The determinants of upper triangular matrices (For any 2x2 and 3x3 matrix)

Question

I am trying hard to figure out what am I supposed to do, if I am supposed to go on write a conjecture about the particular question. How can I go on about to prove it?

Answer 1

Write out formula for determinant of 2x2 and 3x3 matrix. Then put elements under the diagonal to zero. What do you get? The determinant is the product of the elements on the diagonal.

For 2x2 case, this means following:

$$ \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - bc $$

In upper triangular matrix, $c=0$. Thus the determinant is equal to $ad$.

You can use induction on the dimension of the matrix and expansion by the first column to prove this result for any square matrix.

Answer 2

I get the feeling this is homework, so I'm going to see if I can help sketch an argument.

Since this matrix is upper triangular, and you can take the determinant along any row, just choose the bottom row. Now the determinant in the two by two case is just the product of the element in $(2,2)$, times the resulting 1x1 matrix, which is just a scalar, so it's $a_{22}a_{11}$. But both of these elements are on the diagonal, so you might suspect that the determinant of an upper triangular matri x

Using similar reasoning, you can find that that this is true for all upper triangular matrices of any size, using mathematical induction, since the cofactor expansion reduces the problem of a determinant of an $n$ x $n$ matrix to an $(n-1)$ x $(n-1)$ matrix. I'm not so good at formatting, but you can consider that for a 3x3 matrix, you'd have the $(3,3)$ element times the determinant of the resulting 2x2 matrix, which is also upper triangular. See if you can fill in the rest of the details.