Any square matrix $A$ admits $LUP$ and $PLU$ factorizations. If $A$ is invertible, then it admits an $LU \;\text{(or)}\; LDU$ factorization if and only if all its leading principal minors are nonzero.
$\textbf{(or)}$
An invertible matrix $A$ has an $LU$ decomposition provided that all its leading submatrices have non-zero determinants. The $k^{\text{th}}$ leading submatrix of $A$ is denoted $A_{k}$ and is the $k \times k$ matrix found by looking only at the top $k$ rows and leftmost $k$ columns.
If a square, invertible matrix has an $LDU$ (factorization with all diagonal entries of $L$ and $U$ equal to $1$), then the factorization is unique. In that case, the $LU$ factorization is also unique if we require that the diagonal of $L\; \text{(or)}\;U $ consists of ones.
If $A$ is a singular matrix of rank $k$, then it admits an $LU$ factorization if the first $k$ leading principal minors are nonzero, although the converse is not true.
Matrix $A_{k \times k}$ has $LU$ factorization if: $$\mathsf Rank(A_{11})+k\geq Rank\left(\begin{bmatrix}
A_{11} & A_{12}
\end{bmatrix}\right) +Rank\left(\begin{bmatrix}
A_{11} \\ A_{21}
\end{bmatrix}\right)$$
this must hold for each $k = 1,\dots, n − 1.$
We can also write these conditions as $\text{rank}\left(A[{1 \dots k}]\right) + k \geq \text{rank}\left(A[{1 \dots k}, {1 \dots n}] \right)+ \text{rank}\left(A[{1\dots n}, {1 \dots k}]\right)$ for all $k = 1, \dots , n.$
For option $A:A_{1} = \begin{bmatrix} 1 \end{bmatrix},\;A_{2} = \begin{bmatrix} 1 & 2 \\ 3 & 8 \end{bmatrix},\;\text{and}\; A_{3} = \begin{bmatrix} 1 & 2 & 4 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{bmatrix}$
Now, $\mid A_{1} \mid = 1,\;\mid A_{2} \mid = 2,\;\text{and}\; \mid A_{3} \mid = 6$
$\therefore LU$ – decomposition is possible.
For option $B:A_{1} = \begin{bmatrix} 1 \end{bmatrix},\;A_{2} = \begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix},\;\text{and}\; A_{3} = \begin{bmatrix} 1 & 2 & 4 \\ 4 & 8 & 14 \\ 2 & 6 & 13 \end{bmatrix}$
Now, $\mid A_{1} \mid = 1,\;\mid A_{2} \mid = 0,\;\text{and}\; \mid A_{3} \mid = 4$
$\therefore LU$ – decomposition is not possible.
For option $C:A_{1} = \begin{bmatrix} 1 \end{bmatrix},\;A_{2} = \begin{bmatrix} 1 & 2 \\ 3 & 8 \end{bmatrix},\;\text{and}\; A_{3} = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 8 & 14 \\ 2 & 6 & 13 \end{bmatrix}$
Now, $\mid A_{1} \mid = 1,\;\mid A_{2} \mid = 2,\;\text{and}\; \mid A_{3} \mid = 4$
$\therefore LU$ – decomposition is possible.
For option $D:$ $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \implies \det(A) = 0\quad \text{(Singular matrix)}$
Here, $\text{rank(A)} = 2,$ and $A_{1} = [1] \implies \mid A_{1} \mid \neq 0,A_{2} = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \implies \mid A_{2} \mid = 5 – 8 = -3 \neq 0$
$\therefore LU$ – decomposition is possible.
So, the correct answer is $(B).$
