Page 10 - M A T R I K S

a. AB b. BA c. BC d. AC e. (AB)C f. A(BC)
g. B + C h. A(B + C) i. AB + AC j. AI k. IA

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1 2   4  0  − 4 0 +10   0 10 
Jawab : a. AB =     =   =  
 0 3   − 2 5   0 − 6 0 +15   − 6 15 
 4 0   1  2  4 + 0 8 + 0   4 8 
b. BA =     =   =  
 − 2 5   0 3   − 2 + 0 − 4 +15   − 2 11 
 4 0   3 −  2 12 + 0 − 8 +  0 12 −  8
c. BC =     =   =  
 − 2 5   1 4   − 6 + 5 4 + 20   −1 24 

1 2   3 −  2  + 2 − 2 +  8 5 6 
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d. AC =     =   =  
 0 3   1 4   0 + 3 0 +12   3 12 

 0 10   3 −  2  0 +10 0 + 40  10 40 
e. (AB)C =     =   =  
 − 6 15   1 4   −18 +15 12 + 60   − 3 72 

1 2   12 −  8 12 − 2 − 8 + 48  10 40 
f. A(BC) =     =   =  
 0 3   −1 24   0 − 3 0 + 72   − 3 72 
 4  0 3 −  2  4 + 3 0 + ( −2  )  7 −  2
g. B + C =   +   =   =  
 − 2 5   1 4   − 2 +1 5 + 4   −1 9 

1 2   7 −  2  − 4 − 2 +18   3 16 
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f. A(B + C) =     =   =  
 0 3   −1 9   0 − 3 0 + 27   − 3 27 
 0 10   5 6  15 +14 18 +16  29 34 
g. AB + AC =     =   =  
 − 6 15   3 12   5 + 28 6 + 32   33 38 

1 2   1  0  + 01 0 +  2 1  2
g. AI =     =   =   …
 0 3   0 1   0 + 0 0 + 3   0 3 

1 0   1  2  + 01 2 +  0 1  2
h. IA =     =   =   …
 0 1   0 3   0 + 0 0 + 3   0 3 

Sifat-sifat perkalian matriks :
1. Umumnya tidak komutatif (AB  BA)
2. Asosiatif : (AB)C = A(BC)
3. Distributif kiri : A(B + C) = AB + AC
Distributif kanan : (B + C)A = BA + CA
4. Identitas : IA = AI = A
5. k(AB) = (kA)B

LATIHAN SOAL

1. Sederhanakan !
5  3 1   4 6   1  1
a. − 3  4   b.  1 −  2   c.    − 
 2   0 − 4   − 8 − 9   2  3
0 3    5 3 −  4  3  5 2 4   2 −1  4
d.     e.     f.    
 4 −1   1   1 0   − 2 1   1 3   3 0 2 

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