Counter example to the converse of the special property of triangular matrices

Question

Let $A = (a_{ij})$ be a square matrix of order $n$ over $\mathbb R$. Give a counter-example of:

$det(A) = \prod_{t = 1}^n a_{tt}$ $\implies$ $A$ is a triangular matrix.

Any help? Thanks.

Answer 1

A less trivial example: $$ A=\pmatrix{2&4&2\\3&4&5\\5&3&4}. $$

You can also create more less trivial examples by taking an arbitrary triangular matrix and symmetrically permuting its rows and columns, e.g., $$ A=\pmatrix{5&0&0\\2&1&3\\4&0&4}. $$

Answer 2

Try an example (at least $3 \times 3$) where one diagonal entry is $ 0$ and two rows are equal.

Answer 3

In a similar spirit to Robert Israel's answer, consider an arbitrary non-triangular matrix with an entire row/column zero. This clearly has determinant zero, and there is a zero on the diagonal so the product of the diagonal is zero. Obviously, this only works for $n > 2$, which is for good reason (the claim is true for $n = 2$)