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Vectors and Matrices Watch

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    (Original post by r3l3ntl3ss)
    I'm still really confused, so if \alpha = 1 then \beta = -1; would you just say that x = \frac{2 + 2\beta}{\alpha - 1} as the relation?
    Yeah a=1 and b=-1 is one answer.

    I think the relation is a=3+2b.
    (a-1=2+2b add 1 to both sides)

    That's giving you all of the values of a and b such that the matrix is consistent and has at least one solution, so I think that is the final answer.
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    (Original post by rayquaza17)
    Yeah a=1 and b=-1 is one answer.

    I think the relation is a=3+2b.
    (a-1=2+2b add 1 to both sides)

    That's giving you all of the values of a and b such that the matrix is consistent and has at least one solution, so I think that is the final answer.
    ah okay thanks; this question dumbfounded me for a long time... doesn't help when there's no solutions to papers at uni haha.
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    (Original post by r3l3ntl3ss)
    ah okay thanks; this question dumbfounded me for a long time... doesn't help when there's no solutions to papers at uni haha.
    No problem.
    I know what you mean. Lecturers that give solutions, or even just numerical answers are my faves.
 
 
 
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