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Math Help - For what values of k does A^-1 exist?

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    For what values of k does A^-1 exist?



    Another hw question i'm having trouble with How would I go about doing it?
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    Re: For what values of k does A^-1 exist?

    Quote Originally Posted by kmjt View Post


    Another hw question i'm having trouble with How would I go about doing it?
    A square matrix has an inverse if its determinant is nonzero...
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    Re: For what values of k does A^-1 exist?

    Normally I am good with inverses and determinants however the letters are confusing me. All the k in there is what confuses me, how would I approach it?
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    Re: For what values of k does A^-1 exist?

    Quote Originally Posted by kmjt View Post
    Normally I am good with inverses and determinants however the letters are confusing me. All the k in there is what confuses me, how would I approach it?
    The same way you would with any other 3 x 3 square matrix. k is just some number...

    \displaystyle \left|\begin{matrix}a&b&c\\d&e&f\\g&h&j\end{matrix  }\right| = a\left|\begin{matrix}e&f\\h&j\end{matrix}\right| - b\left|\begin{matrix}d&f\\g&j\end{matrix}\right| + c\left|\begin{matrix}d&e\\g&h\end{matrix}\right|
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    Re: For what values of k does A^-1 exist?

    det(A) = (k)(4)(k) + (k)(k^2)(0) + (0)(k^2)(k) - (0)(4)(0) - (k^2)(k)(k) - (k)(k)(k^2)

    you do the rest.
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    Re: For what values of k does A^-1 exist?

    I calculated det(A) = -2k^4 + 4k^2

    But what would I do from there? sorry
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    Re: For what values of k does A^-1 exist?

    Quote Originally Posted by kmjt View Post
    I calculated det(A) = -2k^4 + 4k^2

    But what would I do from there? sorry
    What values of k will make the determinant = 0?
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    Re: For what values of k does A^-1 exist?

    you might find it helpful to factor 4k^2 - 2k^4 = 0 as

    2k^2(2 - k^2) = 0, first. this polynomial in k has 3 distinct roots, which we cannot allow k to be, if our matrix is to be invertible.
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