I've been given this question in class, with the 3×3 matrix:
$$\begin{bmatrix}
2& 1& 0\\
1 & 2 & 1\\
0 & -3 & k\end{bmatrix}.$$
My job here is to find the values of $k$ for which this matrix has an inverse. Now in class the only method I've been taught how to do this is Gaussian Elimination, placing this matrix alongside the identity matrix (all entries are zero, except for ones along the diagonal) and performing row operations on both in order to find the inverse.
My attempts at this usually result in the last row being $k$ plus or minus some number, like $k+2$. Is not assigning a number to $k$ until I've "isolated" it in it's row correct?
Any advice would be appreciated.
Best Answer
If you want to do it by row reduction, then you just go ahead: $$ \begin{bmatrix}2&1&0\\1&2&1\\0&-3& k\end{bmatrix}\xrightarrow{} \begin{bmatrix}1&2&1\\2&1&0\\0&-3& k\end{bmatrix}\xrightarrow{} \begin{bmatrix}1&2&1\\0&-3&-2\\0&-3& k\end{bmatrix}\xrightarrow{} \begin{bmatrix}1&2&1\\0&-3&0\\0&0& k+2\end{bmatrix} $$ So for the matrix to be invertible we need $k+2\ne0$, that is $k\ne-2$.