The problem states: Use elimination to find the determinant of the matrix $$ \left[ \begin{matrix}{} 1&1/2&0&0\\ 0&4&4&8\\ -2&2&3&0\\ 0&0&1&-5\\ \end{matrix} \right] $$
So am I supposed to use elimination, get it to a triangular form and multiply the numbers on the diagonal? If so, which row operations do I use for accomplishing that? Thank you in advance! (EDIT) After elimination the upper triangular matrix I get is: $$ \left[ \begin{matrix}{} 1&1/2&0&0\\ 0&4&4&8\\ 0&0&1&-5\\ 0&0&0&-6\\ \end{matrix} \right] $$ However, the determinant of this is -24 and the answer that I am supposed to get is 24 (found out via calculator), or so I think. Am I doing this right?
Yes, it's $24$.
$$ \begin{bmatrix}{} 1&1/2&0&0\\ 0&4&4&8\\ -2&2&3&0\\ 0&0&1&-5\\ \end{bmatrix} = \begin{bmatrix}{} 1&1/2&0&0\\ 0&4&4&8\\ 0&3&3&0\\ 0&0&1&-5\\ \end{bmatrix}= \begin{bmatrix}{} 1&1/2&0&0\\ 0&4&4&8\\ 0&0&0&-6\\ 0&0&1&-5\\ \end{bmatrix}=\begin{bmatrix}{} 1&1/2&0&0\\ 0&4&4&8\\ 0&0&1&-11\\ 0&0&1&-5\\ \end{bmatrix}=\begin{bmatrix}{} 1&1/2&0&0\\ 0&4&4&8\\ 0&0&1&-11\\ 0&0&0&6\\ \end{bmatrix}.$$
Alternatively, when you get to $$ \begin{bmatrix}{} 1&1/2&0&0\\ 0&4&4&8\\ 0&0&0&-6\\ 0&0&1&-5\\ \end{bmatrix}, $$ swap the two last lines, but take note of a change of sign.