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    Solution Du TP02

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    Naceredine Nacer
    The document contains the solution to a math problem involving matrix operations. It provides the definitions of matrices A, B, C and builds a larger matrix M by combining A, B and C. It then works through various operations on the matrices including multiplication, addition, powers, transposes and extracting submatrices.

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    Solution Du TP02

    Uploaded by

    Naceredine Nacer
    14 views2 pages
    The document contains the solution to a math problem involving matrix operations. It provides the definitions of matrices A, B, C and builds a larger matrix M by combining A, B and C. It then works through various operations on the matrices including multiplication, addition, powers, transposes and extracting submatrices.

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    Solution du TP02

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    The document contains the solution to a math problem involving matrix operations. It provides the definitions of matrices A, B, C and builds a larger matrix M by combining A, B and C. It then works through various operations on the matrices including multiplication, addition, powers, transposes and extracting submatrices.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    14 views2 pages

    Solution Du TP02

    Uploaded by

    Naceredine Nacer
    The document contains the solution to a math problem involving matrix operations. It provides the definitions of matrices A, B, C and builds a larger matrix M by combining A, B and C. It then works through various operations on the matrices including multiplication, addition, powers, transposes and extracting submatrices.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 2

    Solution du TP N° 02

    1 3 
     
    1 2 3 2  0 1 
    A    , B   , C   
    7 2 0 1  1 1
     
     4 8 

    Partie 1

    1 2 3 2  0 3 
    a) A * B  3      3   
    7 2 0 1  18 15

    1 2 3 2 3 4  0 7


    b) A .* B  3    
     3   
     3   
    7 2 0 1  0 2  3 1

    1 2 1 2 1 1 14 5 


    c) A ^ 2 ones (2)       
    7 2 7 2 1 1 20 17

    1 3  0 0 2 6


         
     0 
    1  3 2
      
    0 0  1 2
    d) C * B 1  zeros (4, 2)     1  0 0  2 
    1 1 0 1     2
       
     4 8  
    0 0  13
     1

    1 7 1 7 1 49


    e) A '. ^ 2     
    2 2 2 2 4 4 

    1 3   
      1 3 
    0 1  1 0  0 1 
    f) C *eye (2)       
    1 1 0 1 1 1
       
     4 8   4 8 

    1 3 1 3 1 0


    g) C (1: 2,:) ^ 2     
     0 1  0 1 0 1

    0 1   0 1   0 1 
         
    h) C (2 : end ,:). ^ 2  1 11 1   1 1 
         
     4 8   4 8  16 64

    1
    Partie 2

    1 2 1 3 
     
    7 2 0 1 
    M   
    3 2 1 1
     
    0 1 4 8 

    1) M  [[A ; B ],C ]

    2)
    1 2 1 3 
     
    7 2 0 1 
    a) M(3,2)=3 → M   
    3 3 1 1
     
    0 1 4 8 

    b) M(2,:)-7*M(1,:)= 7 2 0 1 - 71 2 1 3 =

    0 12 7 20

    2 1 3 
     
    2 0 1 
    c) M(:,1)=[] → M   
    3 1 1
     
    1 4 8 

    2 1 3 
     
    d) M(end,:)=[] → M  2 0 1 
     
    3 1 1

     0  1 0 0 2 0 0 
    2 0     
    e) tril(M).*eye(3)= 2 0 0 0 1 0 → 0 0 0 
    
        
    3 1 1 0 0 1 0 0 1

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