determinant=-9+4k.
\begin{bmatrix}<br />-3 & -k\\ <br />4 & 3<br />\end{bmatrix})
just multiply the top column only to perhaps save time and equate it to 3 since 3 is what's on the top right column of the matrix it's meant to be the same as.
27-12k=3
so k=2.
edit: beaten..
Your method is good, but you've found the inverse wrong - should be one over the determinant. Granted, since you have

, you can do this:
\begin{bmatrix}<br />3 & k\\ <br />-4 & -3<br />\end{bmatrix}<br />=<br />\begin{bmatrix}<br />-3 & -k\\ <br />4 & 3<br />\end{bmatrix}<br />\implies<br />12k-27=-3 \implies k=2)
You still came up with the right answer, however you would've lost a mark for calculating the inverse wrong.
Otherwise, don't be worried about having been beaten to putting up an answer - looking at a different way of doing things is always great, and I certainly encourage you to post a second answer to something I've answered if you answer it in a different way (if any of that made sense

)