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lapo4ka [179]
3 years ago
9

The system of equation, if a b are arbitrary numbers x+2y-3z- a 2x+4y-6z 2a+2 has (A) No solutions regardless of values of a and

b (B) Infinitely many solutions regardless of values of a and b (C) a unique solution if a b-0 D) a unique solution regardless of values of a and b
Mathematics
1 answer:
zepelin [54]3 years ago
8 0

Answer:

(A) No solutions regardless of values of a and b.

Step-by-step explanation:

Asumming that the system of equations is x+2y-3z=a\\ 2x+4y-6z=2a+2, the corresponding augmented matrix of the system is \left[\begin{array}{cccc}1&2&-3&a\\2&4&-6&2a+2\end{array}\right].

If two time the row 1 is subtracted to row 1, we get the following matrix

\left[\begin{array}{cccc}1&2&-3&a\\0&0&0&2a+2-2a\end{array}\right].

Then the system has no solutions regardless of values of a and b.

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