Determinant of Triangular Matrix

Theorem
Let $$\mathbf{T}_n$$ be a triangular matrix (either upper or lower) of order $$n$$.

Let $$\det \left({\mathbf{T}_n}\right)$$ be the determinant of $$\mathbf{T}_n$$.

Then $$\det \left({\mathbf{T}_n}\right)$$ is equal to the product of all the diagonal elements of $$\mathbf{T}_n$$.

That is, $$\det \left({\mathbf{T}_n}\right) = \prod_{k=1}^n a_{kk}$$.

Proof
In the light of Determinant of Transpose, we only need to show this for either the upper or lower triangular matrix, as one is the transpose of the other.

So, let's prove this for the upper triangular matrix.

Let $$\mathbf{T}_n = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix}$$ be an upper triangular matrix.

Let $$A_{pq}$$ be the cofactor of an element $$a_{pq} \in \mathbf{T}_n$$.

Then by the Expansion Theorem for Determinants, $$\det \left({\mathbf{T}_n}\right) = \sum_{k=1}^n a_{nk} A_{nk}$$.

As $$a_{nk} = 0$$ when $$k \ne n$$, this reduces to:

$$\det \left({\mathbf{T}_n}\right) = a_{nn} A_{nn}$$.

But from the definition of the cofactor, $$A_{nn} = \left({-1}\right)^{n+n} D_{nn} = D_{nn}$$, where $$D_{nn}$$ is the order $$n-1$$ determinant obtained from $$D$$ by deleting row $$n$$ and column $$n$$.

But $$D_{nn}$$ is simply the upper triangular matrix $$\mathbf{T}_{n-1}$$.

Thus, without being too formal about it, the result follows by induction.

The base case $$\det \left({\mathbf{T}_1}\right)$$ is trivial, while $$\det \left({\mathbf{T}_2}\right)$$ calculates directly:

$$\det \left({\mathbf{T}_2}\right) = a_{11} a_{22} - a_{12} a_{21} = a_{11} a_{22}$$.