Discussion
Consider the following two statements:
P: The matrix \(\begin{bmatrix} 0 & 5 \\ 0 & 7 \end{bmatrix}\) has infinitely many LU factorizations, where \(L\) is lower triangular with each diagonal entry 1 and \(U\) is upper triangular.
Q: The matrix \(\begin{bmatrix} 0 & 0 \\ 2 & 5 \end{bmatrix}\) has no LU factorization, where \(L\) is lower triangular with each diagonal entry 1 and \(U\) is upper triangular.
Then which one of the following options is correct?
Answer: B
Both P and Q are TRUE.
Write \(L = \begin{bmatrix} 1 & 0 \\ l & 1 \end{bmatrix}\) and \(U = \begin{bmatrix} u_{11} & u_{12} \\ 0 & u_{22} \end{bmatrix}\), so
\[ LU = \begin{bmatrix} u_{11} & u_{12} \\ l u_{11} & l u_{12} + u_{22} \end{bmatrix} \]
For P, matching with \(\begin{bmatrix} 0 & 5 \\ 0 & 7 \end{bmatrix}\) gives \(u_{11} = 0\), \(u_{12} = 5\), \(l \cdot 0 = 0\) and \(5l + u_{22} = 7\). The value of \(l\) is free and \(u_{22} = 7 - 5l\), so there are infinitely many factorizations.
For Q, matching with \(\begin{bmatrix} 0 & 0 \\ 2 & 5 \end{bmatrix}\) gives \(u_{11} = 0\) and then \(l u_{11} = 2\), that is \(0 = 2\): no factorization exists. Both statements hold, so every option calling one of them FALSE is wrong.
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