arian487 said:
Thanks frederick, that definitely helps a lot. Let me make sure I understand:
If A = [2 0 0; 1 5 4; 1 2 4] (arranged by column) then my determinant is simply 2 * 5 * 4?
No, that's incorrect. Expanding on the first row the determinant is
\left|\begin{array}{ccc}2 & 0 & 0\\1 & 5 & 4\\ 1 & 2 & 4\end{array}\right|= 2\left|\begin{array}{cc}5 & 4 \\ 2 & 4\end{array}\right|= 2(20- 8)= 24
because all of the other multipliers, the other numbers in the first row, are 0.
If a matrix is "upper triangular" (all numbers below the main diagonal are 0), you can expand on the first column so you have only the number in "first row first column" times a smaller determinant, which can be expanded on the first column, etc. giving just the product of the numbers on the diagonal:
\left|\begin{array}{ccc}a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{array}\right|= a\left|\begin{array}{cc}d & e \\ 0 & f\end{array}\right|= a(d f)= adf
If a matrix is "lower diagonal" (all numbers
above the main diagonal are 0), then expanding on the first row repeatedly gives the same thing.
Finally, you can always "row reduce" a matrix until it is "upper triangular" remembering that:
1) Swapping two rows multiplies the determinant by -1
2) Multiplying an entire row by a number multiplies the determinant by that same number (so you have to
divide the determinant resulting triangular matrix by that number to recover the determinant of the original matrix).
3) Adding or subtracting a multiple of one row from another does
not change the determinant.
