Pure Math - Matrix

1997年7月1日

2013-10-25 2:14 pm

Determine, with reasons, those values of a (if any) such that the augmented system

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has i) a unique solution, ii) no solution, iii) infinitely many solutions.

Thank you very much!

回答 (1)

用戶10012

2013-10-25 3:14 pm

✔ 最佳答案

= ( 1 - 2 a | 1 ) R2 - R1 to R2
( 0 a + 2 1 - a | 0 )
( a 4 10 | 4 )

= ( 1 - 2 a | 1 ) R3 - a x R1 to R3
( 0 a + 2 1 - a | 0 )
( 0 4 + 2a 10 - a^2 | 4 - a )

= ( 1 -2 a | 1 ) R3 - 2 x R2 to R3
( 0 a + 2 1 - a | 0 )
( 0 0 8 - a^2 + 2a | 4 - a )

= ( 1 - 2 a | 1 )
( 0 a + 2 1 - a | 0 )
( 0 0 (4 - a)(a + 2) | 4 - a )
When a = 4 , the last row becomes ( 0 0 0 | 0), there will be infinitely many solutions.
When a = - 2, the last row becomes (0 0 0 | 6 ), so there is no solution.
Other than a = 4 or - 2, the system has unique solution.

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