1     0
     1     0     0
     0     0     1

l*u

ans =

    0.9058    0.6324    0.5469
    0.8147    0.9134    0.2785
    0.1270    0.0975    0.9575


p*A

ans =

    0.9058    0.6324    0.5469
    0.8147    0.9134    0.2785
    0.1270    0.0975    0.9575

l*u

ans =

    0.9058    0.6324    0.5469
    0.8147    0.9134    0.2785
    0.1270    0.0975    0.9575

[v,d] = eig(A)

v =

    0.6752   -0.7134   -0.5420
   -0.7375   -0.6727   -0.2587
   -0.0120   -0.1964    0.7996


d =

   -0.1879         0         0
         0    1.7527         0
         0         0    0.8399

v*d

ans =

   -0.1269   -1.2503   -0.4552
    0.1386   -1.1790   -0.2173
    0.0023   -0.3443    0.6715

A*v

ans =

   -0.1269   -1.2503   -0.4552
    0.1386   -1.1790   -0.2173
    0.0023   -0.3443    0.6715

inv(v)*A*v

ans =

   -0.1879    0.0000    0.0000
   -0.0000    1.7527   -0.0000
   -0.0000   -0.0000    0.8399

% Exploiting A*v = v*d, we obtain
% a diagonalization procedure for A.
diary off