I'm in a linear algebra class and we're doing determinants right now. I got this matrix to do:
$\begin{matrix} 2 & 1 & 0 & 0 & 0 \\ 3 & -1 & 2 & 0 & 0 \\ 0 & 4 & 1 & -1 & 2 \\ 0 & 0 & -3 & 2 & 4 \\ 0 & 0 & 0 & -1 & 3 \end{matrix}$
It wouldn't be hard to solve via row reduction but I hate to see so many 0s in such a near symetric form go to waste... I had problems close to the one I posted above where I figured out through googling about shcurs, which is convenient if you can cancel the ugly term! This one doesn't have the proper 0 positioning for that though, although it's very close. What's a nice way to solve this without rowreduction?
This one isn't going to be very friendly, but aside from row-reduction (really the best method) a not-too-terrible way to proceed would be as follows:
Start by cofactor expansion along the first row, so that
$$\left|\begin{matrix} 2 & 1 & 0 & 0 & 0 \\ 3 & -1 & 2 & 0 & 0 \\ 0 & 4 & 1 & -1 & 2 \\ 0 & 0 & -3 & 2 & 4 \\ 0 & 0 & 0 & -1 & 3 \end{matrix}\right|=2\cdot\left|\begin{matrix} -1 & 2 & 0 & 0 \\ 4 & 1 & -1 & 2 \\ 0 & -3 & 2 & 4 \\ 0 & 0 & -1 & 3 \end{matrix}\right|-1\cdot\left|\begin{matrix} 3 & 2 & 0 & 0 \\ 0 & 1 & -1 & 2 \\ 0 & -3 & 2 & 4 \\ 0 & 0 & -1 & 3 \end{matrix}\right|$$ The second $4\times 4$ cofactor matrix is very friendly, and we find that $$\begin{align}\left|\begin{matrix} 3 & 2 & 0 & 0 \\ 0 & 1 & -1 & 2 \\ 0 & -3 & 2 & 4 \\ 0 & 0 & -1 & 3 \end{matrix}\right| &= 3\cdot\left|\begin{matrix}1 & -1 & 2 \\ -3 & 2 & 4 \\ 0 & -1 & 3 \end{matrix}\right|\\ &= 3\cdot\bigl(6+0+6-0-9-(-4)\bigr)\\ &=3\cdot 7\\ &= 21,\end{align}$$ using the general formula for determinant of a $3\times 3$ matrix, so that $$\left|\begin{matrix} 2 & 1 & 0 & 0 & 0 \\ 3 & -1 & 2 & 0 & 0 \\ 0 & 4 & 1 & -1 & 2 \\ 0 & 0 & -3 & 2 & 4 \\ 0 & 0 & 0 & -1 & 3 \end{matrix}\right|=2\cdot\left|\begin{matrix} -1 & 2 & 0 & 0 \\ 4 & 1 & -1 & 2 \\ 0 & -3 & 2 & 4 \\ 0 & 0 & -1 & 3 \end{matrix}\right|-21.$$ The first $4\times 4$ cofactor isn't as nice, but again using cofactor expansion along the first row, we have $$\begin{align}\left|\begin{matrix} -1 & 2 & 0 & 0 \\ 4 & 1 & -1 & 2 \\ 0 & -3 & 2 & 4 \\ 0 & 0 & -1 & 3 \end{matrix}\right| &= -1\cdot\left|\begin{matrix} 1 & -1 & 2 \\ -3 & 2 & 4 \\ 0 & -1 & 3 \end{matrix}\right|-2\left|\begin{matrix} 4 & -1 & 2 \\ 0 & 2 & 4 \\ 0 & -1 & 3 \end{matrix}\right|\\ &= -1\cdot7-2\cdot 4\cdot\left|\begin{matrix} 2 & 4 \\ -1 & 3 \end{matrix}\right|\\ &= -7-2\cdot 4\cdot\bigl((2)(3)-(4)(-1)\bigr)\\ &=-7-80\\ &=-87,\end{align}$$ so $$\left|\begin{matrix} 2 & 1 & 0 & 0 & 0 \\ 3 & -1 & 2 & 0 & 0 \\ 0 & 4 & 1 & -1 & 2 \\ 0 & 0 & -3 & 2 & 4 \\ 0 & 0 & 0 & -1 & 3 \end{matrix}\right|=2\cdot-87-21=-174-21=-195.$$