Аналитическая геометрия и линейная алгебра. Умнов А.Е. - 340 стр.

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                                            azqwj (1,1)          xësnuntzjuq α ij qujzéq|np                      4 5 6 
                                                                                                                           7 8 9
                                                                                            2
                                            bzqwj (1,0) xësnuntzjuq β qujzéq|np − 3  j

                                                                                            4
                                            czqwj (0,1) xësnuntzjuq γ i qujzéq|np 2 − 3 4 
                         Ëjpzqxknézrq α ij β j q α ij γ i 
       
 cËËÓÒË                ° º º¹¯ËËãËÓÒ  º¹Ë¯ÈÒÒ °m˯ˆ©mÈÓÒ« α ij β j   ˆËÓϺ¯ ˆÒ¹È
                                                                               3
                                  (1,0) °}ºä¹ºÓËӈÈäÒ δ =             i
                                                                              ∑ α ij β j ºªˆºä‚
                                                                             j =1

                                               δ = α1 β + α 2 β + α 3 β = 1⋅ 2 + 2 ⋅ (−3) + 3 ⋅ 4 = 8
                                                 1    1 1       1 2        1 3

                                               δ 2 = α12 β 1 + α 22 β 2 + α 32 β 3 = 4 ⋅ 2 + 5 ⋅ (−3) + 6 ⋅ 4 = 17 
                                               δ 3 = α13 β 1 + α 23 β 2 + α 33 β 3 = 7 ⋅ 2 + 8 ⋅ (−3) + 9 ⋅ 4 = 26.
                                                                                    
                         ° kÓÈãºÒÓº α ij γ i   ˆËÓϺ¯ ˆÒ¹È                                 (0,1)  ° }ºä¹ºÓËӈÈäÒ
                                           3
                                  ϕ j = ∑ α ijγ i ‘ºÈ
                                          i =1
                                           ϕ1 = α11γ 1 + α12 γ 2 + α13γ 3 = 1 ⋅ 2 + 4 ⋅ ( −3) + 7 ⋅ 4 = 18
                                           ϕ 2 = α 21γ 1 + α 22 γ 2 + α 23γ 3 = 2 ⋅ 2 + 5 ⋅ ( −3) + 8 ⋅ 4 = 21                
                                           ϕ 3 = α 31γ 1 + α 32 γ 2 + α 33γ 3 = 3 ⋅ 2 + 6 ⋅ ( −3) + 9 ⋅ 4 = 26 .