Information:
The computational cost of Gaussian elimination is approximately flops (floating-point operations). LU decomposition has the same cost for factorization, but each subsequent solve (forward + backward substitution) costs only flops.
For a system with and different right-hand sides, at what value of does LU decomposition become more efficient than repeated Gaussian elimination?

Question

Information: 
 The computational cost of Gaussian elimination is approximately    flops (floating-point operations). LU decomposition has the same cost for factorization, but each subsequent solve (forward + backward substitution) costs only    flops. 
For a system with    and    different right-hand sides, at what value of    does LU decomposition become more efficient than repeated Gaussian elimination?

Answer

, because factorization reuse saves significant work.

Answer

, because substitution cost must match the matrix dimension for crossover to occur.

Answer

, because LU is always more efficient due to better numerical stability properties.

Answer

, because the storage overhead of L and U requires multiple solves to amortize.

Answer

, because the breakeven point scales linearly with the matrix size parameter value.

Answer

, because the factorization cost only becomes negligible at very large values of   .

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