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Example III.19 (Feldman’s notes) 1 1 Let A = . We look for a matrix B that obeys AB = I. The 1 2 1 0 columns of I are b e1 = and b e2 = . Denote the columns of B by 0 1 ~ 1 and B ~ 2 . That is B ′ X X′ X ~1 B ~2 ~1 = ~ 2 = X′ = B B= B B ′ Y Y Y Y We are to find a B obeying X +Y X′ + Y ′ ~ ~ b b = = A B A B AB = e e 1 2 1 2 X + 2Y X ′ + 2Y ′ ~1 = b ~2 = b ~ 1 and or, equivalently, AB e1 , AB e2 . We wish to solve for B ~ 2 . The augmented matrices for these two systems of equations are B Ab e1 and A b e2 . For efficiency, we combine them into a single larger agumented matrix A b e1 b e2 = 1 1 1 1 2 0 0 1 We can apply row operations to bring this to reduced row echelon form 1 1 1 0 1 0 2 −1 (1) − (2) 0 1 −1 1 (2) − (1) 0 1 −1 1 The augmented on the right matrix represents the two systems of equa~ 1 = 2 and I B ~ 2 = −1 . Thus tions I B −1 1 2 −1 2 −1 ~1 = ~2 = ~1 B ~2 = B B B= B −1 1 −1 1 c Joel Feldman. 2011. All rights reserved.