Sincepermutation matrices all satisfy the condition Pk = I P k = I (for some k k) the existence of a PLU P L U -decomposition for A A naturally suggests that Pk−1A = LU P k − 1 A = L U. Therefore, even when a LU L U decomposition is not available we can just flip a few rows to find a LU L U -decomposable matrix.
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can you add a 2x3 and a 3x2 matrix
