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

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• Oct 23rd 2011, 11:47 PM
kmjt
For what values of k does A^-1 exist?
http://img.photobucket.com/albums/v1...4at34549AM.png

Another hw question i'm having trouble with (Headbang) How would I go about doing it?
• Oct 23rd 2011, 11:50 PM
Prove It
Re: For what values of k does A^-1 exist?
Quote:

Originally Posted by kmjt
http://img.photobucket.com/albums/v1...4at34549AM.png

Another hw question i'm having trouble with (Headbang) How would I go about doing it?

A square matrix has an inverse if its determinant is nonzero...
• Oct 24th 2011, 12:05 AM
kmjt
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?
• Oct 24th 2011, 12:20 AM
Prove It
Re: For what values of k does A^-1 exist?
Quote:

Originally Posted by kmjt
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...

$\displaystyle \left|\begin{matrix}a&b&c\\d&e&f\\g&h&j\end{matrix }\right| = a\left|\begin{matrix}e&f\\h&j\end{matrix}\right| - b\left|\begin{matrix}d&f\\g&j\end{matrix}\right| + c\left|\begin{matrix}d&e\\g&h\end{matrix}\right|$
• Oct 24th 2011, 12:20 AM
Deveno
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.
• Oct 24th 2011, 02:10 AM
kmjt
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
• Oct 24th 2011, 02:24 AM
Prove It
Re: For what values of k does A^-1 exist?
Quote:

Originally Posted by kmjt
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?
• Oct 24th 2011, 02:28 AM
Deveno
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.