Hướng dẫn giải bài tập toán cao cấp

LỜI GIẢI MỘT SỐ BÀI TẬP TOÁN CAO CẤP Lời giải số tập tài liệu dùng để tham khảo Có số tập số sinh viên giải Khi học, sinh viên cần lựa chọn phương pháp phù hợp đơn giản Chúc anh chị em sinh viên học tập tốt BÀI TẬP VỀ HẠNG CỦA MA TRẬN Bài 1: Tính hạng ma trận:  −4      −2 −4  η1↔ η2  → 1) A =   −1    −7 −4   −2 −4 −4 1 −1 −7 −4        →     −2 −4    0 −4  η 2↔ η3 → −1     −5 3  −2 −4   −1  0 −4  −5 3    η2(5)+η4  →    −2 −4    −1  η3( 2)+ η4   → 0 −4     0 −2 15  −2 −4   −1  0 −4  0 33  h1(−2)+η2 η1(−1)+η4 ⇒ ρ( Α) = 2)    A=     −4    −1 −4   1↔ η2  η →  −10         1 η2    2  →     −1 −4    −4  η1(3)+ η3  η1( )+ η4   →   −10      −1 −4   η2(11)+ η3  −2  η2( −5)+ η4  η2(5)+ η5 −11 22   →   −10   −5 10   −1 −4   −4  −11 22  −10   −5 10  −1 −4   −2  0  ⇒ ρ( Α) = 0   0   −1 −2  η1(−2)+η2  −1 −2  1)+η3   2) A =  −2  η1(−   →  0 −1 −1   −1   0 −2 10 −2   −1 −2    →  0 −1 −1  ⇒ ρ( Α) =  0 0  h2(-2)+η3 3)   A=   −1 −1 −5 1 7 −1    →    η1( −2)+η2  ( −5)+η3  η1  η1( −7 )+η4   →     −1 −7 −15 0 0 −6  1 η3   6  −1  ( −2 )+η3  η2  −7 −15  η2( −2)+η4   → −14 −24 12     −14 −26     η4 −4 +η4  ( )   →     −1 −7 −15 0 0 0 −6 −1 −7 −15 0 0 −6         ⇒ ρ( Α) =   4)   A=   −1 −3 −3 −5 −5     η3  η1↔  →     −3 −5  η1( −5)+η2  )+η3  η1( −3  −3  η1 ( −7 )+η4 → −1     −5    →     −3 −5  ( −3)+η3  η2  −8  η2 ( −4 )+η4 → 12 27 −31    16 36 −48     →   −3 −5   −8  ⇒ ρ Α = ( ) 0 0 −7  0 0   1 η3  ↔ η2  2  16  + η4   η3 − 5) −3 −5 12 27 −31 18 −16 16 36 −48 −3 −5   −8  0 0 −7  0 0 −16          A=      2 −1    −2   −2  η1↔ η2  →  −1 −2 −6    −3 −1 −8 −1    −3 −2  −2   2 −1  −2  −1 −2 −6   −3 −1 −8 −1  −3 −2    η1( −2)+ η2  η1( −2)+ η3 η1+ η4 η → 1(3)+ η5  η1( −1)+ η6     −2    −7 −3   −3 −1  η 2↔ η3 →   −2 −8   −1 −5    −7 −3     η2( −2)+ η3 η2( 2)+ η4 η → 2+ η5  η2(−2)+ η6    0 0 0 −2 1 −3 −1 −1 −1 0 −4 0 −3 −1 −1       η3+ η5   →  η3( −1)+ η6        −2   −3 −1  −7 −3  −2 −8   −1 −5  −7 −3  0 0 0 −2   −3 −1  −1 −1  ⇒ ρ Α = ( ) 0 −4   0 0  0 0  6)    A=     −1    −1  η1(−2)+ η2  1+ η3 −1 1  ηη → 1(−1)+ η4 η1(−3)+ η5   −8 −5 −12   −7 13      2↔ η3 η →    −1   −5 −4 −8  1  −10 −8 −16   −4   −1    1  η2(−3)+ η3  2(−6)+ η4 −5 −4 −8  η → η2( 4)+ η5   −10 −8 −16   −4     h3( −1)+ η4  3+ η5 η →     −1    1   5( −4)+ η3 0 −8 −13 −29  η →  0 0    0 −2 −1   −1   1  0 −8 −13 −29  0 −16 −26 −58   0 12 29  −1   1  0 −9 −29  0 0   0 −2 −1     η4↔ η3 h5↔  →    −1   1  0 −2 −1  ⇒ ρ( Α) = 0 −9 −29   0 0  7)   A=    −3 −7    −1 −8  η1↔ η2  → −2        2(−1)+ η3 η →   −1 0 0 −8 −22 32 0 0 −1  −1 −8  η 1(−3)+ η   −3 −7  η1(4)+ η3   → η1+ η4 −2          3↔ η4  η →     −1 0 −1 −8   −22 32  −2 22 −32  0 −1  −8   −22 32  ⇒ ρ( Α) = −1  0  8)   A=    1  −1 3 −4   η1( 4)+ η2  −7 −2  η 1( −3)+ η3 → η1( −2)+ η4 −3    −2    2+ η3  ηη → 2+ η4   −1 0 0 η2    5 −1 3 −4   1   3  10 −15  η  4  →  1 −4 −8 12  η4 3   −3 −6  −1 3 −4   −3  −1 −2  −1 −2  −4   −3  ⇒ ρ( Α) = 0  0  9)   A=   17 3 1   2(−1)+ η3  η → η2(−1) η4   10)  −1   η1(−7)+ η2  −3 10  η1(−17 )+ η3   → η1(−3)+ η4 −7 22     −2 10   −1   η2   −20 −32   4  →  1 η −50 10 −80    10    −5 −8  −1   −5 −8  ⇒ ρ( Α) = 0 0  0 0  −1   −5 −8  −5 −8  −5 −8    A=   10 −1 16 52 −1 −7     1↔ η2  η →       η2( −4 )+ η3  10  → η2+ η4  0 −20  0 Bài 2: Biện luận theo tham số λ  1   λ 10 A = 1)  17  2 −1 −1   10  20 17  −1 −10 −3     ⇒ ρ( Α) =   hạng ma trận:      η2 ↔ η4  2 →     λ    η2 h1↔  →     ↔ η3 η2 →   −5 −7 −15 −5 λ−2 1 17 10 λ    η1 −8 + η3  ( )  η → 1( −4 )+ η4     −1 10 16 52 −1 −7  η1( −4)+η2  ( −3)+η3  η1  η1( −1)+η4   →     17 10     χ4  χ1↔  →     −7 −15 −5 −5 λ−2   η2 +η3 ( )  η2 −2 +η4  ( )   →     0 1 17 10 λ      −5 20 −40 −4 λ +          −5   →  0 20 −40   0 λ  Vậy : - Nếu λ = r(A) = - Nếu λ ≠ r(A) =  1 η3  + η4  5   2) A =    λ  1    10  η2 ↔ η4  → 17      λ 1 4 17 10     χ4  χ1↔  →     3 1 17 10 λ             χ2 c1↔  →   4 3 17 1 10 λ  η1( −2)+η2  ( −7 )+η3  η1  η1( −4)+η4   →       −5 −4  →  0 0  0 λ Vậy: - Nếu λ = r(A) = - Nếu λ ≠ r(A) = η2( −5)+η3 η2( −3)+η4   3) A =     3    10  Χ2 ↔ Χ4  →     λ −8     →   h1( −4 )+η3 η1( −6 )+η4   4) A =    −3    →       η3  η1↔  →        1 η2       →         η4  η3↔  →          10 3 −8 λ −5 −25 −15 −10 −50 λ − 24 −1 −5 −3 0 λ +6 0 0           r(A) = r(A) =  14    10  Χ2 ↔ Χ4  →     λ  h1(3)+η3 η1( −3)+η4   −5 −4  0 λ  0 0  3 10 −8 λ 10 −5 −25 −15 −10 −50 λ − 24    →  0  0 λ + Vậy: - Khi λ + = ⇔ λ = −6 - Khi λ + ≠ ⇔ λ ≠ −6 η2(5)+ η3 η2(10 )+ η4     η4  η3↔  →          −5 −4 −25 10 −20 −15 λ − 12 −3 10 35 21 −4 −20 λ − 12  14    10  η1↔ η3   →     λ     1  η2     →     −3   10  14  λ  7 35 21 −4 −20 λ − 12         →   h2( −7 )+η3 η2( )+η4 Vậy : - 0 0      η3↔ η4  → 0     λ  Nếu λ = r(A) = Nếu λ ≠ r(A) = 0 0    λ  0  BÀI TẬP VỀ MA TRẬN NGHỊCH ĐẢO VÀ PHƯƠNG TRÌNH MA TRẬN Bài 1: Tìm ma trận nghịch đảo ma trân sau:   1) A =     Ta có:  1  5  η1 3  η1 −  + η2     3   η2(3) →  (A I)=   →  −         3    3  −5   η2 −  + η1      3  →  −4  ⇒ Α−1 =  −4   −5   −5   −2  2) A =    −9  Ta có: −1  −2   −9   −2   δ −β  A =  = αδ − βχ   = 1.(−9) − (−2).4   =   −9   −χ α   −4   −1  −1  3) A =    Ta có:   (A I) =   −4 −3 −5 −1     −4 0 −3 1 −5 −1 0   −1 −1   η2(−1) + η1   →  −3 1      −5 −1 0   −1  −1 −1  −1  η1( −2)+η2    2) + η3  η1  →  −1 −7 −2  η2(−  →  −1 −7 −2  ( −3)+η3  −2 −13 −3   0 1 −3   −1 −1   −1 −3 11 −4  η3 −7 +η2    ( ) 1)  η2(−  →  −3  η3  →  −5 18 −7  ( −4)+η1  0 1 −3   0 1 −3   0 −8 29 −11    η2+η1  →  −5 18 −7   0 1 −3   − 29 − 11    Vậy ma trận A ma trận khả nghịch A =  − 18 −   −3    -1  4) A =    Ta có:   (A I)=        0 0     η3↔η1  → 0   0  0   →  −6 −5 −3  −3 −3 −2 η1( −3)+η2 η1( −2 )+η3       0  η3↔η2  →  −3 −3 −2     −6 −5 −3   0  η2 −   0  3   h2(-2)+η3  →  −3 −3 −2  →  1 −   0 −2 1   0 −2  −3 −2   → −1 −  3  0 −2 1 h3( −1)+η2 η3( −3)+η1   0 −    5)+η1   η2(−  →     0 −2   − −       ⇒ Α−1 =  −1 −  3     −2  2  5) A =  −2     −2  Ta có:          −1 − 1 −         2 0  h1( −2 ) + h  2 0   ÷ h1( −2 ) + h  ÷ A =  −2 ÷  →  −3 −6 −2 ÷  −2 0 ÷  −6 −3 −2 ÷       1 h 2 − ÷ 1 2  3 1 2 0  1 h 3 ÷ h 2( −2 ) + h  ÷ 9 →  −3 −6 −2 ÷→  −  3  0 −2 ÷    2 0 − 9  2   1 9 − ÷ 1 0 9  ÷  h 3( −2 ) + h 2 ÷ h 2( −2 ) + h1  h 3( −2 ) + h1  → −  →  ÷  9 9  ÷   0 − ÷  0 − ÷ 9  9    1 9 9 ÷  ÷ 2÷ −1  ⇒A = − 9 9÷  ÷  − ÷ ÷ 9  9 Bài Giải phương trình ma trận sau 1 2  5 1)  ÷X =  ÷  4  9 1 2  5 Đặt A =  ÷; B =  ÷ 3 4 5 9 Ta có: AX = B ⇔ X = A−1 B −1  −2 1 2  −2    d −b  A = ÷ =  ÷=  ÷= ad − bc  −c a  1.4 − 2.3  −3   3 4   −2     −1 −1 ⇒ X =  −1 ÷ = ÷  ÷ ÷  2 3  2   −2   −1  2) X  ÷=  ÷  −4   −5  −1 10  −1 ÷ ÷   0÷ ÷ 0÷ ÷ 1÷ ÷ 9  ÷ ÷ 2÷ − 9÷ ÷ ÷ ÷  −5 −3 −1 D= −9 −6 −5 c1↔c −1 −3 ‡ˆ ˆˆ ˆˆ ˆ† ˆˆ − −9 −6 −5 h1+ h h1( −2) + h −1 ‡ˆ ˆˆh1(ˆˆ−1)ˆˆ+ˆhˆ4ˆ†ˆˆ − −1 −1 2 −1 −1 = −1 × −1 − −1 −1 = − ( + 12 − − 12 ) = 4) −3 −5 −3 −6 h 4+ h1 D= ‡ˆ ˆˆ ˆˆ ˆ† ˆˆ −5 −7 −4 −6 −1 0 h1( −3) + h −3 −6 h1( 2) + h ˆ ˆ ‡ ˆh1ˆ(ˆ−4ˆˆ) +ˆhˆ4†ˆˆ −5 −7 −4 −6 −12 −6 = −1× −5 −7 = −1× × −5 −7 −5 −7 −14 −14 = −2 ( 98 + 54 + 150 − 126 − 45 − 140 ) = −2 × ( −9 ) = 18 5) −3 h 3+ h1 −5 h 3+ h ˆ ˆ ˆ D= ‡ ˆh 3(ˆ−ˆ1)ˆ+ˆˆh 4ˆ†ˆˆ −5 −3 −2 −8 −4 −5 4 −1 −1 −5 −3 −2 −3 −1 −3 4 −1 −1 −1 = −21 −3 −18 −21 −3 ‡ˆ ˆˆh1(ˆˆ−3)ˆˆ+ˆhˆ4ˆ†ˆ −21 −3 −18 −15 −1 −15 −15 −1 −15 −1 −15 h1+ h h1( −4) + h = 315 − 270 + 189 − 405 − 126 + 315 = 18 6) 35 −1 0 2 −12 −5 −7 −14 D= −1 −1 1 −1 2 −1 h1( −1)+ h3 h1+ h 1 ‡ˆ ˆˆ ˆhˆ1+ˆˆh 5ˆˆ †ˆˆˆ 1 1 0 0 1 −1 1 −1 −1 = 1× −1 −1 −1 −1 1 −1 −2 −3 −2 −3 −2 −3 h1( −2) + h = 1× −1 −1 ‡ˆ ˆˆh1(ˆˆ−1)ˆˆ+ˆhˆ3ˆ†ˆ −1 −2 −2 −2 = − 12 − + − 16 + 24 = 7) 0 0 −2 D = 18 −6 17 −15 19 20 24 3 18 −6 0 −2 h1↔ h ‡ˆ ˆˆ ˆˆ ˆˆ†ˆ − 0 17 −15 19 20 24 3 18 −6 2 −2 h1( −4) + h ˆˆ ˆˆ ˆ† 0 = −1 × ‡ˆ ˆˆh1(ˆˆ−19) ˆˆ − 0 +h5 −63 −6 −37 −318 117 −33 h2 ‡ˆ ˆˆ ˆˆ†ˆ −( −1) +1 −2 0 0 −63 −6 −37 −318 117 −33 −2 −2 − −6  M 22  = −5 × −37 117 −33 −37 117 = −5 ( −594 − 444 + 1404 − 330 ) = −5 × 36 = −180 8) 36 0 −2 D= 0 0 h1( −1) + h ‡ˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆ†ˆ 0 10 h1( −1) + h ‡ˆ ˆˆ ˆˆ ˆˆ ˆˆ †ˆˆ 15 2 −2 5 0 0 −7 16 h 2( −1)+ h ‡ˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆ†ˆˆ 15 0 1 10 −7 = 1× 15 1 −3 −7 16 h 3+ h ‡ˆ ˆˆ ˆˆ ˆ† ˆˆ −9 15 −7 15 0 1 −7 16 −3 −9 0 = × × (−3) × = −180 9) −9 D= −2 −3 hh1(1( −−1)1) ++ hh 32 ‡ˆ ˆˆ ˆˆ ˆˆ ˆˆ †ˆˆ h1( −1) + h 4 −12 −5 −2 −6 −12 h1↔ h ‡ˆ ˆˆ ˆˆ ˆˆ†ˆ − −3 −5 −2 −2 −6 −3 −2 −12 87 −12 −15 29 −4 −5 29 −4 87 −12 −15 h1( −7) + h ˆˆ ˆˆ ˆˆ ˆ†ˆ − = −1 × − 5 − = −3 × − 5 − −5 ‡ˆ ˆˆh1(2) +h −5 −3 −30 −30 −30 −30 = −3 ( 580 − 360 + 125 − 750 + 435 − 80 ) = −3 × (−50) = 150 37 BÀI TẬP VỀ HỆ PHƯƠNG TRÌNH KRAMER Giải hệ phương trình phương pháp Kramer: + x3 = −1 2 x1  1)  x + x + x =  x2 + x3 =  Ta có: * D = = + – 20 = -7 −1 * Dx1 = = - + 35 – 20 + 10 = 21 −1 * Dx2 = = 14 + – 20 +1 = −1 * Dx3 = = 40 – -70 = -35 Vì D ≠ nên hệ có nghiệm nhất: Dx 21  x = D = − = −3  Dx  =− =0 x = D  Dx  − 35 x = D = − =   x − x + 3x =  4x − 5x = −13 2)  3x − x3 =  Ta có: −1 * D= − = - +15 – 36 = -29 −2 38 * Dx1= − 13 1 * −1 − = - 48 +5 -12 + 26 = -29 −2 Dx2 = − 13 − = 26 – 90 + 117 +5 = 58 −2 −1 * Dx3 = − 13 = + 39 – 72 = -29 Vì D ≠ nên hệ có nghiệm nhất: Dx − 29  x = D = − 29 =  Dx 58  =− = −2 x = D 29  Dx − 29  x = D = − 29 =  =2  x1 + x2 − x3  x2 − x3 − x4 = −8  3)  − x3 =5  x1  x1 − x2 − 3x4 = Ta có: −1 −1 −3 −5 −3 −3 −5 h1( −2) + h −3 −5 D= ====== = × − −8 −1 h1( −1) + h −8 −6 −3 −6 1 −2 −3 −6 −3 = −6 + 40 − 30 + 72 = 76 −1 −8 −3 −5 c1↔c Dx1 = ====− −1 0 −2 −3 −1 −1 −3 −8 −5 h1( −3) + h2 −10 −14 −5 =====− −1 h1( −1) + h3 −4 0 −2 −3 −2 −3 −10 −14 −5 −10 −14 = −1× −4 −4 −2 −3 −2 = − ( 90 − 30 + 168 ) = −228 39 −1 −1 −8 −3 −5 −8 −3 h1( −2) + h3 −8 −3 −5 −8 −3 −5 Dx2 = ====== = 1× 1 1 −1 h1( −1) + h4 1 −2 −3 −2 1 0 −3 −2 −3 = 24 − − 10 − = 4 2 −8 −5 −8 h1( −2 ) + h3 −8 −5 −8 −5 Dx3 = ====== = × − −8 h1( −1) + h4 −8 −6 −2 −3 −6 −2 −2 −3 −6 −2 −3 = −6 − 80 − 30 + 192 = 76 −1 −1 2 −3 −8 −3 h1( −2 ) + h3 −3 −8 −3 −8 Dx4 = ====== = 1× −8 1 −8 −1 h1( −1) + h4 −8 1 −6 −2 −6 1 −2 0 −6 −2 = −4 + 18 + 64 − 48 − + 48 = 76 Vì D ≠ nên hệ có nghiệm nhất: Dx1 228   x1 = D = 76 =   x = Dx2 = =  D 76 hay (3,0,1,1)   x = Dx3 = 76 =  D 76  Dx4 76  x4 = = =1 D 76  − x3 + x4 =  x1 2 x − x − x4 =  4)  x2 − x3 + x4 =   3x2 − x4 = Ta có: −3 1 −3 −1 −3 −1 −1 h1( −2) + h −1 −3 D=  ====== = 1× −5 2 −5 2 −5 −1 −1 −1 = −5 + 36 – 45 + 12 = −2 40 −3 1 −3 −1 −1 c1↔c −1 −1 Dx1 = ======− −5 2 −5 −1 −1 h1+ h2 h1( −2 ) + h3 ======− h1+ h4 −3 −1 −3 −3 −1 −3 = −1× 1 = − ( −6 − − 12 − − + 36 ) = −3 Dx2 = 0 −3 1 −3 −4 −3 0 −1 −4 −3 h1(−2) + h = 1× −5 −5 −5 −1 −1 −1 = −20 + 48 – 60 + 30 = −2 1 −1 −4 −3 −1 −1 h1( −2) + h −1 −4 −3 Dx3 = ====== = 1× 2 2 −1 −1 −1 = – 24 Dx4 = 0 – 24 + 45 + – = −3 −3 −1 −4 −1 0 h1( −2) + h −1 −4 ====== = 1× −5 −5 −5 4 =  20 + 90 − 60 − 48 = Vì D ≠ nên hệ có nghiệm nhất: Dx  x = D = − =  x = Dx = − =  D −2  x = Dx = − = −1  D  Dx x = = − = −1 D  41 2 BAØI TẬP BIỆN LUẬN THEO THAM SỐ Bài 1: Giải biện luận: 3x1 + x2 + x3 + x4 = 2 x + x + x + x =   x − x − x − 20 x = −11  4 x1 + x2 + x3 + λ x4 = Giaûi:  3  −6 −9 −20 −11  ÷  ÷ ÷ h1↔ h  ÷  ( A B ) =  −6 −9 −20 −11÷→  ÷  ÷  ÷ λ  λ  4 4 −11 1 ÷ h 2   27 ÷  ÷ 3 0  → 20 32 64 36 ÷ h 3 ÷  ÷  25 40 λ + 80 46  0  −6 −9 −20 −11 1  ÷  16 ÷ h 3↔ h  h 2( −1) + h   → → h 2( −5) + h 0 0 0 0 ÷  ÷  λ  0 0 0  x1 − x2 − x3 − 20 x4 = −11  (1) ⇔  x2 + x3 + 16 x4 = (2)  λ x4 =  1  →  0  0 h1( −2) + h h1( −3) + h h1( −4) + h −6 −9 15 24 −20 48 − −9 −11 ÷ ÷ 16 ÷ ÷ 25 40 λ + 80 46  −6 −9 −20 −11 ÷ 16 ÷ 0 λ ÷ ÷ 0 0  λ + 3λ t −   x1 = − × λ   x = − × −9λ + 8λ t + 16  1) Khi λ ≠ : (2) ⇔  λ ( t ∈ R)  x3 = t  x =  λ 1x1 − x2 − x3 − 20 x4 = −11  2) Khiλ = : (3) ⇔ 15 x2 + 24 x3 + 48 x4 = 27 : hệvônghiệ m 0 =  Bài 2: Cho hệ phương trình: 42 −20 16  x1 − x2 + x3 + x4 = 4 x − x + x + x =   6 x1 − x2 + x3 + x4 =  mx1 − x2 + x3 + 10 x4 = 11 a) Tìm m để hệ phương trình có nghiệm b) Giải hệ phương trình m = 10 Giải: a) Ta coù:  −1   −1   ÷  ÷ −2 ÷ c1↔c ↔c1  −2 ÷  ( A B ) =  −3 ÷→  −3 ÷  ÷  ÷ ÷ ÷  m −4 10 11  −4 10 m 11  −1 4 5  −1 4 5  ÷  ÷ h1( −2) + h −2 −1 −3 ÷ h 2( −2) + h  −2 −1 −3 ÷ h1( −3) + h   → → h1( −4) + h  −4 − −6 ÷ h 2( −3) + h  0 0 0÷  ÷  ÷  0 m −8 ÷ ÷  −6 −3 m − −9     −1 4 5  ÷ −2 −1 −3 ÷ h 3↔ h  →  0 m −8 ÷  ÷ 0÷ 0 0  Ta thaáy: ∀m ∈ R : r ( A B ) = r ( A ) < Suy hệ có nghiệm với giá trò m b) Giải hệ m = 10: Biến đổi sơ cấp hàng ta coù:   −1   −1  ÷  ÷ −2 ÷ −6 −10 −14 ÷   → → ( A / B) =   0 −2 −4 −6 ÷ −3 ÷  ÷  ÷ 0  10 −4 10 11 0 0  x1 = 2 x1 − x2 + x3 + x4 =  x = − 2t   (1) ⇔  x2 − x3 − 10 x4 = −14 ⇔  ( t ∈ R) x = − t  −2 x − x = −    x4 = t Baøi Giải biện luận hệ phương trình sau theo tham soá λ : 43 ( λ + 1) x1 + x2 + x3   x1 + ( λ + 1) x2 + x3   x1 + x2 + ( λ + 1) x3 Giải: Ta có λ +1 D= 1 =1 =λ = λ2 λ +3 λ +3 λ +3 1 λ + 1 ‡ˆ ˆˆ ˆˆ ˆˆ ˆˆ†ˆ λ +1 = ( λ + 3) λ + 1 λ +1 1 λ +1 1 λ +1 h 3+ h + h1 1 †ˆˆ ( λ + 3) λ = ( λ + 3) λ ‡ˆ ˆˆh1(ˆˆ−1)ˆˆ+ˆhˆ3ˆ 0 λ h1( −1) + h Dx1 = λ λ2 1 1 h1( − λ ) + h λ + 1 ‡ˆ ˆˆh1(ˆˆ− λˆˆ2 )ˆˆ+hˆ† ˆˆ λ +1 1− λ 1 1− λ = 1× 1− λ −λ + λ + 1− λ −λ + λ + = −λ + λ + − ( − λ ) ( − λ ) = −λ + λ + − + λ + λ − λ = −λ + 2λ = λ ( − λ ) λ +1 1 1 c1↔ c Dx2 = λ ‡ˆ ˆˆ ˆˆ ˆ† λ ˆˆ − 1 λ2 λ +1 λ +1 λ λ +1 1 1 λ +1 λ −1 −λ − λ − − λ = − × ‡ˆ ˆˆh1(ˆˆ− (ˆˆλ +ˆˆ1))ˆ+ˆhˆ† ˆˆ λ − λ − −λ − 2λ 2 λ − λ − −λ − 2λ h1( −1) + h = − ( λ − 1) ( −λ − 2λ ) − ( λ − λ − 1) ( −λ )  = −  −λ − 2λ + λ + 2λ + λ − λ − λ  = 2λ − λ = λ ( 2λ − 1) λ +1 1 λ +1 c1↔ c Dx3 = λ + λ ‡ˆ ˆˆ ˆˆ ˆ† λ ˆ ˆ − λ +1 1 λ 1 λ2 1 −λ − 2λ −1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ† −1 = − × ‡ ˆ hˆ1(ˆ−1)ˆ+ hˆ3 ˆˆ − −λ − 2λ −λ λ −1 −λ λ −1 h1( − ( λ +1)) + h = λ× λ+2 −1 λ −1 = λ ( λ + ) ( λ − 1) + 1 = λ ( λ + 2λ − λ − 1) Ta thaáy: 44 λ ≠ −3 (1) D = ( λ + 3) λ ≠ ⇔  Khi hệ có nghiệm nhất: λ ≠  Dx1 λ ( − λ ) − λ2  x1 = = = D ( λ + ) λ ( λ + 3) λ   Dx2 λ ( 2λ − 1) 2λ −  = =  x2 = D ( λ + 3) λ ( λ + ) λ   Dx λ + 2λ − λ −  x3 = = D ( λ + 3) λ  (2) Neáu λ = −3 Dx1 = 3(2 − 9) = −21 ≠ : Hệ vô nghiệm (3) Nếu λ = hệ trở thành:  x1 + x2 + x3 =   x1 + x2 + x3 = x + x + x =  Hệ vô nghiệm Bài Giải biện luận hệ phương trình sau theo tham số λ : 5 x1 − x2 + x3 + x4 = 4 x − x + 3x + x =   8 x1 − x2 − x3 − x4 = 7 x1 − 3x2 + x3 + 17 x4 = λ Giaûi   −3   −1 −1 −3  ÷  ÷ −2 ÷ h 2( −1)+ h1  −2 ÷  → ( A B ) =  −6 −1 −5 ÷ h 2( −2) + h  −2 −7 −19 ÷ h 2( −1) + h  ÷  ÷  −3 17 λ   −1 10 λ − 1   −1 −1 −3  −1 −1 −3   ÷  ÷ 19 −7 ÷ 19 −7 ÷ h1( −4) + h h 2+ h3    → → h1( −3) + h  −2 −7 −19 ÷ h 2( −1) + h  0 0 ÷  ÷  ÷  19 λ −  0 0 λ   −1 −1 −3   19 −7 ÷ h 4↔ h ÷ →  0 0 λ ÷  ÷ 0 0 0  Hệ phương trình tương đương với heä: 45 x3 − x4 =  x1 − x2 −  x2 + x3 + 19 x4 = −7   0= λ  Ta thấy: (1)Khi λ ≠ hệ vô nghiệm (2)Khi λ = hệ trở thành: (1)  x1 − x2 − x3 − 3x4 =  x2 + x3 + 19 x4 = −7 (2)  19 (2) : x2 = − x3 − x4 − 2 19 13 (1) ⇔ x1 + x3 + x4 + − x3 − x4 = ⇔ x1 = − x3 − x4 − 2 2 Vậy nghiệm hệ là: 13   x1 = − x3 − x4 −  19   x2 = − x3 − x4 − 2  y yù  x3 , x4 tù   Bài Giải biện luận hệ phương trình sau theo tham số λ 3x1 + x2 + x3 + x4 = 2 x + x + x + x =    x1 − x2 − x3 − 20 x4 = −11 4 x1 + x2 + x3 + λ x4 = Giải Ta có: 3   −6 −9 −20 −11  ÷  ÷ ÷ h 3↔ h1  ÷  ( A B ) =  −6 −9 −20 −11÷→  ÷  ÷  ÷ λ ÷ λ ÷ 4  4  1  h1( −2) + h h1( −3) + h  → h1( −4) + h 0  0 −6 15 20 25 −9 −20 −11  −6 −9 −20 −11 ÷ h 3 ↔ h  ÷ 24 48 27 ÷ 16 ÷  ÷ 4  →  15 24 32 64 36 ÷ 48 27 ÷ ÷   25 40 λ + 80 46 ÷ ÷ 40 λ + 80 46 ÷    46  −6  h 2( −3) + h  → h 2( −5) + h 0  0 Khi đó: (1) Nếu λ ≠ −9 −20 −11  −6 −9 −20 −11 ÷  ÷ 16 ÷ h3↔ h  16 ÷ → 0 0 0 ÷ λ ÷ ÷  ÷ 0 0 λ ÷ 0 ÷    r ( A B ) = r ( A ) = < : hệ có vơ số nghiệm (tìm nghiệm trên) (2) Nếu λ = : r ( A B ) = 3  ⇒ r ( A B ) ≠ r ( A ) : hệ vô nghiệm r ( A ) =  Bài Giải biện luận hệ phương trình sau theo tham số λ ( λ + 1) x1 + x2 + x3 = λ + 3λ    x1 + ( λ + 1) x2 + x3 = λ + 3λ    x1 + x2 + ( λ + 1) x3 = λ + 3λ Giải Ta có: λ +1 1 λ +3 λ +3 λ +3 1 h 3+ h + h1 D= λ + 1 ‡ˆ ˆˆ ˆˆ ˆˆ ˆˆ†ˆ λ +1 = ( λ + 3) λ + 1 1 λ +1 1 λ +1 1 λ +1 1 ˆˆˆ ( λ + 3) λ ‡ˆ ˆˆh1(ˆˆ−1)ˆˆ+ˆhˆ3† 0 h1( −1) + h = ( λ + 3) λ λ λ + 3λ 1 λ ( λ + 3) Dx1 = λ + 3λ λ + 1 = λ ( λ + 3) λ + 3λ λ + λ ( λ + 3) 1 ‡ˆ ˆˆh1(ˆˆ− λˆˆ2 )ˆˆ+ hˆ† ˆˆ λ ( λ + 3) 1− λ2 h1( − λ ) + h 1 λ + 1 = λ ( λ + 3) λ λ +1 λ2 1 1− λ = λ ( λ + 3) × 1− λ2 −λ + λ + 1 λ +1 1 λ +1 1− λ −λ + λ + = λ ( λ + 3)  −λ + λ + − ( − λ ) ( − λ )  = λ ( λ + 3) −λ + λ + − + λ + λ − λ  = λ ( λ + 3)  −λ + 2λ  = λ ( λ + 3) ( − λ ) 47 λ + λ + 3λ λ + λ ( λ + 3) Dx2 = λ + 3λ = λ ( λ + 3) λ + 3λ λ + 1 λ ( λ + 3) 1 λ ‡ˆ ˆˆ ˆˆ ˆ† ˆˆ −λ ( λ + 3) λ +1 λ c1↔ c = −λ ( λ + ) × 1 λ +1 1 = λ ( λ + 3) λ λ +1 λ2 1 λ +1 λ +1 1 λ +1 h1( −1) + h ‡ˆ ˆˆh1(ˆˆ− (ˆˆλ +ˆˆ1))ˆˆ+ hˆ† λ −1 −λ ˆˆ −λ ( λ + 3) λ − λ − −λ − 2λ λ −1 −λ λ − λ − − λ − 2λ = −λ ( λ + 3) ( λ − 1) ( −λ − 2λ ) − ( λ − λ − 1) ( −λ )  = −λ ( λ + 3) −λ − 2λ + λ + 2λ + λ − λ − λ  = −λ ( λ + 3) ( −2λ + λ ) = λ ( λ + 3) ( 2λ − 1) λ +1 λ + 3λ λ +1 λ ( λ + 3) λ +1 1 2 Dx3 = λ + λ + 3λ = λ + λ ( λ + 3) = λ ( λ + 3) λ +1 λ 3 1 λ + 3λ 1 λ ( λ + 3) 1 λ2 λ +1 1 h1( − ( λ +1)) + h 2 λ ‡ˆ ˆˆ ˆˆh1(ˆˆ−1)ˆˆ+ hˆˆ3 ˆ† ‡ˆ ˆˆ ˆˆ ˆ† ˆ ˆ −λ ( λ + ) λ + 1 ˆˆ −λ ( λ + 3) −λ − 2λ 1 λ −λ c1↔ c −1 λ −1 λ+2 −1 −λ − 2λ −1 = −λ ( λ + ) × = λ λ + ( ) λ −1 −λ λ −1 = λ ( λ + 3) ( λ + ) ( λ − 1) + 1 = λ ( λ + 3) ( λ + 2λ − λ − 1) Ta thaáy: λ ≠ ⇒ D ≠ Suy hệ có nghiệm nhaát: (1)Khi:   λ ≠ −3 2  Dx1 λ ( λ + 3) ( − λ )  x1 = = = − λ2 D ( λ + 3) λ   Dx2 λ ( λ + 3) ( 2λ − 1)  x = = = 2λ −  2 D λ + λ ( )    x = Dx3 = ( λ + 3) λ ( λ + 2λ − λ − 1) = λ + 2λ − λ −  D ( λ + 3) λ  λ = ⇒ D = vaø Dx1 = Dx2 = Dx3 = suy hệ có vô số (2)Khi   λ = −3 nghiệm 48 49 ...  43 50   2  2 −1 12 BÀI TẬP VỀ HỆ PHƯƠNG TRÌNH TUYẾN TÍNH Bài 1: Giải hệ phương trình sau: 7 x1 + x2 + x3 = 15  1) 5 x1 − x2 + x3 = 15 10 x − 11x + x = 36  Giải: Ta có: 7 15  2 ... ⇔  ( t, s ∈ R )  x3 = t  − x2 − x3 + x4 =  x ,x tuø   y yù  x4 = s BÀI TẬP VỀ ĐỊNH THỨC Bài 30 Tính định thức cấp 2: 1) D = = 5.3 – 7.2 = 15 – 14 = 3 2) D = = 3.5 – 8.2 = 15 – 16 = -1... − x3 =  x tuø x = t   yý Bài 2: Giải hệ phương trình sau: 2 x1 + x2 − x3 + x4 = 4 x + 3x − x + x =  1)  8 x1 + x2 − x3 + x4 = 12 3x1 + x2 − x3 + x4 = Giải: Ta có: 15 ( 2  A B) =