matricne jednacine

Kemal Halilović,profesor matematike Brčko

1

 Matrične jednačine

1. Rije{iti matri~nu jednad`bu  A  X 

gdje je

=  B  A−1 B,

1 2 1    A= 1 1 0   2 0 − 1

RJE[ENJE:

1 1 1    B= 0 0 0   0 0 0

−1

−1

A X = B A − B, −1

/ A−

−1

−1

A − AX = A − BA − B

(

X = A −1 B

)

2

-1  Na|imo prvo inverznu matricu matricu A-1

A−

1

 1   0  2 1 = adjA = −1 − det A  0  2  1 

 − 1 − 2 − 1 − 1 1 − 3 1  = − 2 4 − 1

0

−

−1

1

0

2 −1 1 1

1

−1 1

2

−

0

−1 1 1 1 0

1

1 

T



2 0  1 2

1

2

1

det A = 1

1

0 = −1

2

0

−1 T

 − 1 1 − 2   −  = −1 − 2 − 3 4  = 2 0  − 1 1 − 1 1 2   1 1 

2 1  1 − 1 3  1 −    2 − 4 1 

Sada je 2

2

      1   1 2 1  1 1 1  1 1   1 1  1 1 1  1             X =  −1 3 −1 ⋅ 0 0 0  =  −1 −1 −1  = −1 −1 −1 . −1 −1 −1 =           2   2 − 4 1   0 0 0        2 2 2 2 2 2 2 2                 2 2   2  − 2 − 2 − 2    4 4 4 

Kemal Halilović,profesor matematike Brčko

2

2. Rije{iti matri~nu jednad`bu: (2B − X)A = I − 3A ,gdje je  1 0 1   A = 2 −1 1 ,    3 2 2

 1 0 2   B= −3 1 0    0 1 2

RJE[ENJE RJE[ENJE 

 ,Ijedini~na matrica.  ,I - jedini~na

(2B − X )A = I − 3A (2B − X )AA−1 = (I − 3A )A −1 −1 2B − X = (I − 3A )A

(

−1 −1 X = 2B − IA − 3AA

)

−

1 X = 2B − A − + 3I

−

X = 2B + 3I − A − 1

Uvrstimo date matrice u jedna~inu:  1 0 2   1 0 0  1 0 1        X = 2 − 3 1 0 + 3 0 1 0 − 2 −1 1        0 1 2 0 0 1  3 2 2  2 0 4  3 0 0   1 0 1        X − 6 2 0 + 0 3 0 − 2 −1 1        0 2 4 0 0 3  3 2 2   5 0 4  1 0 1      X = −6 5 0 − 2 −1 1      0 2 7   3 2 2   −1   2  5 0 4   1 0 X = − 6 5 0 − −  3   2  0 2 7   0  −1 

1 2 1 2 1 1

−

2

1

3 2 1 1 3 2 1 1

−

2 1

−1

−1

−1

 5 0 4 1 0 1  1   =  − 6 5 0 − adj 2 − 1 1   det A  0 2 7  3 2 2 T

−1  T 3 2   5 0 4 − 4 − 1 7  1 0   1  −  =  − 6 5 0  −  2 − 1 − 2 3 2 3  0 2 7  1 1 − 1 1 0   2 −1 2

 19  3 1   5 0 4 − 4 2    1  − 1 − 1 1  = − 17 X= −6 5 0 −  3   3   0 2 7   7 − 2 − 1  7  − 3

−

2

3 16 3 8 3

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11  3  1

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