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

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

Originally Posted by kmjt

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...

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?

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...

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.

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

But what would I do from there? sorry

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?

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.