<Dd> (4 3 6 3) = (l 11 0 l 21 l 22) (u 11 u 12 0 u 22). (\ displaystyle (\ begin (bmatrix) 4&3 \ \ 6&3 \ end (bmatrix)) = (\ begin (bmatrix) l_ (11) &0 \ \ l_ (21) &l_ (22) \ end (bmatrix)) (\ begin (bmatrix) u_ (11) &u_ (12) \ \ 0&u_ (22) \ end (bmatrix)).) </Dd> <P> One way to find the LU decomposition of this simple matrix would be to simply solve the linear equations by inspection . Expanding the matrix multiplication gives </P> <Dl> <Dd> l 11 ⋅ u 11 + 0 ⋅ 0 = 4 l 11 ⋅ u 12 + 0 ⋅ u 22 = 3 l 21 ⋅ u 11 + l 22 ⋅ 0 = 6 l 21 ⋅ u 12 + l 22 ⋅ u 22 = 3 . (\ displaystyle (\ begin (aligned) l_ (11) \ cdot u_ (11) + 0 \ cdot 0& = 4 \ \ l_ (11) \ cdot u_ (12) + 0 \ cdot u_ (22) & = 3 \ \ l_ (21) \ cdot u_ (11) + l_ (22) \ cdot 0& = 6 \ \ l_ (21) \ cdot u_ (12) + l_ (22) \ cdot u_ (22) & = 3. \ end (aligned))) </Dd> </Dl> <Dd> l 11 ⋅ u 11 + 0 ⋅ 0 = 4 l 11 ⋅ u 12 + 0 ⋅ u 22 = 3 l 21 ⋅ u 11 + l 22 ⋅ 0 = 6 l 21 ⋅ u 12 + l 22 ⋅ u 22 = 3 . (\ displaystyle (\ begin (aligned) l_ (11) \ cdot u_ (11) + 0 \ cdot 0& = 4 \ \ l_ (11) \ cdot u_ (12) + 0 \ cdot u_ (22) & = 3 \ \ l_ (21) \ cdot u_ (11) + l_ (22) \ cdot 0& = 6 \ \ l_ (21) \ cdot u_ (12) + l_ (22) \ cdot u_ (22) & = 3. \ end (aligned))) </Dd>

Advantages of matrix factorization in linear algebra computations