Bài tập ⎛ −2 ⎞ ⎛ 2⎞ Cho A = ⎜ ⎟ vaø B = ⎜ ⎟ Tìm 2A − 3B ⎝ −6 ⎠ ⎝ −7 ⎠ ⎛ −7 −4 ⎞ ÑS : ⎜ ⎟ ⎝ 29 −36 ⎠ ⎛ −1 ⎞ ⎛ −2 −3 ⎞ ⎛ −2 ⎞ Cho A = ⎜ ⎟, B = ⎜ ⎟ vaø C = ⎜ ⎟ Tìm 3A + 4B − 2C ⎝ −4 ⎠ ⎝ −1 ⎠ ⎝ −1 −1 ⎠ ⎛ 10 −11 ⎞ ÑS: ⎜ ⎟ ⎝ −2 10 ⎠ ⎛ 2⎞ ⎛2 ⎛1 3⎞ ⎜ ⎟ ⎜ ⎜ ⎟ Cho A = ⎜ −1 ⎟ , B = ⎜ ⎟ vaø C = ⎜ ⎜ 1⎟ ⎜4 ⎜ −3 − ⎟ ⎝ ⎠ ⎝ ⎝ ⎠ ⎛ 11 ⎞ ⎜ ⎟ ÑS: ⎜ −11 ⎟ ⎜ 27 15 ⎟ ⎝ ⎠ ⎛ −1 ⎞ ⎛ −2 ⎞ ⎜ ⎟ Cho A = ⎜ ⎟ vaø B = ⎜ ⎟ Tìm ⎝ ⎠ ⎜ −3 ⎟ ⎝ ⎠ 5⎞ ⎟ ⎟ Tìm 5A − 3B + 2C ⎟⎠ AB , BA ⎛ −1 −8 10 ⎞ ⎛ −15 19 ⎞ ⎜ ⎟ ÑS: AB = ⎜ −2 ⎟ ; BA = ⎜ ⎟ − 10 ⎝ ⎠ ⎜ 22 −15 ⎟ ⎝ ⎠ ⎛ −3 ⎞ ⎛2 ⎜ ⎟ ⎜ Cho A = ⎜ −4 ⎟ vaø B = ⎜ ⎜ −5 ⎟ ⎜1 ⎝ ⎠ ⎝ ⎛ −5 ⎞ ⎛ 29 ⎜ ⎟ ⎜ ÑS: AB = ⎜ 10 ⎟ ; BA = ⎜ 17 ⎜ −7 ⎟ ⎜ 14 ⎝ ⎠ ⎝ ⎛ −4 ⎞ ⎛3 ⎜ ⎟ ⎜ Cho A = ⎜ −5 ⎟ vaø B = ⎜ ⎜ −3 ⎟ ⎜9 ⎝ ⎠ ⎝ 6⎞ ⎟ ⎟ Tìm AB , BA ⎟⎠ −56 27 ⎞ ⎟ −36 19 ⎟ −25 11 ⎟⎠ 5⎞ ⎟ −1 ⎟ Tìm AB , BA ⎟⎠ ⎛ 11 −22 29 ⎞ ⎛ 47 77 −37 ⎞ ⎜ ⎟ ⎜ ⎟ ÑS: AB = ⎜ −27 32 ⎟ ; BA = ⎜ 26 44 −20 ⎟ ⎜ 13 −17 26 ⎟ ⎜ 101 161 −81 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ −1 ⎞ ⎛ 0⎞ T T Cho hai ma traän A = ⎜ ⎟ Tính 3A ± 2B , A A vaø AA ⎟, B = ⎜ − 2 − ⎝ ⎠ ⎝ ⎠ ⎛ −2 ⎞ ⎛ 10 −3 ⎞ ⎛2 ⎛6 ⎞ −3 ⎞ ⎜ ⎟ T T ÑS: A + B = ⎜ ⎟ ⎟; A −B = ⎜ ⎟ ; A A = ⎜ 2 −5 ⎟ ; AA = ⎜ ⎝ −6 −8 ⎠ ⎝ −1 −16 ⎠ ⎝ 17 ⎠ ⎜ −2 −5 17 ⎟ ⎝ ⎠ ⎛ −2 ⎞ ⎛ −1 ⎞ ⎛ 1⎞ ⎜ ⎟ Cho A = ⎜ ⎟ ⎟ , B = ⎜ ⎟ vaø C = ⎜ ⎝ ⎠ ⎝ 2⎠ ⎜ −1 ⎟ ⎝ ⎠ a) Có thể thành lập tích cặp ma trận ma trận b) Tính AB , ABC c) Tính ( AB ) , Cn với n ∈ ÑS: a) AB, BA, BC, CA ⎛ −1 −3 ⎞ ⎛ −1 −4 ⎞ ⎛1 n⎞ n b) AB = ⎜ ⎟ (Chứng minh : C = C Giả sử ⎟ ; ABC = ⎜ ⎟; C = ⎜ ⎝0 1⎠ ⎝2 0⎠ ⎝2 2⎠ ⎛1 n⎞ ⎛ n ⎞ ⎛ 1⎞ ⎛ n + 1⎞ n +1 Cn = ⎜ = Cn C = ⎜ ⎟ Ta coù C ⎟⎜ ⎟=⎜ ⎟ đúng) ⎠ ⎝0 1⎠ ⎝ ⎠ ⎝ 1⎠ ⎝ ⎛ 0⎞ ⎜ ⎟ Cho A = ⎜ 0 ⎟ Tính A A ⎜ 0 0⎟ ⎝ ⎠ ⎛0 1⎞ ⎛ 0 0⎞ ⎜ ⎟ ⎜ ⎟ ÑS: A = ⎜ 0 ⎟ ; A = ⎜ 0 ⎟ ⎜ 0 0⎟ ⎜ 0 0⎟ ⎝ ⎠ ⎝ ⎠ 10 Cho ma traän ⎛ −2 ⎞ ⎜ ⎟ A = ⎜ −8 ⎟ ⎜ −2 ⎟ ⎝ ⎠ Tìm ma trận X cho 3A + 2X = I3 ⎛ −3 ⎞ ⎜ ⎟ ÑS: X = ( 3A − I3 ) = ⎜ −12 ⎟ ⎜ −3 ⎟ ⎝ ⎠ k k 11 Xác định k cho = 2k ÑS: k k 2k = 2k − 4k = 2k ( k − ) = ⇔ k = ∨ k = 12 Tính định thức cấp sau sin x cos x a) b) − cos x sin x ÑS: a) 5; b) 13 Tính định thức cấp sau 1 1 a) −1 −1 − c) −2 −3 b) 1 1 −2 −4 d) −1 −2 −1 e) −3 −2 1 f) −2 3 g) −2 −1 h) 0 j) −4 −3 −1 − i) ÑS: a) 1; b) 2; c) 21; d) −11 ; e) 100; f) 0; g) 55; h) 27; i) −5 ; j) 10 14 Tính a) d định thức cấp sau a b b) c 0 1 c) x 1 1 x 1 1 x 1 1 x 1 0 1 d) 1 1 1 1 0 1 0 1 ĐS: a) abcd Khai triển theo a b c d 0 0 a =d0 b c = d ( −c ) a b ⎡⎣ hàng ⎤⎦ ⎡⎣ cột1⎤⎦ = d ( −c )( −ab ) = abcd ( b) x − ) ( + x ) Biến đổi : x 1 1 x 1 1 x 1 1 x =− 1 x x 1 ⎡ (1 ) ∼ ( ) ⎤ ⎣ ⎦ 1 x x 1 1 =− 1 x 1− x x −1 1− x x −1 0 ⎡( 2) := ( ) − (1) ⎤ ⎢ ⎥ ⎢( 3) := ( 3) − (1) ⎥ ⎢ ⎥ ⎢⎣( ) := ( ) − x (1) ⎥⎦ − x − x − x2 =− 1 x 1− x x −1 1− x 1− x x −1 0 0 =− x −1 (1 − x )( + x ) x 1− x 1− x 0 x −1 0 ⎡( ) := ( ) + ( ) ⎤ ⎣ ⎦ ⎡( ) := ( ) + ( 3) ⎤ ⎣ ⎦ (1 − x )( + x ) = − ( x − ) (1 − x ) ( + x ) = ( x − ) ( + x ) c) −3 ; d) −1 15 Tính định thức cấp x 0 g sau a b y e z h k 0 c d f u l v ÑS: xyzuv Khai trieån theo x a b c y 0 d z e f =v g h k u 0 l 0 v x a b 0 y 0 z e ⎡⎣ haøng 5⎤⎦ g h k u x a b = vu y 0 e = vux z y e z ⎡⎣ coät ⎤⎦ ⎡⎣ coät 1⎤⎦ = vuxyz 16 Tìm ma trận nghịch đảo ma trận A, có, ⎛ 7⎞ ⎜ ⎟ a) A = ⎜ 2 ⎟ ⎜ −7 ⎟ ⎝ ⎠ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ 22 −53 −12 ⎟ ⎜ −9 22 ⎟⎠ ⎝ ÑS: A −1 ÑS: A −1 ÑS: A −1 ÑS: A −1 ÑS: A −1 ÑS: A −1 ÑS: A −1 ÑS: A −1 ⎛ −1 ⎞ ⎜ ⎟ i) A = ⎜ ⎟ ⎜0 1⎟ ⎝ ⎠ ÑS: A −1 ⎛ 0⎞ ⎜ ⎟ j) A = ⎜ ⎟ ⎜ 6⎟ ⎝ ⎠ ÑS: A = , A −1 ⎛ −1 ⎞ ⎜ ⎟ b) A = ⎜ −1 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎛1 ⎞ ⎜ ⎟ c) A = ⎜ 2 ⎟ ⎜1 3⎟ ⎝ ⎠ ⎛ −3 ⎞ ⎜ ⎟ d) A = ⎜ ⎟ ⎜0 ⎟ ⎝ ⎠ ⎛ −1 ⎞ ⎜ ⎟ e) A = ⎜ −1 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎛ −3 ⎞ ⎜ ⎟ f) A = ⎜ −2 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎛ 2 3⎞ ⎜ ⎟ g) A = ⎜ −1 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎛ −1 −1 ⎞ ⎜ ⎟ h) A = ⎜ −1 −1 ⎟ ⎜2 0⎟ ⎝ ⎠ 17 Tìm ma trận X cho XA = ⎛ −3 ⎞ ⎛ −2 ⎜ ⎟ ⎜ A = ⎜ 1 ⎟ vaø B = ⎜ ⎜ −3 −4 ⎟ ⎜ −2 ⎝ ⎠ ⎝ ⎛ 11 −2 −5 ⎞ 1⎜ ⎟ = ⎜ −3 ⎟ 3⎜ ⎟ ⎝ −1 1 ⎠ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ −1 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎛ −2 ⎞ ⎜ ⎟ = ⎜ −2 ⎟ ⎜0 ⎟ ⎝ ⎠ ⎛ −1 ⎞ ⎜ ⎟ = ⎜ −1 −2 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎛ −2 −1 ⎞ 1⎜ ⎟ = ⎜ −4 −4 ⎟ 2⎜ ⎟ ⎝ −4 −3 ⎠ ⎛ −4 −3 ⎞ ⎜ ⎟ = ⎜ −5 −3 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎛ −1 ⎞ ⎟ 1⎜ = ⎜ −1 1 ⎟ 4⎜ ⎟ ⎝ −2 − ⎠ ⎛ 1 −4 ⎞ ⎜ ⎟ = ⎜ −2 ⎟ ⎜0 ⎟ ⎝ ⎠ B , với 6⎞ ⎟ −8 ⎟ ⎟⎠ ⎛ −2 ⎞ ⎛ ⎜ ⎟⎜ ÑS: X = B ⋅ A = ⎜ −8 ⎟ ⎜ −2 ⎜ −2 ⎟ ⎜ ⎝ ⎠⎝ 18 Giải phương trình ma traän ⎞ ⎛ 17 ⎟ ⎜ −3 ⎟ = ⎜ −18 ⎟⎠ ⎜⎝ 16 sau ⎛ −2 ⎞ ⎛ ÑS: X = ⎜ 1⎟ ⎜ ⎟⎜ − ⎠ ⎝5 ⎝ −1 ⎛1 2⎞ ⎛ 5⎞ ⎟X = ⎜ ⎟ ⎝3 4⎠ ⎝ 9⎠ a) ⎜ ⎛ −2 ⎞ ⎛ − ⎞ ⎟=⎜ ⎟ ⎝ −4 ⎠ ⎝ −5 ⎠ b) X ⎜ 26 ⎞ ⎟ −25 ⎟ 25 ⎟⎠ ⎞ ⎛ −1 −1 ⎞ ⎟=⎜ ⎟ 9⎠ ⎝ ⎠ ⎛ −1 ⎞ ⎛ −1 ⎞ ⎛ −2 ⎞ = ⎟ ⎜⎜ ⎟ 3⎟ ⎟ ⎜ ⎝ −5 ⎠ ⎝ − ⎠ ⎝ −4 ⎠ ÑS: X = ⎜ ⎛ −1 ⎞ ⎛ ⎞ ⎛ 14 16 ⎞ ⎟X⎜ ⎟=⎜ ⎟ ⎝ −2 ⎠ ⎝ ⎠ ⎝ 10 ⎠ ⎛ −1 ⎞ ⎛ 14 16 ⎞ ⎛ −4 ⎞ ⎛ ⎞ = ⎟⎜ ⎟ ⎜⎜ ⎟ 5⎟ ⎟ ⎜ ⎝ −3 ⎠ ⎝ 10 ⎠ ⎝ − ⎠ ⎝ ⎠ c) ⎜ ÑS: X = ⎜ ⎛ −3 ⎞ ⎛ −3 ⎞ ⎜ ⎟ ⎜ ⎟ d) ⎜ −4 ⎟ X = ⎜ 10 ⎟ ⎜ −1 ⎟ ⎜ 10 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ −4 −2 ⎞ ⎛ −3 ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ÑS: X = ⎜ −8 −5 ⎟ ⎜ 10 ⎟ = ⎜ 2 ⎟ ⎜ −7 −4 ⎟ ⎜ 10 ⎟ ⎜ 3 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ 13 −8 −12 ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ e) X ⎜ 12 −7 −12 ⎟ = ⎜ ⎟ ⎜ −4 − ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ 13 −8 −12 ⎞ ⎛ 55 −34 −51 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ÑS: X = ⎜ ⎟ ⎜ 12 −7 −12 ⎟ = ⎜ 148 −91 −138 ⎟ ⎜ ⎟ ⎜ −4 −5 ⎟ ⎜ 241 −148 −225 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ −2 ⎞ ⎛7 0⎞ ⎛ −2 ⎞ ⎛ ⎞ ⎛ 20 −4 12 ⎞ ⎜ ⎟ ⎜ ⎟ 1⎜ ⎟⎜ ⎟ 1⎜ ⎟ f) ⎜ −4 ⎟ X = ⎜ ⎟ ÑS: X = ⎜ −4 ⎟ ⎜ ⎟ = ⎜ 34 −8 −6 ⎟ 6⎜ ⎜ −1 ⎟ ⎜1 5⎟ ⎟⎜ ⎟ ⎜ 23 −19 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ −5 ⎠ ⎝ ⎠ ⎝ ⎠ 19 Tìm hạng ⎛1 ⎜ a) ⎜ −1 ⎜ −3 −2 ⎝ ma trận sau −1 ⎞ ⎟ 2⎟ ⎟⎠ ⎛1 ⎜ −1 − b) ⎜ ⎜5 ⎜⎜ ⎝ 10 1⎞ ⎟ 0⎟ 1⎟ ⎟ ⎟⎠ ⎛ −2 ⎞ ⎜ ⎟ −1 −2 −3 ⎟ ⎜ c) ⎜ −1 ⎟ ⎜⎜ ⎟⎟ ⎝ −3 −8 ⎠ ⎛ −1 −1 ⎞ ⎜ ⎟ −1 − ⎟ ⎜ d) ⎜1 −6 −9 ⎟ ⎜⎜ ⎟⎟ ⎝ 12 −2 −2 −10 ⎠ ⎛ −1 −1 ⎞ ⎜ ⎟ −1 ⎟ ⎜ e) ⎜ −1 −2 −7 ⎟ ⎜⎜ ⎟⎟ ⎝ −1 −1 ⎠ ⎛0 ⎜ f) ⎜ ⎜3 ⎜⎜ ⎝4 −3 −5 ⎞ ⎟ −2 − ⎟ −5 12 ⎟ ⎟ 5 ⎟⎠ ĐS: a) Biến đổi ⎛ −1 ⎞ ⎛ −1 ⎞ ⎛ −1 ⎞ 2):= ( 2) + (1) ): = ( ) − ( ) ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ( ( → ⎜ ⎟ ⎯⎯⎯⎯⎯⎯ → ⎜ 2⎟ ⎜ −1 ⎟ ⎯⎯⎯⎯⎯⎯ : 3 = + ( ) ( ) () ⎜ −3 − ⎟ ⎜ 2⎟ ⎜ 0 0⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ b) Biến đổi ⎛1 ⎜ ⎜ −1 − ⎜5 ⎜⎜ ⎝ 10 ⎛1 1⎞ 1⎞ ( ) : = ( ) − (1 ) ⎜ ⎟ ⎟ 0⎟ ( 3):= ( 3) − 5(1) ⎜ −11 −6 −7 −2 ⎟ ⎯⎯⎯⎯⎯⎯ → ( ):= ( ) − (1) ⎜ −22 −12 −14 −4 ⎟ 1⎟ ⎟ ⎜⎜ ⎟⎟ ⎟⎠ ⎝ −11 −6 −7 −2 ⎠ ⎛1 1⎞ ⎜ ⎟ ( 3):= ( 3) − 2( 2) ⎜ −11 −6 −7 −2 ⎟ ⎯⎯⎯⎯⎯⎯ → ( ): = ( ) − ( ) ⎜0 0 0⎟ ⎜⎜ ⎟ 0 ⎟⎠ ⎝0 c) Biến đổi ⎛ −2 ⎞ ⎛1 −2 ⎞ ( ) : = ( ) − (1 ) ⎜ ⎜ ⎟ ⎟ ( ) : = ( ) − ( ) ⎜ −5 − − ⎟ ⎜ −1 −2 −3 ⎟ ⎯⎯⎯⎯⎯⎯ → ( ):= ( ) − 2(1) ⎜ −4 −10 −14 ⎟ ⎜ −1 ⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎝ −3 −8 ⎠ ⎝ −7 −4 −20 ⎠ ⎛1 ⎜ ( 3):= ( 3) − 45 ( 2) ⎜ −5 −8 ⎯⎯⎯⎯⎯⎯→ ( ):= ( ) − 75 ( 2) ⎜ 0 − 18 ⎜ 36 ⎜0 ⎝ −2 36 18 ⎞ ⎛1 −2 ⎞ ⎟ ⎜ ⎟ −4 ⎟ ( ):= ( ) + 2( 3) ⎜ −5 −8 −4 ⎟ ⎟ ⎯⎯⎯⎯⎯⎯→ ⎜ 0 − 18 36 − 54 ⎟ − 54 ⎟ 5 ⎟ ⎜ ⎜0 ⎟ ⎟ − 72 − 18 36 ⎝ ⎠ ⎠ d) Biến đổi ⎛3 ⎜ ⎜1 ⎜1 ⎜⎜ ⎝ 12 ⎛1 ⎜ ⎜0 ⎜0 ⎜⎜ ⎝0 −1 −1 −1 −2 −6 −2 −2 −1 −2 2 5 −10 −10 10 25 −50 ⎛ −1 ⎞ ⎟ ⎜ ⎟ (1 ) ∼ ( ) ⎜ − ⎯⎯⎯⎯→ ⎜1 −9 ⎟ ⎟⎟ ⎜⎜ −10 ⎠ ⎝ 12 −2 ⎛1 ⎞ ⎟ ⎜ −14 ⎟ ): = ( ) − ( ) ( ⎜0 ⎯⎯⎯⎯⎯⎯→ ( ): = ( ) − ( ) ⎜ −14 ⎟ ⎟⎟ ⎜⎜ −70 ⎠ ⎝0 −2 −1 −6 −2 −1 − 2 0 ⎞ ( ) : = ( ) − (1 ) ⎟ ⎟ ( ) : = ( ) − (1 ) ⎯⎯⎯⎯⎯⎯⎯ → ):= ( ) −12(1) ( −9 ⎟ ⎟ −10 ⎟⎠ ⎞ ⎟ −10 −14 ⎟ 0 ⎟ ⎟ 0 ⎟⎠ e) Biến đổi ⎛ − 1 −1 ⎞ ⎛ −1 − ⎞ ( 2):= ( 2) − (1) ⎜ ⎟ ⎜ ⎟ −2 −2 ⎟ ( 3):= ( 3) + (1) ( 3) := ( ) − ( ) ⎜ −1 ⎟ ⎯⎯⎯⎯⎯⎯ ⎜ → ⎯⎯⎯⎯⎯⎯ → ( ):= ( ) − 2(1) ⎜ −1 −5 ⎟ ( ):= ( ) − ( 2) ⎜ −1 −2 −7 ⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎝ − −1 ⎠ ⎝ −3 −1 ⎠ ⎛ −1 −1 ⎞ ⎛ −1 −1 ⎞ ⎜ ⎟ ⎜ ⎟ ( ):= ( ) + ( 3) → ⎜ −2 −2 ⎟ ⎜ −2 −2 ⎟ ⎯⎯⎯⎯⎯⎯ ⎜ 0 −3 ⎟ ⎜ 0 −3 ⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎝ 0 −1 −1 ⎠ ⎝ 0 −2 ⎠ f) Biến đổi ⎛0 ⎜ ⎜1 ⎜3 ⎜⎜ ⎝4 ⎛1 ⎜ ⎜0 ⎜0 ⎜⎜ ⎝0 −3 −2 −5 0 −2 −3 12 −22 22 −24 20 Bieän ⎛1 ⎜ a) ⎜ ⎜1 ⎝ ⎛ −2 −4 ⎞ −5 ⎞ ⎟ ⎜ ⎟ −4 ⎟ (1) ∼ ( 2) ⎜ −3 −5 ⎟ ( 3):= ( 3) − 3(1) ⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯⎯ → ( ) : = ( ) − (1 ) ⎜ −5 12 ⎟ 12 ⎟ ⎟ ⎜⎜ ⎟⎟ ⎟⎠ ⎝4 5 ⎠ ⎛ −2 −4 ⎞ −4 ⎞ ⎜ ⎟ ⎟ −5 ⎟ −5 ⎟ ( ):= ( ) − 1222 ( 3) ⎜ −3 ⎯⎯⎯⎯⎯⎯⎯ →⎜ 0 12 −22 34 ⎟ 34 ⎟ ⎜ ⎟ ⎟⎟ 49 79 ⎟ ⎜0 0 − 36 ⎠ 3 ⎠ ⎝ luận theo m hạng ma trận sau −3 ⎞ ⎟ m⎟ m ⎟⎠ ⎛ m 5m − m ⎞ ⎜ ⎟ m 10m ⎟ b) ⎜ 2m ⎜ − m −2m −3m ⎟ ⎝ ⎠ ⎛3 ⎜ m c) ⎜ ⎜1 ⎜⎜ ⎝2 1 4⎞ ⎟ 10 ⎟ 17 ⎟ ⎟ ⎟⎠ ⎛ −1 ⎞ ⎜ ⎟ −1 2 ⎟ ⎜ d) ⎜ 1 2⎟ ⎜⎜ ⎟⎟ ⎝ −2 − 1 m − ⎠ ÑS: a) m = hay m = −5 : rank = 2; m ≠ 0, −5 : rank = ⎛ 1 −3 ⎞ ⎛1 −3 ⎞ ⎜ ⎟ ⎟ ( ) : = ( ) − (1 ) ⎜ ( 3):= ( 3) + ( m −1)( 2) →⎜0 −1 → m + ⎟ ⎯⎯⎯⎯⎯⎯⎯⎯ ⎜ m ⎟ ⎯⎯⎯⎯⎯⎯ ( 3):= ( 3) − (1) ⎜1 m ⎟ ⎜0 m − ⎟⎠ ⎝ ⎠ ⎝ ⎛1 ⎞ ⎛1 −3 −3 ⎞ ⎜ ⎟ ⎜ ⎟ m+6 m+6 ⎟ ⎜ −1 ⎟ = ⎜ −1 ⎜ 0 + ( m − 1)( m + ) ⎟ ⎜ 0 m2 + 5m ⎟ ⎠ ⎝ ⎠ ⎝ b) m = : rank = 0; m ≠ : rank = ⎛ m 5m −m ⎞ ⎛ m 5m −m ⎞ ) : = ( ) − (1 ) ⎜ ⎟ ⎜ ⎟ ( 3):= ( 3) + 13 ( 2) ( → − ⎯⎯⎯⎯⎯⎯ → m 10m ⎟ ⎯⎯⎯⎯⎯⎯ 9m 12m ⎜ 2m ⎜ ⎟ ( ) : = ( ) + (1 ) ⎜ − m −2m −3m ⎟ ⎜ 3m −4m ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ m 5m − m ⎞ ⎜ ⎟ ⎜ −9m 12m ⎟ ⎜0 0 ⎟⎠ ⎝ c) m = : rank = 3; m ≠ : rank = ⎛3 ⎜ ⎜m ⎜1 ⎜⎜ ⎝2 ⎛1 ⎜ ⎜0 ⎜0 ⎜⎜ ⎝0 ⎛ 17 ⎞ 1 4⎞ ( 2):= ( 2) − 2(1) ⎟ ⎜ ⎟ 10 ⎟ (1 ) ∼ ( ) ⎜ 2 ⎟ ( ) : = ( ) − (1 ) ⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯⎯⎯ → ( ) ∼ ( ) ⎜ 1 ⎟ ( ) : = ( ) − m (1 ) 17 ⎟ ⎟ ⎜⎜ ⎟⎟ ⎟⎠ ⎝ m 10 ⎠ 17 ⎞ ⎟ −12 −30 −5 ⎟ −20 −50 −5 ⎟ ⎟ − 7m 10 − 17m − 3m ⎟⎠ ⎛1 ⎛1 17 ⎞ 17 ⎞ ⎜ ⎟ ⎜ ⎟ −5 ⎟ −5 ⎟ 3) ∼ ( ) ⎜ −12 −30 ⎜ −12 −30 ( 3):= ( 3) − 1220 ( 2) ( ⎯⎯⎯⎯⎯⎯⎯→ 10 m ⎟ ⎯⎯⎯⎯→ ⎜ 0 ⎟ − 23 − m ( ):= ( ) + −127m ( 2) ⎜ 0 ⎟ ⎜ ⎟ ⎜ 10 m ⎜0 ⎟ ⎜0 ⎟ − 23 − m ⎠ ⎝ ⎝ ⎠ ⎛1 17 ⎞ ⎜ ⎟ −12 −30 −5 ⎟ ) := ( ) + ( ) ( ⎜ m = : ⎯⎯⎯⎯⎯⎯→ ⎜ 0 − 23 ⎟ ⎜ ⎟ ⎜0 0 ⎟⎠ ⎝ d) m = −2 : rank = 3; m ≠ −2 : rank = ⎛ −1 ⎞ ⎛ −1 ( ) : = ( ) + (1 ) ⎜ ⎜ ⎟ ( 3):= ( 3) + (1) → ⎜ 0 ⎜ −1 2 ⎟ ⎯⎯⎯⎯⎯⎯ : = − ( ) ( ) () ⎜0 ⎜ 1 2⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎝ −2 − 1 m − ⎠ ⎝ −1 −3 m − ⎛ −1 ⎛ −1 0⎞ ⎜ ⎟ ⎜ 2⎟ ( ): = ( ) + ( ) → ⎜ ⎜0 ⎯⎯⎯⎯⎯⎯ ⎜0 ⎜0 2⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎝ −1 −3 m − −2 ⎠ ⎝ 0 m+2 10 0⎞ ⎟ 2⎟ ( 3) ∼ ( ) ⎯⎯⎯⎯→ 2⎟ ⎟ −2 ⎟⎠ 0⎞ ⎟ 2⎟ 2⎟ ⎟ ⎠⎟