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BÀI TẬP TOÁN CAO CẤP 2
LỜI GIẢI MỘT SỐ BÀI TẬP TOÁN CAO CẤP 2 Lời giải một số bài tập trong tài liệu này dùng để tham khảo. Có một số bài tập do một số sinh viên giải. Khi học, sinh viên cần lựa chọn những phương pháp phù hợp và đơn giản hơn. Chúc anh chị em sinh viên học tập tốt BÀI TẬP VỀ HẠNG CỦA MA TRẬN Bài 1: Tính hạng của ma trận: 1) A = 2 −4 3 1 0 1 −2 1 −4 2 0 1 −1 3 1 1 −7 4 −4 5 η1↔η2 → 1 −2 1 −4 2 2 −4 3 1 0 0 1 −1 3 1 1 −7 4 −4 5 h1(−2)+η2 η1(−1)+η4 → 1 −2 1 −4 2 0 0 1 9 −4 0 1 −1 3 1 0 −5 3 0 3 η 2↔ η 3 → 1 −2 1 −4 2 0 1 −1 3 1 0 0 1 9 −4 0 −5 3 0 3 η2(5)+η4 → 1 −2 1 −4 2 0 1 −1 3 1 0 0 1 9 −4 0 0 −2 15 8 η 3(2)+ η 4 → 1 −2 1 −4 2 0 1 −1 3 1 0 0 1 9 −4 0 0 0 33 0 ⇒ ρ Α ( ) = 4 2) A = 0 2 −4 −1 −4 5 3 1 7 0 5 −10 2 3 0 η 1↔ η 2 → −1 −4 5 0 2 −4 3 1 7 0 5 −10 2 3 0 η 1 3 ( ) + η 3 η 1 2 ( ) + η 4 → −1 −4 5 0 2 −4 0 −11 22 0 5 −10 0 −5 10 η 2 1 2 → −1 −4 5 0 1 −2 0 −11 22 0 5 −10 0 −5 10 η 2 11 ( ) + η 3 η 2 −5 ( ) + η 4 η 2 5 ( ) + η 5 → −1 −4 5 0 1 −2 0 0 0 0 0 0 0 0 0 ⇒ ρ Α ( ) = 2 1 2) A = 2 −1 3 −2 4 4 −2 5 1 7 2 −1 1 8 2 η1(−2)+η2 η1(−1)+η3 → 2 −1 3 −2 4 0 0 −1 5 −1 0 0 −2 10 −2 h2(-2)+η3 → 2 −1 3 −2 4 0 0 −1 5 −1 0 0 0 0 0 ⇒ ρ Α ( ) = 2 3) A = 1 3 5 −1 2 −1 −5 4 5 1 1 7 7 7 9 −1 η1 −2 ( ) +η2 η1 −5 ( ) +η3 η1 −7 ( ) +η4 → 1 3 5 −1 0 −7 −15 6 0 −14 −24 12 0 −14 −26 6 η2 −2 ( ) +η3 η2 −2 ( ) +η4 → 1 3 5 −1 0 −7 −15 6 0 0 6 0 0 0 4 −6 η3 1 6 → 1 3 5 −1 0 −7 −15 6 0 0 1 0 0 0 4 −6 η4 −4 ( ) +η4 → 1 3 5 −1 0 −7 −15 6 0 0 1 0 0 0 0 −6 ⇒ ρ Α ( ) = 4 4) A = 3 −1 3 2 5 5 −3 2 3 4 1 −3 −5 0 7 7 −5 1 4 1 η1↔ η3 → 1 −3 −5 0 7 5 −3 2 3 4 3 −1 3 2 5 7 −5 1 4 1 η1 −5 ( ) +η2 η1 −3 ( ) +η3 η1 −7 ( ) +η4 → 1 −3 −5 0 7 0 12 27 3 −31 0 8 18 2 −16 0 16 36 4 −48 η3 1 2 ↔ η2 → 1 −3 −5 0 7 0 4 9 1 −8 0 12 27 3 −31 0 16 36 4 −48 η2 −3 ( ) +η3 η2 −4 ( ) +η4 → 1 −3 −5 0 7 0 4 9 1 −8 0 0 0 0 −7 0 0 0 0 −16 η3 − 16 7 + η4 → 1 −3 −5 0 7 0 4 9 1 −8 0 0 0 0 −7 0 0 0 0 0 ⇒ ρ Α ( ) = 3 5) 2 A = 2 2 1 5 −1 1 0 4 −2 1 2 1 5 −2 1 −1 −2 2 −6 1 −3 −1 −8 1 −1 1 2 −3 7 −2 η 1↔ η 2 → 1 0 4 −2 1 2 2 1 5 −1 2 1 5 −2 1 −1 −2 2 −6 1 −3 −1 −8 1 −1 1 2 −3 7 −2 η 1(−2)+ η 2 η 1(−2)+ η 3 η 1+ η 4 η 1(3)+ η 5 η 1(−1)+ η 6 → 1 0 4 −2 1 0 2 −7 9 −3 0 1 −3 2 −1 0 −2 6 −8 2 0 −1 4 −5 2 0 2 −7 9 −3 η 2↔ η 3 → 1 0 4 −2 1 0 1 −3 2 −1 0 2 −7 9 −3 0 −2 6 −8 2 0 −1 4 −5 2 0 2 −7 9 −3 η 2(−2)+ η 3 η 2(2)+ η 4 η 2+ η 5 η 2(−2)+ η 6 → 1 0 4 −2 1 0 1 −3 2 −1 0 0 −1 3 −1 0 0 0 −4 0 0 0 1 −3 1 0 0 −1 3 −1 η 3+ η 5 η 3(−1)+ η 6 → 1 0 4 −2 1 0 1 −3 2 −1 0 0 −1 3 −1 0 0 0 −4 0 0 0 0 0 0 0 0 0 0 0 ⇒ ρ Α ( ) = 4 6) A = 1 −1 2 3 4 2 1 −1 2 0 −1 2 1 1 3 1 5 −8 −5 −12 3 −7 8 9 13 η 1(−2)+ η 2 η 1+ η 3 η 1(−1)+ η 4 η 1(−3)+ η 5 → 1 −1 2 3 4 0 3 −5 −4 −8 0 1 1 3 7 0 6 −10 −8 −16 0 −4 2 0 1 η 2↔ η 3 → 1 −1 2 3 4 0 1 1 3 7 0 3 −5 −4 −8 0 6 −10 −8 −16 0 −4 2 0 1 η 2(−3)+ η 3 η 2(−6)+ η 4 η 2(4)+ η 5 → 1 −1 2 3 4 0 1 1 3 7 0 0 −8 −13 −29 0 0 −16 −26 −58 0 0 6 12 29 h3(−1)+ η 4 η 3+ η 5 → 1 −1 2 3 4 0 1 1 3 7 0 0 −8 −13 −29 0 0 0 0 0 0 0 −2 −1 0 η 5(−4)+ η 3 → 1 −1 2 3 4 0 1 1 3 7 0 0 0 −9 −29 0 0 0 0 0 0 0 −2 −1 0 3 h5↔ η 4↔ η 3 → 1 −1 2 3 4 0 1 1 3 7 0 0 −2 −1 0 0 0 0 −9 −29 0 0 0 0 0 ⇒ ρ ( Α ) = 4 7) A = −3 2 −7 8 −1 0 5 −8 4 −2 2 0 1 0 3 7 η 1↔ η 2 → −1 0 5 −8 −3 2 −7 8 4 −2 2 0 1 0 3 7 η 1(−3)+ η 2 η 1(4)+ η 3 η 1+ η 4 → −1 0 5 −8 0 2 −22 32 0 −2 22 −32 0 0 8 −1 η 2(−1)+ η 3 → −1 0 5 −8 0 2 −22 32 0 0 0 0 0 0 8 −1 η 3↔ η 4 → −1 0 5 −8 0 2 −22 32 0 0 8 −1 0 0 0 0 ⇒ ρ ( Α ) = 3 8) A = −1 3 3 −4 4 −7 −2 1 −3 5 1 0 −2 3 0 1 η 1(4)+ η 2 η 1(−3)+ η 3 η 1(−2)+ η 4 → −1 3 3 −4 0 5 10 −15 0 −4 −8 12 0 −3 −6 9 η 2 1 5 η 3 1 4 η 4 1 3 → −1 3 3 −4 0 1 2 −3 0 −1 −2 3 0 −1 −2 3 η 2+ η 3 η 2+ η 4 → −1 3 3 −4 0 1 2 −3 0 0 0 0 0 0 0 0 ⇒ ρ ( Α ) = 2 9) A = 1 3 −1 6 7 1 −3 10 17 1 −7 22 3 4 −2 10 η 1(−7)+ η 2 η 1(−17)+ η 3 η 1(−3)+ η 4 → 1 3 −1 6 0 −20 4 −32 0 −50 10 −80 0 −5 1 −8 η 2 1 4 η 3 1 10 → 1 3 −1 6 0 −5 1 −8 0 −5 1 −8 0 −5 1 −8 η 2(−1)+ η 3 η 2(−1) η 4 → 1 3 −1 6 0 −5 1 −8 0 0 0 0 0 0 0 0 ⇒ ρ ( Α ) = 2 10) 4 A = 0 1 10 3 2 0 4 −1 16 4 52 9 8 −1 6 −7 η 1↔ η 2 → 2 0 4 −1 0 1 10 3 16 4 52 9 8 −1 6 −7 η 1 −8 ( ) + η 3 η 1 −4 ( ) + η 4 → 2 0 4 −1 0 1 10 3 0 4 20 17 0 −1 −10 −3 η 2 −4 ( ) + η 3 η 2+ η 4 → 2 0 4 −1 0 1 10 3 0 0 −20 5 0 0 0 0 ⇒ ρ ( Α ) = 3 Bài 2: Biện luận theo tham số λ hạng của các ma trận: 1) A = 3 1 1 4 λ 4 10 1 1 7 17 3 2 2 4 1 η2↔ η4 → 3 1 1 4 2 2 4 1 1 7 17 3 λ 4 10 1 χ1↔ χ 4 → 4 1 1 3 1 2 4 2 3 7 17 1 1 4 10 λ h1↔ η2 → 1 2 4 2 4 1 1 3 3 7 17 1 1 4 10 λ η1 −4 ( ) +η2 η1 −3 ( ) +η3 η1 −1 ( ) +η4 → 1 2 4 2 0 −7 −15 −5 0 1 5 −5 0 2 6 λ − 2 η2↔ η3 → 1 2 4 2 0 1 5 −5 0 −7 −15 −5 0 2 6 λ − 2 η2 7 ( ) +η3 η2 −2 ( ) +η4 → 1 2 4 2 0 1 5 −5 0 0 20 −40 0 0 −4 λ +8 η3 1 5 + η4 → 1 2 4 2 0 1 5 −5 0 0 20 −40 0 0 0 λ Vậy : - Nếu λ = 0 thì r(A) = 3 - Nếu λ ≠ 0 thì r(A) = 4 2) A = 3 1 1 4 λ 4 10 1 1 7 17 3 2 2 4 3 η2↔ η4 → 3 1 1 4 2 2 4 3 1 7 17 3 λ 4 10 1 χ1↔ χ 4 → 4 1 1 3 3 2 4 2 3 7 17 1 1 4 10 λ 5 c1↔ χ 2 → 1 4 1 3 2 3 4 2 7 3 17 1 4 1 10 λ η1 −2 ( ) +η2 η1 −7 ( ) +η3 η1 −4 ( ) +η4 → 1 4 1 3 0 −5 2 −4 0 −25 10 −20 0 −15 6 λ −12 η2 −5 ( ) +η3 η2 −3 ( ) +η4 → 1 4 1 3 0 −5 2 −4 0 0 0 0 0 0 0 λ η3↔ η4 → 1 4 1 3 0 −5 2 −4 0 0 0 λ 0 0 0 0 Vậy: - Nếu λ = 0 thì r(A) = 2 - Nếu λ ≠ 0 thì r(A) = 3 3) A = 4 1 3 3 0 6 10 2 1 4 7 2 6 λ −8 2 Χ2↔Χ4 → 4 3 3 1 0 2 10 6 1 2 7 4 6 2 −8 λ η1↔ η3 → 1 2 7 4 0 2 10 6 4 3 3 1 6 2 −8 λ h1 −4 ( ) +η3 η1 −6 ( ) +η4 → 1 2 7 4 0 2 10 6 0 −5 −25 −15 0 −10 −50 λ − 24 η2 1 2 → 1 2 7 4 0 1 5 3 0 −5 −25 −15 0 −10 −50 λ − 24 η 2 5 ( ) + η 3 η 2 10 ( ) + η 4 → 1 2 7 4 0 1 5 3 0 0 0 0 0 0 0 λ + 6 η3↔ η4 → 1 2 7 4 0 −1 −5 −3 0 0 0 λ + 6 0 0 0 0 Vậy: - Khi λ + 6 = 0 ⇔ λ = −6 thì r(A) = 2 - Khi λ + 6 ≠ 0 ⇔ λ ≠ −6 thì r(A) = 3 4) A = −3 9 14 1 0 6 10 2 1 4 7 2 3 λ 1 2 Χ2↔Χ4 → −3 1 14 9 0 2 10 6 1 2 7 4 3 2 1 λ η1↔η3 → 1 2 7 4 0 2 10 6 −3 1 14 9 3 2 1 λ h1 3 ( ) +η3 η1 −3 ( ) +η4 → 1 2 7 4 0 2 10 6 0 7 35 21 0 −4 −20 λ −12 η2 1 2 → 1 2 7 4 0 1 5 3 0 7 35 21 0 −4 −20 λ −12 6 h2 −7 ( ) +η3 η2 4 ( ) +η4 → 1 2 7 4 0 1 5 3 0 0 0 0 0 0 0 λ η3↔ η4 → 1 2 7 4 0 1 5 3 0 0 0 λ 0 0 0 0 Vậy : - Nếu λ = 0 thì r(A) = 2 - Nếu λ ≠ 0 thì r(A) = 3 7 BÀI TẬP VỀ MA TRẬN NGHỊCH ĐẢO VÀ PHƯƠNG TRÌNH MA TRẬN Bài 1: Tìm ma trận nghịch đảo của các ma trân sau: 1) A = 3 4 5 7 Ta có: A I ( ) = 3 4 1 0 5 7 0 1 η1 − 5 3 + η2 → 3 4 1 0 0 1 3 − 5 3 1 η 1 1 3 η 2 3 ( ) → 1 4 3 1 3 0 0 1 −5 3 η 2 − 4 3 + η 1 → 1 0 7 −4 0 1 −5 3 ⇒ Α −1 = 7 −4 −5 3 2) A = 1 −2 4 −9 Ta có: A −1 = 1 −2 4 −9 −1 = 1 αδ − βχ δ − β − χ α = 1 1.(−9)− (−2).4 −9 2 −4 1 = 9 −2 4 −1 3) A = 3 −4 5 2 −3 1 3 −5 −1 Ta có: A I ( ) = 3 −4 5 1 0 0 2 −3 1 0 1 0 3 −5 −1 0 0 1 η2(−1)+ η 1 → 1 −1 4 1 −1 0 2 −3 1 0 1 0 3 −5 −1 0 0 1 η1 −2 ( ) +η2 η1 −3 ( ) +η3 → 1 −1 4 1 −1 0 0 −1 −7 −2 3 0 0 −2 −13 −3 3 1 η2(−2)+ η 3 → 1 −1 4 1 −1 0 0 −1 −7 −2 3 0 0 0 1 1 −3 1 η2(−1) → 1 −1 4 1 −1 0 0 1 7 2 −3 0 0 0 1 1 −3 1 η3 −7 ( ) +η2 η3 −4 ( ) +η1 → 1 −1 0 −3 11 −4 0 1 0 −5 18 −7 0 0 1 1 −3 1 η2+η1 → 1 0 0 −8 29 −11 0 1 0 −5 18 −7 0 0 1 1 −3 1 8 Vậy ma trận A là ma trận khả nghịch và A -1 = − −− −− 131 7185 11298 4) A = 2 7 3 3 9 4 1 5 3 Ta có: A I ( ) = 2 7 3 1 0 0 3 9 4 0 1 0 1 5 3 0 0 1 η3↔η1 → 1 5 3 0 0 1 3 9 4 0 1 0 2 7 3 1 0 0 η1 −3 ( ) +η2 η1 −2 ( ) +η3 → 1 5 3 0 0 1 0 −6 −5 0 1 −3 0 −3 −3 1 0 −2 η3↔η2 → 1 5 3 0 0 1 0 −3 −3 1 0 −2 0 −6 −5 0 1 −3 h2(-2)+η3 → 1 5 3 0 0 1 0 −3 −3 1 0 −2 0 0 1 −2 1 1 η2 − 1 3 → 1 5 3 0 0 1 0 1 1 − 1 3 0 2 3 0 0 1 −2 1 1 h3 −1 ( ) +η2 η3 −3 ( ) +η1 → 1 5 0 6 −3 −2 0 1 0 5 3 −1 − 1 3 0 0 1 −2 1 1 η2(−5)+η1 → 1 0 0 − 7 3 2 − 1 3 0 1 0 5 3 −1 − 1 3 0 0 1 −2 1 1 ⇒ Α −1 = − 7 3 2 − 1 3 5 3 −1 − 1 3 −2 1 1 5) A = 1 2 2 2 1 −2 2 −2 1 Ta có: 9 ( ) ( ) ( ) 1 2 2 1 2 3 1 2 3 1 3 2 2 3 9 1 2 2 1 0 0 1 2 2 1 0 0 2 1 2 0 1 0 0 3 6 2 1 0 2 2 1 0 0 1 0 6 3 2 0 1 1 2 2 1 0 0 1 2 2 1 0 0 2 1 0 3 6 2 1 0 0 1 2 0 3 3 0 0 9 2 2 1 2 2 1 0 0 1 9 9 9 h h h h h h h h A − + − + − ÷ ÷ − + ÷ ÷ = − → − − − ÷ ÷ ÷ ÷ − − − − ÷ ÷ ÷ ÷ → − − − → − ÷ ÷ ÷ − ÷ ÷ − ( ) ( ) ( ) 3 2 2 3 2 1 2 2 1 5 4 2 1 2 2 1 2 0 1 0 0 9 9 9 9 9 9 2 1 2 2 1 2 0 1 0 0 1 0 9 9 9 9 9 9 2 2 1 2 2 1 0 0 1 0 0 1 9 9 9 9 9 9 h h h h h h − + − + − + − ÷ ÷ ÷ ÷ ÷ ÷ → − → − ÷ ÷ ÷ ÷ ÷ ÷ − − ÷ ÷ 1 1 2 2 9 9 9 2 1 2 9 9 9 2 2 1 9 9 9 A − ÷ ÷ ÷ ⇒ = − ÷ ÷ ÷ − ÷ Bài 2 Giải các phương trình ma trận sau 1) 1 2 3 5 3 4 5 9 X = ÷ ÷ Đặt 1 2 3 5 ; 3 4 5 9 A B = = ÷ ÷ Ta có: 1 AX B X A B − = ⇔ = 1 1 2 1 1 2 4 2 1 1 3 1 3 4 3 1 1.4 2.3 2 2 2 1 3 5 1 1 3 1 5 9 2 3 2 2 d b A c a ad bc X − − − − − ÷ = = = = − ÷ ÷ ÷ ÷ − − − − − − − ÷ ⇒ = = − ÷ ÷ ÷ 2) 3 2 1 2 5 4 5 6 X − − = ÷ ÷ − − 10 [...]... 35 1 0 0 2 0 2 4 12 0 −5 −7 9 0 3 5 14 1 0 D= 1 1 1 0 1 2 0 1 1 1 2 1 2 1 h1( 1) + h3 4 ˆ h1+ hˆ ˆˆ 1 0 1 ‡ ˆˆ ˆ 1+ ˆ 5ˆˆ † ˆ h h 1 0 2 1 1 1 1 0 0 0 0 0 1 2 0 1 1 1 2 1 1 2 1 2 0 1 1 = 1 0 2 1 4 1 2 0 3 1 2 1 0 1 1 2 1 4 2 0 3 1 1 2 1 2 −3 1 2 −3 0 2 −3 1 h1( 2) + h 2 ˆ ˆ ˆˆ † = 1 2 1 4 2 1 ‡ ˆˆh1(ˆ 1) +ˆˆ3ˆˆ ˆ h 0 2 1 4 1 2 4 1 2 0 1 2 4 = 8 − 12 − 4 + 1 − 16 + 24 = 1 7)... 1 1 1 1 1 0 2 0 0 = 1 ( 2) × ( 2) × ( 2) = −8 0 0 2 0 0 0 0 2 2) 0 1 D= 1 1 1 0 1 1 1 1 0 1 1 1 0 1 c1↔ c 2 0 1 ˆ ‡ ˆˆ ˆˆ ˆˆ † − ˆ 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 h1( 1) + h 3 0 1 1 1 ˆ ˆ ˆˆ ˆ ‡ ˆˆh1(ˆ 1) +ˆˆ4† − ˆ h 1 0 1 1 0 0 0 1 0 1 1 1 1 1 1 = − 1 × 1 − 1 0 1 1 1 0 1 1 0 = − ( 1 + 1 + 1) = −3 3) 34 2 −5 1 −3 7 1 D= 3 −9 2 4 −6 1 2 1 −5 2 4 c1↔c 3 1 7 −3 ˆ ‡ ˆˆ ˆˆ ˆˆ † − ˆ 7 2 −9 3 2. .. 2 2 x1 − x2 + x3 − x4 = 1 2 x − x − 3x = 2 1 2 4 12 ) 3 x1 − x3 + x4 = −3 2 x1 + 2 x2 − 2 x3 + 5 x4 = −6 Giải 2 1 1 1 1 2 1 1 1 1 ÷ h1( 1) + h 2 ÷ h1( 1) + h 3 2 1 0 −3 2 ÷ 0 0 1 2 1 ÷ ( A B ) = 3 0 1 1 −3 ÷ h1( 1) +h 4→ 1 1 2 2 −4 ÷ 2 2 2 5 −6 ÷ ÷ 0 3 −3 6 −7 ÷ ÷ 1 1 2 2 −4 1 1 2 2 −4 ÷ ÷ 0 0 1 2 1 ÷ h1 ( 2) + h 2 ... trình: x1 − 13 x2 = 15 x1 = 2 x2 + 7 x3 = 6 ⇔ x2 = 1 36 x = 36 x = 1 3 3 2 x1 + x2 − 2 x3 = 10 2) 3x1 + 2 x2 + 2 x3 = 1 5 x + 4 x + 3 x = 4 1 2 3 Giải: Ta có: 2 1 2 10 h1( 1) + h 2 2 1 2 10 1 1 4 −9 ÷ h1( 2) + h 3 ÷ h1↔ h 2 → ( A B ) = 3 2 2 1 ÷ 1 1 4 −9 ÷→ 2 1 2 10 ÷ ÷ 5 4 3 4 ÷ 1 2 7 16 ÷ 1 2 7 16 ÷ 1 1 4 −9 1 1 4 −9... 1 2 x1 + 3x2 + 11 x3 + 5 x4 = 2 x + x + 5x + 2 x = 1 1 2 3 4 2) 2 x1 + x2 + 3x3 + 2 x4 = −3 x1 + x2 + 3x3 + 4 x4 = −3 Giải: Ta có: 2 3 11 5 2 1 ÷ h1↔ h2 1 1 5 2 1 ÷ 2 → ( A / B) = 2 2 1 3 2 −3 ÷ ÷ 1 1 3 4 −3 1 1 5 2 1 ÷ 3 11 5 2 ÷ 1 3 2 −3 ÷ ÷ 1 3 4 −3 16 1 0 → 0 0 1 1 ÷ 1 1 1 0 ÷ h2+ h3 0 → 0 1 −7 2 −5÷ ÷ 0 2 2 −4 0 1 ... 1 5 40 ÷ h1( 2) +h 4→ 1 2 5 1 16 ÷ 6 4 5 3 41 ÷ ÷ 0 −6 11 11 7 ÷ ÷ 1 2 5 1 16 1 2 5 1 16 ÷ ÷ 0 1 ÷ h1↔ h 3 h1( 1) + h 2 1 −7 8 1 15 ÷ 0 −5 3 → → 3 5 −3 2 12 ÷ h1( −3) + h 3 0 11 18 1 −36 ÷ 0 −6 11 11 7 ÷ ÷ 0 −6 11 1 17 ÷ ÷ 1 2 5 1 16 1 ÷ 0 1 ÷ h 2 ↔ h 4 0 h 2( 2) + h 3 0 −5 3 → → h 2( 1) + h 4 0 1 12 1. .. x = 2 1 2 3 4 18 ) 2 x1 + 3x2 + 5 x3 + 9 x4 = 2 x1 + x2 + 2 x3 + 7 x4 = 2 Giải: 1 1 1 1 2 1 ÷ h1( 1) + h 2 1 2 3 42 0 h1( 2) + h 3 → ( A B ) = 2 3 5 9 2 ÷ 0 h1( 1) + h 4 ÷ ÷ 1 1 2 7 2 0 1 0 h 3( 1) + h 4 → 0 0 1 1 0 0 1 2 1 0 1 1 1 0 1 2 3 1 1 2 1 ÷ 3 0 ÷ h 2( 1) + h 3 0 → 0 7 2 ÷ ÷ 0 6 0÷ 1 2 ÷ 3 0÷ 4 2 ÷ ÷ ÷ 2 2 Hệ... 1 2 3 4 1) 8 x1 + 5 x2 − 3 x3 + 4 x4 = 12 3x1 + 3 x2 − 2 x3 + 2 x4 = 6 Giải: Ta có: 15 ( 2 4 A B) = 8 3 2 3 5 3 2 h2( −3) + h3 0 → 0 0 1 4 h1(( 2) ) + h2 2 2 1 1 h1 −4 + h3 ÷ h1 − 3 + h4 2 6÷ 0 1 1 0 ÷ 2 → 0 −3 4 12 ÷ 1 0 ÷ 2 6 0 0 1/ 2 1/ 2 2 1 1 4 2 2 1 ÷ 1 1 0 2 ÷ h3( 1/ 4)+ h4 0 1 1 → 0 0 2 0 2 0 2 ÷ 0 1/ 2 1/ 2. .. 3 3 2 2) D = = 3.5 – 8 .2 = 15 – 16 = -1 8 5 n +1 n 3) D = = (n +1) (n -1) – n2 = n2 - 1 - n2 = -1 n n 1 cos α − sin α 4) D = = cos2 α +sin2 α = 1 sin α cos α Bài 2: Tính các định thức cấp 3: 2 1 3 1) D = 5 3 2 = 18 +2+ 60-9 -16 -15 = 40 1 4 3 3 2 1 2) D = 2 5 3 = 30 +18 +8 -15 -36-8 = -3 3 4 2 4 −3 5 3) D = 3 − 2 8 = 40 -24 -10 5 +10 +22 4-45 =10 0 1 −7 −5 3 2 −4 4) D = 4 1 − 2 =-9 -20 - 32+ 20 + 12 +24 = -5 5 2 −3 1 1 1 5)... −9 3 2 1 −6 4 2 1 −5 2 h1+ h 2 4 h1( 2) + h 3 0 2 1 ˆ ˆ ˆˆ † ‡ ˆˆh1(ˆ 1) +ˆˆ4ˆˆ − ˆ h 7 0 1 1 2 0 1 2 2 6 3 0 2 1 6 2 1 = 1 × 1 1 3 1 − 1 1 2 0 1 2 = − ( 3 + 12 − 6 − 12 ) = 3 4) 3 −3 −5 8 1 0 0 2 h1( −3) + h 2 −3 2 4 −6 h 4+ h1 −3 2 4 −6 h1( 2) + h 3 ˆ ˆˆ ˆˆ ˆ† ˆ ˆˆ ˆˆ ˆˆ ˆˆ † D= ‡ ˆˆ ‡ h1( −4) + h 4ˆ ˆ 2 −5 −7 5 2 −5 −7 5 −4 3 5 −6 −4 3 5 −6 2 4 12 1 2 −6 1 2 = 1 −5 −7 9 = 1 2 × −5 . = Giải: Ta có: ( ) 1( 1) 2 1( 2) 3 1 2 1( 2) 2 1( 1) 2 2 3 2 1 2 10 2 1 2 10 1 1 4 9 3 2 2 1 1 1 4 9 2 1 2 10 5 4 3 4 1 2 7 16 1 2 7 16 1 1 4 9 1 1 4. 3 1( 2) 2 2 1 1 1 1 0 0 1 2 1 2 1 0 3 2 2 1 0 3 2 3 0 1 1 3 1 1 1 4 5 3 2 2 5 6 0 3 2 8 8 1 1 1 4 5 1 1 1 4 2 1 0 3 2 0 3 2 11 0 0 1 2 1 0 0 1 2 0 3 2