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    how do i do part b?
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    (Original post by jakaloupe)

    how do i do part b?
    Show that  \text{det} \mathbf{A} \neq 0  for all real values of a ( \forall a, a\in \mathbb{R} ).
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    (Original post by B_9710)
    Show that  \text{det} \mathbf{A} \neq 0  for all real values of a ( \forall a, a\in \mathbb{R} ).
    so if i solve it and find the roots -2+6i and -2-6i then does that prove it? thus has a solution and therefore is non singular
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    (Original post by jakaloupe)
    so if i solve it and find the roots -2+6i and -2-6i then does that prove it? thus has a solution and therefore is non singular
    Well technically the roots are -2\pm \sqrt6i therefore the det(A)\not=0 for all a\in\mathbb{R} thus showing that the matrix is non-singular for all real a.
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    (Original post by RDKGames)
    Well technically the roots are -2\pm \sqrt6i therefore the det(A)\not=0 for all a\in\mathbb{R} thus showing that the matrix is non-singular for all real a.
    oh ok
 
 
 
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