TITLE: Prove the determinant is the product of its diagonal entries
QUESTION [13 upvotes]: Prove that the determinant of an upper triangular matrix is the
product of its diagonal entries.
What I have so far:
We will prove this by induction for an $n$ $\times$ $n$ matrix. For the case of a $2 \times 2$ matrix, let A=$ \left( \begin{array}{ccc}
a_{11} & a_{12} \\
0 & a_{22}  \end{array} \right)$. So det($A$)=$a_{11}a_{22}$ and the statement is true for the case of a $2 \times 2$ matrix. Now suppose that this statement is true for an $n$ $\times$ $n$ matrix. We will show that it also is true for an $(n + 1)$ $\times$ $(n + 1)$ matrix. Let A =$ \left( \begin{array}{ccc}
a_{11} & a_{12} & \cdots & a_{1(n+1)}\\
0 & a_{22} & \cdots & a_{2(n+1)} \\
\vdots & \cdots &  & \vdots\\
0 & 0 & \cdots & a_{(n+1)(n+1)}\end{array} \right)$.
I don't know what to do after this.

REPLY [15 votes]: Using the cofactor expansion along the first column we get 
$$|A|=(-1)^{1+1}a_{11} \left |\begin{matrix} a_{22}&a_{23} & \cdots & a_{2(n+1)} \\
                                 0 &a_{33} & \cdots & a_{3(n+1)} \\
                                \vdots & \vdots & \vdots & \vdots \\
                                0 & 0 & 0 & a_{n+1n+1}
                 \end{matrix}\right |$$ (Note that remaining terms in the expansion are zero.)
Using the induction hypothesis (you know det of the $n\times n$ matrix), we get 
$|A|$=$a_{11}a_{22}a_{33}\cdots a_{n+1n+1}$