The following is a problem from Apostol, Calculus, Volume II (p. 80).
The question requests we compute the determinant of
$$ \begin{pmatrix}
a & 1 & 0 & 0 & 0 \\
4 & a & 2 & 0 & 0 \\
0 & 3 & a & 3 & 0 \\
0 & 0 & 2 & a & 4 \\
0 & 0 & 0 & 1 & a
\end{pmatrix}$$
by transforming it to an upper triangular matrix.
Is there an efficient way to transform this to an upper triangular matrix? I can go about row-reducing which should eventually get me there (my attempts at that have lead to extremely long and tedious calculation), but the form of the matrix suggests there might be a faster way? I can't seem to find it. (One can, of course, compute the determinant by just doing the expansion, but I'm more curious if there is a fast way to get to the upper triangular matrix the problem requests.)
Best Answer
This matrix is called a tridiagonal matrix.
In this case, you should be able to turn this matrix into an upper triangular matrix with just four elementary row operations*, all row additions. Row additions don't effect the determinant, so you just have to multiply out the elements on the diagonal of the resulting upper triangular matrix and get the determinant of the result. The standard method of the LU decomposition people are taught does the same thing... there is something called Crout's method for the LU decomposition as well that is easier to calculate out, but it is most likely simpler just to row reduce it.
*it doesn't look like you'll have to do any pivoting at a quick glance. Pivoting just multiplies the determinant by -1.