I'm asked to prove that the matrix $$A=\begin{pmatrix} a & b & 0\\ c & d & e\\ 0 & 1 & g \end{pmatrix}$$
has a $LU$ decomp iff $a\not=0$ and $ad\not=bc$. However, I do not believe this to be true since we have $$\frac{1}{2}\begin{pmatrix} 1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 0 & \frac{1}{2} & 0\\ 0 & 1 & \frac{1}{2}\\ 0 & 1 & 1 \end{pmatrix}$$ Where the LHS is a $LU$ decomp of the RHS, and the RHS has the form mentioned in the question, with $a=ad=bc=0$, contradicting the iff condition.
So what's going on here? Is this question just wrong, or am I not understanding something?
Thanks
Given: $A=\begin{bmatrix} a &b &0 \\ c&d &e \\ 0 &1 &g \end{bmatrix}$, we use the fact the the LU decomposition is possible if no row exchanges are required.
$\Rightarrow :$ Assuming $A$ has an $LU$ decomposition, we start by doing an elimination step. If $a\ne 0$, then the multipliers are $m_{21}=\frac{c}{a}$ and $m_{31}=0$. Hence the elimination produces
$$A^{(1)}=\begin{bmatrix} a &b &0 \\ 0&d-\frac{bc}{a} &e \\ 0 &1 &g \end{bmatrix}$$
A row exchange will not be required if $d\ne \frac{bc}{a}$, or $ad\ne bc$. If this is the case, we proceed with another elimination step with $m_{32}=\frac{1}{d-\frac{bc}{a}}$, from which we obtain
$$A^{(2)}=\begin{bmatrix} a &b &0 \\ 0&d-\frac{bc}{a} &e \\ 0 &0 &g-\frac{e}{d-\frac{bc}{a}} \end{bmatrix}$$
Therefore the decomposition becomes $$A=LU=\begin{bmatrix} 1 &0 &0 \\ -\frac{c}{a} &1 &0 \\ 0 &-\frac{1}{d-\frac{bc}{a}} & 1 \end{bmatrix} =\begin{bmatrix} a &b &0 \\ 0&d-\frac{bc}{a} &e \\ 0 &0 &g-\frac{e}{d-\frac{bc}{a}} \end{bmatrix}$$
$\Leftarrow$: Let us prove by contrapositive. Assume $a=0$ or $ad=bc$.
If $a=0$, then $A$ will be $$A=\begin{bmatrix} 0 &b &0 \\ c&d &e \\ 0 &1 &g \end{bmatrix}$$
We see that the first pivot is zero, and hence row exchanges are required and the LU decomposition is not possible.
If $ad=bc$, then a similar step of elimination (see above) produces $$A^{(1)}=\begin{bmatrix} a &b &0 \\ 0&\frac{1}{a}0=0 &e \\ 0 &1 &g \end{bmatrix}$$ which shows a zero second pivot, and hence requires a row exchange.
Therefore we conclude that the $LU$ decomposition exists iff $a\ne 0$ and $ad \ne bc$.