Finding a matrix that can be represented with only single LU decomposition

Question

I'm trying to disprove the following statement:

Let $M$ be a singular matrix $3\times 3$ that can be represented with LU decomposition ($M=LU$), then the decomposition is unique (only one decomposition).

I can't think of a basic example to disprove this statement. Can you help me with an example?

Answer 1

Hint.

$$\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}=\begin{pmatrix}1&0&0\\1&1&0\\1&x&1\end{pmatrix}\begin{pmatrix}1&1&1\\0&0&0\\0&0&0\end{pmatrix}.$$

Answer 2

Let $n\geq 2$, $K$ be a field and $M\in GL_n(K)$.

Then $M$ admits a factorization $M=LU$ -with the $l_{i,i}=1$-

IFF its leading principal minors are non-zero (the leading principal minor of $M$ of order $k$ is the determinant of the submatrix of dimension $k$ obtained by deleting the last $n − k$ rows and columns).

In these conditions, the factorization is unique. Ref:

Horn, Roger A.; Johnson, Charles R. (1985), "Matrix Analysis".

EDIT. I read the OP's statement too quickly.

When $M$ is singular, the NSC for the unicity of the decomposition is:

all its leading principal minors are non-zero -except the last one ($\det(M)=0$)-.

In the @jan 's good example, we see that the second minor of $M$ is $0$ and then, there is no uniqueness.