4 This assignment, based on the content of Unit 4 of the Study Guide, is worth 3% of your grade. We recommend that you hand it in after you complete Unit 4. You must show all of your work in order to obtain full marks. For your convenience, each exercise mentions the section of the Study Guide that corresponds to the given problem. Note: Each of the questions below is of equal weight, and each will be marked out of twenty (20) points. To gain full marks, you must show all your work. (Section 4.1) 20 points 1. a. Let ~u = (1, 2, 3, 5, 0), ~v = (0, 4, 1, 1, 2), and w ~ = (7, 1, 4, 2, 3). Find the components of i. ~v + w ~ ii. 3(2~u ~v ) iii. (3~u ~v ) (2~u + 4w) ~ iv. b. 1 (w ~ 5~v + 2~u) + ~v 2 Find scalars c1 , c2 , and c3 for which the equation is satisfied, c1 (1, 0, 2) + c2 (2, 2, 2) + c3 (1, 2, 1) = (6, 12, 4) (Section 4.2) 20 points 2. a. Find the Euclidean distance between ~u and ~v and the cosine of the angle between those vectors. State whether that angle is acute, obtuse, or 90 . i. ~u = (1, 2, 3, 0), ~v = (5, 1, 2, 2) ii. ~u = (0, 1, 1, 1, 2), ~v = (2, 1, 0, 1, 3) b. Verify that the Cauchy-Schwarz inequality holds. i. ~u = (4, 1, 1), ~v = (1, 2, 3) ii. ~u = (1, 2, 1, 2, 3), ~v = (0, 1, 1, 5, 2) (Section 4.3) 20 points 3. a. Find the distance between the given parallel planes 2x y + z = 1; b. i. 2x y + z = 1. Show that ~v = (a, b) and w ~ = (b, a) are orthogonal vectors. ii. Use the result in part i. to find two vectors that are orthogonal to ~v = (2, 3). iii. Find two unit vectors that are orthogonal to ~v = (3, 4). Mathematics 270: Linear Algebra I Assignment 4 1 (Section 4.4) 20 points 4. a. Find the general solution to the linear system and confirm that the row vectors of the coefficient matrix are orthogonal to the solution vectors. x1 + 3x2 4x3 = 0 x1 + 2x2 + 3x3 = 0 b. i. Find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors in IR3 that are orthogonal to ~a = (3, 2, 1) and ~b = (0, 2, 2). ii. What kind of geometric object is the solution space? iii. Find a general solution of the system obtained in part i., and confirm that Theorem 3.4.3 of the textbook holds. (Section 4.5) 20 points 5. a. Find the area of the triangle in 3-space that has vertices P (1, 1, 2), Q(0, 3, 4) and R(6, 1, 8). b. Given ~u = (1, 2, 4), ~v = (3, 4, 2), w ~ = (1, 2, 5), compute the scalar triple product ~u (~v w) ~ c. Given ~u (~v w) ~ = 3, find the following, i. ~v (~u w) ~ ii. (~u w) ~ ~v iii. ~v (w ~ w) ~ 2 Assignment 4 Mathematics 270: Linear Algebra I Assignment 5 This assignment, based on the content of Unit 5 of the Study Guide, is worth 3% of your grade. We recommend that you hand it in after you complete Unit 5. You must show all of your work in order to obtain full marks. For your convenience, each exercise mentions the section of the Study Guide that corresponds to the given problem. Note: Each of the questions below is of equal weight, and each will be marked out of twenty (20) points. To gain full marks, you must show all your work. (Section 5.1) 20 points 1. a. Let V be the set of real-valued functions that are defined at each x in the interval (, ). If f~ = f (x) and ~g = g(x) are two functions in V and if k is any scalar, then define the operations of addition and scalar multiplication by (f~ + ~g ) = f (x) + g(x) k f~ = kf (x) Verify the Vector Space Axioms for the given set of vectors. b. Let V consist of the form ~u = (u1 , u2 , ..., un , ...) in which u1 , u2 , ..., un , ... is an infinite sequence of real numbers. Define two infinite sequences to be equal if their corresponding components are equal, and define addition and scalar multiplication componentwise by ~u + ~v = (u1 , u2 , ..., un , ...) + (v1 , v2 , ..., vn , ...) = (u1 + v1 , u2 + v2 , ..., un + vn , ...) k~u = (ku1 , ku2 , ..., kun , ...). Prove that the given set is a vector space. (Section 5.2) 20 points 2. a. In each part express the vector as a linear combination of p~1 = 2 + x + 4x2 , p~2 = 1 x + 3x2 , and p~3 = 3 + 2x + 5x2 . i. 9 7x 15x2 ii. 6 + 11x + 6x2 iii. 0 iv. 7 + 8x + 9x2 b. Suppose that v~1 = (2, 1, 0, 3), v~2 = (3, 1, 5, 2), and v~3 = (1, 0, 2, 1). Which of the following vectors are in span {v~1 , v~2 , v~3 }? i. (2, 3, 7, 3) ii. (0, 0, 0, 0) iii. (1, 1, 1, 1) iv. (4, 6, 13, 4) Mathematics 270: Linear Algebra I Assignment 5 1 (Section 5.3) 20 points 3. a. i. Show that the vectors v~1 = (1, 2, 3, 4), v~2 = (0, 1, 0, 1), and v~3 = (1, 3, 3, 3) form a linearly dependent set in IRn . ii. Express each vector in part i. as a linear combination of the other two. b. Prove that if {v~1 , v~2 , v~3 } is a linearly independent set of vectors, then so are {v~1 , v~3 }, and {v~2 }. (Section 5.4) 20 points 4. a. i. Use matrix multiplication to find the construction of (a, b) with factor k = 1/, where > 1. ii. Use matrix multiplication to find the construction of (a, b) with factor k = , where > 1. b. Find the standard matrix for the reflection of IR2 about the line that makes an angle of /4 (= 45 ) with the positive x-axis, and then use that matrix to find the reflection of (1, 2) onto this line. (Section 5.5) 20 points 2 5. a. Find the matrix for a shear in the x-direction that transforms the triangle with vertices (0, 0), (2, 1), and (3, 0) into a right triangle with the right angle at the origin. b. Draw a figure that shows the image of the triangle with vertices (0, 0), (1, 0), and (0.5, 1) under a shear by a factor of 2 in the x-direction. Assignment 5 Mathematics 270: Linear Algebra I : 2 9 7 15 6 + 11 + 6 = 2 2 + =4 2+ +4 0 =0 2+ +4 7+8 +9 +4 +1 1 5 1 +0 1 =0 2+ +4 +3 2 1 +3 2 3+2 +5 +3 +1 3+2 +5 +0 3+2 +5 +3 +3 3+2 +5 . 2 3 1 0 = 1 1 0 5 2 3 2 1 5 1 5 5 1 0 2 1 ~ 0 1 0 0 ~ 0 5 1 1 0 0 0 0 1 4 0 0 3 0 3 3 0 2 2 0 1 16 1 6 2 3 1 2 ~ 0 1 0 0 5 1 0 0 1 4 0 0 3 0 1 1 0 2 2 0 1 16 1 6 3 1 1 2 ~ 0 5 1 1 0 0 0 0 1 4 0 0 3 0 1 1 0 0 0 0 1 16 1 6 3 1 5 0 , 4 0 3 0 7 0 6 0 2 3 7 3 , 0 0 0 0 1 4 1 6 1 13 1 4 1 16 1 6 1 13 2 14 [ [ , , , , 2 3 2 3 1 2 3 4 = 0 1 0 1 1 3 3 3 1 ~ 0 0 2 3 4 1 0 1 1 0 1 1 ~ 0 0 2 3 4 1 0 1 0 0 0 ,[ ,,[ 1 1,2,3,4 1 0,1,0, 1 + 1 1,3,3,3 = 0 , , 1,2,3,4 = 1 0,1,0, 1 + 1 1,3,3,3 . 1 0,1,0, 1 = 1 1,2,3,4 + 1 1,3,3,3 1,3,3,3 = 1 1,2,3,4 + 1 0,1,0, 1 , , + => + = 0, =0 = 0, = 0 + => = 0 =0 0. , , => =0 =0 , =0 =0 0 . 4 5 = 0 0 = 0 0 3 3 = => 3 = 3 0 0 2 0 = 1 1 [ =0 =1, =0 => =1 =0 => 2 + 3 = 0 => =1, 2 + 2 + =1 =0 = 2 , = 0 , 1 0 = 2 + =1 =1 2 1 = 0 0 = 0 0 1 2 = => 0 0 =2 1 0.5 = => + 1 1 2 =2 => =0 =2, 2 + + =2 =1, =0, =1 = =2 =0 2 0 2 =1 2 + =1 1 1 0,0 , 1,0 , 0.5,1 . Assignment 4 1 (a) u= (1,2,3,5,0 ) , v =( 0,4,1,1,2 ) w =( 7,1,4,2,3 ) . (i) v + w=( 0,4,1,1,2 )+ ( 7,1,4,2,3 )=(7,5,5,1,5) ( ii ) 3 ( 2 u v ) =3 [ 2 ( 1,2,3,5,0 )( 0,4,1,1,2 ) ] =3 ( 2,0,5,9,2 )=( 6,0,15,27,6 ) ( iii ) ( 3 uv )( 2 u +4 w ) =[ 3 ( 1,2,3,5,0 )( 0,4,1,1,2 ) ] [ 2 ( 1,2,3,5,0 ) +4 ( 7,1,4,2,3 ) ] ( 3 ,2 , 10,14,2 ) ( 30,8,22,2,12 )=(27 ,6,32,12,4 ) ( iv ) 1 1 ( w 5 v +2 u ) + v = [( 7,1,4,2,3 )5 ( 0,4,1,1,2 ) +2 ( 1,2,3,5,0 ) ] + ( 0,4,1,1,2 ) 2 2 1 ( 8,15,5,3,7 ) + ( 0,4,1,1,2 ) 2 7 7 5 3 4 , , , , . 2 2 2 2 ( ) (b) c 1 (1,0,2 ) +c 2 ( 2,2,2 )+ c 3 ( 1,2,1 )=(6,12,4 ) >c 1+ 2 c2 +c 3=6 2 c 22 c3 =12 2 c1 2 c2 +c 3=4 AX =b . Augmented [ 1 [ A : b ]= 0 [ [ 2 2 1 :6 2 2 :12 2 1: 4 1 2 1 :6 0 2 2 :12 ,[R 3 R3 +2 R 1] 0 2 3 :8 1 2 1:6 0 2 2:12 ,[R 3 R 3R1 ] 0 0 5:20 >5 c 3=20= c 3=4 2 c 22 c3 =12= c 2=2 c1 +2 c 2+c 3=6= c1 =6 Thus c 1=6 , c 2=2 , c 3=4 . 2 (a) (i) u= (1,2,3,0 )v =( 5,1,2,2 ) |u|= 1+ 4+ 9+0=14|v|=25+ 1+ 4+ 4= 34 Cos= ( u . v ) 5+ 26+ 0 1 = = |u ||v| 14 34 476 =Co s1 1 ( 476 ) 87.37 Angle is ( ii ) u= ( 0,1,1,1,2 )v =(2,1,0,1,3) |u|= 0+1+1+1+4= 6|v|= 4 +1+ 0+1+9= 15 Cos= ( u . v ) 0+1+ 01+6 6 = = |u ||v| 6 15 90 =Co s1 ( 690 ) 50.77 Angle is (b) (i) u= ( 4,1,1 )v =( 1,2,3 ) Cuachy Schwarz inequality u , v > u , v > 4 +2+3=9 u , u > 16+1+1=18 v , v > 1+4 +9=14 2 Here 9 <18 14 thus satisfied ( ii ) u )v =( 0,1,1,5,2 cuachy schwarz inequality , v> u , v > 0+2+1+106=7 u , u > 1+ 4+ 1+ 4+ 9=19 v , v > 0+1+1+25+ 4=31 Here 72 <19 31 thus satisfied 3 (a) 2 x y+ z=12 y + choose a point on the plane say ( 0,01 ) distance between these two planes is =(b) (i) v , b w |2 0 )0+1+1| 4+1+1 6 . hence are orthogonal ii 2,3 let ,b use i if 2a3>Choose ab Thus two vectors that are orthogonal v are w 1=( 3,2 ) w 2= ( 6 , 4 ) ( iii ) v =(3,4 ) Les us take two o rthogonal vector v w 1=( 4,3 ) w2 =(4,3). n^1= 1 w 1 1 = ( 4,3 )= ( 4,3 ) 5 w1| 16+ 9 | n^2= 2 w 1 1 = (4,3 )= (4,3 ) 5 w2| 16+ 9 | Here n^1 ^ n2 areunit vectors n^1 . v =0^ n2 . v =0 Hence n^1^ n2 are two unit vectors which orthogonal v 4 (a) x 1+3 x 24 x 3=0 x 1+2 x 2+ 3 x 3=0 > > > [ ] [ ][ ] [ ] [ x 1 3 4 1 =0 1 2 3 x2 0 x3 [ x 1 3 4 1 = 0 , [ R 2 R 2R1 ] 0 1 7 x 2 0 x3 [ x 1 0 17 1 = 0 , [ R1 R1 +3 R1 ] 0 1 7 x 2 0 x3 > x 1+ 17 x3 =0x 2+ 7 x3 =0. Setting x3 =k then wehave > x 1=17 k x 2=7 k x 3=k Thus genral solution is (17 k ,7 k , k ) where k R 1 3 4 1 2 3 Now Take raw of [ ]that is (1,3,4 )( 1,2,3 ) ( 1,3,4 ) . (17 k , 7 k , k ) =17 k + 21 k4 k =0 ( 1,2,3 ) . (17 k , 7 k , k )=17 k +14 k +3 k =0 1 3 4 1 2 3 Thus genral solution is (17 k ,7 k , k ) is orthogonal the raw of [] (b) (i) Let x =( x 1 , x 2 , x 3 ) be the three unknowndefine a . x =0 b . x =0 > (3 , 2,1 ) . ( x 1 , x 2 , x 3 ) =0 ( 0,2,2 ) . ( x 1 , x 2 , x 3 )=0 >3 x1 +2 x 2x 3=0 0. x12 x 22 x 3=0 [ ] [ ] x 3 2 1 1 = 0 ... ... ... . (1) 0 2 2 x 2 0 x3 > [ > x 3 0 3 1 =0 0 2 2 x2 0 x3 [ 3 x 13 x 3=02 x 22 x3 =0 setting x 3=k then x 1=k x 2=k solution is ( x 1 , x 2 , x 3 ) =(k ,k , k ) . 3 2 1 0 2 2 Thus genral solution (k ,k , k ) is orthogonal the raw of [ ] >Thus genral solution (k ,k , k ) is orthogonal the a b > solutionspace of ( 1 ) is orthogonal the a b ( ii ) Solution space=k (1 ,1,1 ) where k R Span { (1,1,1 ) } Straight line passing ( 0,0,0 ) (1 ,1,1 ) ( iii ) >3 x1 +2 x 2x 3=0 0. x12 x 22 x 3=0 [ ] [ ] x 3 2 1 1 = 0 ... ... ... . (1) 0 2 2 x 2 0 x3 > [ > x 3 0 3 1 =0 0 2 2 x2 0 x3 [ 3 x 13 x 3=02 x 22 x3 =0 setting x 3=k then x 1=k x 2=k Genral solution is ( x1 , x2 , x3 ) =(k ,k , k ) . . 5 (a) P=( 1,1,2 ) , Q=( 0,3,4 )R=( 6,1,8 ) Let u=QP=(1,4,2 ) v =RP=( 5,2,6 ) [ i j k u v = 1 4 2 5 2 6 ( 244 ) i (610 ) j+ (220 ) k 20 i+16 j22 k=(20,16,22) . 1 Area of triangle PQR is= |u v| 2 1 400+256+ 484 2 1 1 1140= 2 285= 285. 2 2 (b) u= (1 , 2,4 ) , v =( 3,4,2 ) w =(1 ,2,5 ) u . ( v w )=Det ([ 1 2 4 3 4 2 1 2 5 ) 1 ( 20+ 4 ) 2 ( 152 ) +4 ( 6+4 ) 2426+40 10 . (c ) u . ( v w )=3 Since u . ( v w )=v . ( w u )= w . ( u v ) . (i) v . ( u w )=v . ( w u )=3 ( ii ) ( u w ) . v =v . ( u w )=3 ( iii ) v . ( w w )=0. Ans : 1 (a) Let V =Set of all real values functions x ( , ) We have prove that V isVector Space field real line R : the Addition Axiom: ( 0 ) Closure property if f ( x ) , g ( x ) V then f ( x ) +g (x)V , f (x ), g( x ) V ( i ) Associative property If f ( x ), g(x ) ,h ( x)V then f ( x ) + ( g ( x )+ h ( x ) ) =f ( x ) +g ( x ) +h ( x ) ( f ( x ) + g ( x ) ) + h ( x )=( f ( x )+ g ( x ) ) +h ( x ) > f ( x ) + ( g ( x )+ h ( x ) ) =( f ( x ) + g ( x ) ) + h ( x ) , f (x ), g( x ), h(x ) V Thus Associative property hold ( ii ) Commutaive property if f ( x ) , g ( x ) V then f ( x ) +g ( x )=g ( x ) + f ( x ) , f , g V Thus Commutative property hold ( iii ) existence of identity Since zero functionO ( x ) V such that for all x R we have O ( x )=0 f ( x )+ O ( x )=f ( x ) +0=0+ f ( x ) > f ( x ) +O ( x ) =O ( x ) + f ( x )=f ( x ) , f ( x ) V Thus identity of V exist Ois identity for V ( iv ) Existence of inverse For each f ( x ) V we havef ( x ) V such that f ( x ) + (f ( x ) )=f ( x )f ( x )=f ( x )+ f ( x )=(f + f ) ( x ) =O ( x ) > f ( x ) + (f ( x ) )=f ( x ) + f ( x )=O ( x ) Thus inverse element exist for each f ( x ) V f ( x ) is inverse for f ( x ) Multiplication Axiom : ( 0 ) Closure property if k R then cf ( x ) V , k R ( i ) Compatibility of Scaler multiplication with field multiplication if , a , b R an d f ( x ) V then a ( bf ( x )) =abf ( x )=( ab ) f ( x ) >a ( bf ( x ) )= ( ab ) f ( x ) , f ( x ) V ( ii ) Identity element of scaler multiplication since 1 R1 f ( x ) =f ( x )= 1 f ( x )=f ( x ) , f ( x ) V ( iii ) Distributive of scaler multiplication w . r .t . vector addition if a Rf ( x ) , g ( x ) V then a { f ( x ) + g ( x ) }=af ( x )+ bg ( x ) , a Rf ( x ) , g ( x ) V ( iv ) Distributive of scaler multiplication w . r . t . field addition if a , b Rf ( x ) V then ( a+b ) f ( x ) =af ( x ) +bf (x) a , b Rf V Thus V satisfied all condition of vector space hence V is vector space proved (b) Let V =Set of all vector u =(u 1 , u2 , ... ... .u n , ... ... ... .) We have prove that V isVector Space field real line R : the Addition Axiom: ( 0 ) Closure property if u , v V then u + v =( u1 , u2 ,... ... . un , ... ... ... . ) + ( v 1 , v 2 , ... ... . v n , ... ... .... ) (u1 + v n ,u2 + v 2 , ... ... . un +v n , ... ... ... .)V , u , v V ( i ) Associative property If u , v , w V then u + ( v + w ) =( u1 ,u 2 , ... ... . un ,... ) + {( v 1 , v 2 , ... ... . v n ,.. )+ ( w 1 , w 2 , ... w n , .. ) } ( u1 ,u 2 , ... ... . un , ... ) + ( v 1+ w1 , v 2+ w2 , ... ... . v n +w n , .. ) ( u1 + v1 + w1 ,u2 + v 2+ w2 , ... ... . ,un + v n +w n , .. ) ( u1 + v n , u2 +v 2 , ... ... .u n+ v n , ... ... ... . ) + ( w 1 , w 2 , ... w n ,.. ) { u +v }+ w. > u + ( v + w )={ u + v }+ w , , u , v , w V Thus Associative property hold ( ii ) Commutaive property if u , v V then u + v =( u1 , u2 ,... ... . un , ... ... ... . ) + ( v 1 , v 2 , ... ... . v n , ... ... .... ) ( u1 +v 1 , u2+ v n , ... ... .u n+ v n , ... ... ... . ) ( v 1 +u1 , v 2 +u2 , ... ... , v n +un , .... ) v + u =u +v =v + u , u , v V Thus Commutative property hold ( iii ) existence of identity Since zero vector 0 V such that we have 0 + u =( 0,0, ....,0, ... .. ) + ( u1 ,u 2 , ... ... . un , ... ... ... . ) = ( 0+u 1 , 0+u2 , ... ... . 0+un , ... ... ... . ) ( u1 +0 , u2 +0 , ... ... .u n+ 0 , ... ... ... . ) ( u1 ,u 2 , ... ... . un , ... ... .... )=u . u + 0 =( u1 ,u 2 , ... ... . un , ... ... ... . ) + ( 0,0, ....,0, ... .. ) ( u1 +0 , u2 +0 , ... ... .u n+ 0 , ... ... ... . ) ( u1 ,u 2 , ... ... . un , ... ... .... )=u . > 0 +u =u + 0 =u , u V Thus identity of V exist 0 isidentity for V ( iv ) Existence of inverse For each u V we haveu V such that u + (u )=u u=( u1 ,u2 , ... ... .u n , ... ... ... . ) ( u1 ,u 2 , ... ... . un , ... ... ... . ) ( u1u 1 , u2u2 , ... . ,u nun , ... ... ) ( 0,0, ....,0, ... .. )=0 . u + u=( u1 , u2 , ... ... . un , ... ... ... . ) + ( u1 ,u 2 , ... ... . un , ... ... ... . ) (u1+ u1 ,u2 +u2 , ... ... ,un +u n , ... . ) ( 0,0, ....,0, ... .. )=0 . > u + (u ) =u + u =0 Thus inverse element exist for each u V u isinverse for u Multiplication Axiom : ( 0 ) Closure property if k R then k u V , k R ( i ) Compatibility of Scaler multiplication with field multiplication if , a , b Ru V thena ( b u ) =a ( b u1 , b u2 , ... ... , b un , ... ... ... . ) ( a b u1 , a b u2 ,... ... ,a b u n , ... ... ... . ) ( ab) ( u1 , u2 ,... ... . un , ... ... ... . )=( ab ) u >a ( b u )=( ab ) u , u V ( ii ) Identity element of scaler multiplication since 1 R1 u =( 1 u1 , 1 u2 ,... ... . , 1u n , ... ... ... . ) =( u1 ,u 2 , ... ... . un , ... ... ... . ) >1 u=u , u V ( iii ) Distributive of scaler multiplication w . r .t . vector addition if a Ru , v V then a { u + v }=a ( u1 + v 1 , u2 +v n , ... ... . un +v n , ... ... ... . ) ( a ( u 1+ v 1 ) , a ( u 2+ v 2 ) , ... ... .a ( u n+ v n ) , ... ... ... . ) ( a u1 +a v 1 , a u2 +a v n , ... ... . a un +a v n , ... ... ... . ) ( a u1 , a u2 , ... ... . , a un , ... ... ... . ) + ( a v1 , a v 2 , ... ... . a v n , ... ... ... . ) a u +a v . a { u + v }=a u +a v , a Ru , v V ( iv ) Distributive of scaler multiplication w . r . t . field addition if a , b Ru V then ( a+ b ) u=( ( a+b ) u1 ,(a+ b) u2 , ... ... . ,(a+b)un , ... ... ... . ) ( a u1 +b u1 , a u2+ b u2 , ... .... , a un +b u n , ... ... ... . ) ( a u1 , a u2 , ... ... . , a un , ... ... ... . ) + ( b u1 , b u2 ,... ... . , b un , ... ... ... . ) a u +b u > ( a+ b ) u=a u + b u , a , b Ru V Thus V satisfied all condition of vector space hence V is vector space proved , 2 (a) (i) 97 x15 x 2=2 ( 2+ x + 4 x 2 )+ 1 ( 1x+3 x 2 )2 ( 3+2 x +5 x2 ) ( ii ) 6+11 x +6 x 2=4 ( 2+ x+ 4 x 2 )5 ( 1x +3 x 2 ) +1 ( 3+2 x +5 x 2 ) ( iii ) 0=0 ( 2+ x+ 4 x 2 ) +0 ( 1x +3 x 2 )+ 0 ( 3+2 x+5 x 2 ) ( iv ) 7+8 x +9 x 2=0 ( 2+ x +4 x 2 )2 ( 1x+3 x 2 ) +3 ( 3+2 x+5 x 2 ) . (b) [ 2 3 1 : 2 0 14 0: 3 01 6 Define= 1 1 0 5 2 :7 0113 3 2 1:3 0 1 4 [ [ [ [ [[ 0 5 1 :4 0116 R 1 R12 R2 1 1 0:3 0 1 6 R4 R 4 3 R 2 0 5 2:7 0 113 05 1 :6 0214 0 5 1 :4 0116 R 3 R 3R1 1 1 0:3 0 1 6 R4 R 4 R 1 0 0 3:3 0 2 3 0 0 2:2 01 2 0 5 1 :4 0116 1 1 0:3 0 1 6 [ R 3 R3R 4 ] 0 0 1:1 0 3 1 0 0 2:2 01 2 0 5 1 :4 0116 1 1 0:3 0 1 6 [ R 4 R 42 R3 ] 0 0 1:1 0 3 1 0 0 0 :0 0 5 0 ( i ) , ( ii ) , (iv ) span { v1 , v2 , v3 } Only ( iii ) is not span { v1 , v2 , v3} 3 (a) (i) [ 1 2 34 Define a = 0 1 01 1 3 33 [ [ 1 2 34 0 1 01 ,[ R3 R3R 1] 0 1 01 1 2 34 0 1 01 , , [ R3 R3R 2 ] 0 0 00 Thus we have1 ( 1,2,3,4 )1 ( 0,1,0,1 )+1 ( 1,3,3,3 )=0 Hence { v1 , v 2 , v 3 } is linearly dependent ( ii ) ( 1,2,3,4 ) =1 ( 0,1,0,1 ) +1 ( 1,3,3,3 ) . 1 ( 0,1,0,1 )=1 ( 1,2,3,4 ) +1 ( 1,3,3,3 ) ( 1,3,3,3 )=1 ( 1,2,3,4 ) +1 ( 0,1,0,1 ) (b) Suppose that if { v 1 , v2, v3 } is linearly independent thenif a v 1 +b v 2+ c v 3=0 >a=0, b=0, c=0 Now if a1 v 1+ a3 v 3=0 >a 1 v1 =a3 v3 If a1 0a3 0 . Then v 1 v 3 are hence { v 1 , v2 , v3 } is linearly dependent We reach contradiction Thus a1=0a 2=0 > { v1 , v3 } linearly independent Similarly if b2 v 2=0 then b 2=0 since v2 0 Hence { v 2 } islinearly independent . 4 (a) (i) Define A= [ a11 a12 then a21 a22 A a =k a b b > [ [ ] [ ] a11 a12 a =k a a21 a22 b b >a 11 a+a12 b=ka a21 a+a22 b=kb LHS=RHS whena 11=k=a 22 a12=a 21=0 [ ] A= k 0 . 0 k 5 (a) [ ] Let A= a b then c d this send a vector which is on xaxis is xaxis [ ][ ] [ ] a b 0=0 c d 0 0 [ ][ ] [ ] a b 3 = 3 = 3 a=33 c=0= a=1c=0 c d 0 0 Because we make a b 2 = 0 [angle triangle] c d 1 1 [ ][ ] [ ] >2 a+b=02 c+ d=1 Thus we have a=1 , c=02 a+b=02c +d =1 >a=1 , b=2 , c=0 , d=1 [ A= 1 2 0 1 (b) [ ] Let A= a b then question c d [ ][ ] [ ] a b 0=0 c d 0 0 [ ][ ] [ ] a b 1 = 2 = a=2c=0 c d 0 0 [ ][ ] [ ] a b 0.5 = 1 = a +b= 2c + d=1 c d 1 1 2 2 Thus we have a=2c= 0a 2c + b= +d=1 2 2 >a=2 , b=1 , c=0 , d =1 [ ] A= 2 1 0 1 The send the triangle ( 0,0 ) , ( 1,0 ) , ( 0.5,1 ) the triangle see graph . 5.4 Introduction to Linear Transformations: Basic Matrix Transformations in R2 and 1&3 To introduce the idea of a linear transformation, consider, for example, an object located at point {2, 1] of the siscoordinate system. Assume that, under a transformation, this object is moved to the point {3, 6]. Now, observe that (i i)(i)=()- (5-1) As you can see, the displacement {transformation} is described by the matrix 3 2 A=(1 4). [5.2) We will call A a mah'ix nnsfonndon or mam"): operator. In this section we will study some basic types of matrix transformations, which can be interpreted geometrically in two and threedimensional spaces. These transformations are important in computer graphics, physics and other areas of applied mathematics. In the next section we analyze one of these applications in more detail. 4 (b) Let us suppose that reflection xaxis is B then [ ] B x = x B 0 = 0 = 0 ... ... ... ... ... ....(1) 0 0 y y y [ ] > B= 10 01 Now define a rotation whenany coordinate rotate degree anticlockwise [ [ [ A= Cos Sin Sin Cos A x = xCos A 0 = ySin 0 xSin y yCos Let L be the reflection theline y =tan x +k then. [ [ LA x = xCos L A 0 = ySin 0 xSin y yCos > LA x =A x LA 0 = A 0 0 0 y y A1 LA= x A1 LA= 0 ... ... ... ( 2 ) 0 y Compare ( 1 )( 2 ) A1 LA=B [ [ ] [ [ > L= AB A1= Cos Sin 10 Cos Sin = cos 2 sin2 Sin Cos 01 Sin Cos sin 2 cos 2 Now Sinceline y=tan x +k makes angle degree with xaxis so put = then 4 [ ] L= 0 1 . 1 0 : 2 9 7 15 6 + 11 + 6 = 2 2 + =4 2+ +4 0 =0 2+ +4 7+8 +9 +4 +1 1 5 1 +0 1 =0 2+ +4 +3 2 1 +3 2 3+2 +5 +3 +1 3+2 +5 +0 3+2 +5 +3 +3 3+2 +5 . 2 3 1 0 = 1 1 0 5 2 3 2 1 5 1 5 5 1 0 2 1 ~ 0 1 0 0 ~ 0 5 1 1 0 0 0 0 1 4 0 0 3 0 3 3 0 2 2 0 1 16 1 6 2 3 1 2 ~ 0 1 0 0 5 1 0 0 1 4 0 0 3 0 1 1 0 2 2 0 1 16 1 6 3 1 1 2 ~ 0 5 1 1 0 0 0 0 1 4 0 0 3 0 1 1 0 0 0 0 1 16 1 6 3 1 5 0 , 4 0 3 0 7 0 6 0 2 3 7 3 , 0 0 0 0 1 4 1 6 1 13 1 4 1 16 1 6 1 13 2 14 [ [ , , , , 2 3 2 3 1 2 3 4 = 0 1 0 1 1 3 3 3 1 ~ 0 0 2 3 4 1 0 1 1 0 1 1 ~ 0 0 2 3 4 1 0 1 0 0 0 ,[ ,,[ 1 1,2,3,4 1 0,1,0, 1 + 1 1,3,3,3 = 0 , , 1,2,3,4 = 1 0,1,0, 1 + 1 1,3,3,3 . 1 0,1,0, 1 = 1 1,2,3,4 + 1 1,3,3,3 1,3,3,3 = 1 1,2,3,4 + 1 0,1,0, 1 , , + => + = 0, =0 = 0, = 0 + => = 0 =0 0. , , => =0 =0 , =0 =0 0 . 4 5 = 0 0 = 0 0 3 3 = => 3 = 3 0 0 2 0 = 1 1 [ =0 =1, =0 => =1 =0 => 2 + 3 = 0 => =1, 2 + 2 + =1 =0 = 2 , = 0 , 1 0 = 2 + =1 =1 2 1 = 0 0 = 0 0 1 2 = => 0 0 =2 1 0.5 = => + 1 1 2 =2 => =0 =2, 2 + + =2 =1, =0, =1 = =2 =0 2 0 2 =1 2 + =1 1 1 0,0 , 1,0 , 0.5,1 . Assignment 4 1 (a) u= (1,2,3,5,0 ) , v =( 0,4,1,1,2 ) w =( 7,1,4,2,3 ) . (i) v + w=( 0,4,1,1,2 )+ ( 7,1,4,2,3 )=(7,5,5,1,5) ( ii ) 3 ( 2 u v ) =3 [ 2 ( 1,2,3,5,0 )( 0,4,1,1,2 ) ] =3 ( 2,0,5,9,2 )=( 6,0,15,27,6 ) ( iii ) ( 3 uv )( 2 u +4 w ) =[ 3 ( 1,2,3,5,0 )( 0,4,1,1,2 ) ] [ 2 ( 1,2,3,5,0 ) +4 ( 7,1,4,2,3 ) ] ( 3 ,2 , 10,14,2 ) ( 30,8,22,2,12 )=(27 ,6,32,12,4 ) ( iv ) 1 1 ( w 5 v +2 u ) + v = [( 7,1,4,2,3 )5 ( 0,4,1,1,2 ) +2 ( 1,2,3,5,0 ) ] + ( 0,4,1,1,2 ) 2 2 1 ( 8,15,5,3,7 ) + ( 0,4,1,1,2 ) 2 7 7 5 3 4 , , , , . 2 2 2 2 ( ) (b) c 1 (1,0,2 ) +c 2 ( 2,2,2 )+ c 3 ( 1,2,1 )=(6,12,4 ) >c 1+ 2 c2 +c 3=6 2 c 22 c3 =12 2 c1 2 c2 +c 3=4 AX =b . Augmented [ 1 [ A : b ]= 0 [ [ 2 2 1 :6 2 2 :12 2 1: 4 1 2 1 :6 0 2 2 :12 ,[R 3 R3 +2 R 1] 0 2 3 :8 1 2 1:6 0 2 2:12 ,[R 3 R 3R1 ] 0 0 5:20 >5 c 3=20= c 3=4 2 c 22 c3 =12= c 2=2 c1 +2 c 2+c 3=6= c1 =6 Thus c 1=6 , c 2=2 , c 3=4 . 2 (a) (i) u= (1,2,3,0 )v =( 5,1,2,2 ) |u|= 1+ 4+ 9+0=14|v|=25+ 1+ 4+ 4= 34 Cos= ( u . v ) 5+ 26+ 0 1 = = |u ||v| 14 34 476 =Co s1 1 ( 476 ) 87.37 Angle is ( ii ) u= ( 0,1,1,1,2 )v =(2,1,0,1,3) |u|= 0+1+1+1+4= 6|v|= 4 +1+ 0+1+9= 15 Cos= ( u . v ) 0+1+ 01+6 6 = = |u ||v| 6 15 90 =Co s1 ( 690 ) 50.77 Angle is (b) (i) u= ( 4,1,1 )v =( 1,2,3 ) Cuachy Schwarz inequality u , v > u , v > 4 +2+3=9 u , u > 16+1+1=18 v , v > 1+4 +9=14 2 Here 9 <18 14 thus satisfied ( ii ) u )v =( 0,1,1,5,2 cuachy schwarz inequality , v> u , v > 0+2+1+106=7 u , u > 1+ 4+ 1+ 4+ 9=19 v , v > 0+1+1+25+ 4=31 Here 72 <19 31 thus satisfied 3 (a) 2 x y+ z=12 y + choose a point on the plane say ( 0,01 ) distance between these two planes is =(b) (i) v , b w |2 0 )0+1+1| 4+1+1 6 . hence are orthogonal ii 2,3 let ,b use i if 2a3>Choose ab Thus two vectors that are orthogonal v are w 1=( 3,2 ) w 2= ( 6 , 4 ) ( iii ) v =(3,4 ) Les us take two o rthogonal vector v w 1=( 4,3 ) w2 =(4,3). n^1= 1 w 1 1 = ( 4,3 )= ( 4,3 ) 5 w1| 16+ 9 | n^2= 2 w 1 1 = (4,3 )= (4,3 ) 5 w2| 16+ 9 | Here n^1 ^ n2 areunit vectors n^1 . v =0^ n2 . v =0 Hence n^1^ n2 are two unit vectors which orthogonal v 4 (a) x 1+3 x 24 x 3=0 x 1+2 x 2+ 3 x 3=0 > > > [ ] [ ][ ] [ ] [ x 1 3 4 1 =0 1 2 3 x2 0 x3 [ x 1 3 4 1 = 0 , [ R 2 R 2R1 ] 0 1 7 x 2 0 x3 [ x 1 0 17 1 = 0 , [ R1 R1 +3 R1 ] 0 1 7 x 2 0 x3 > x 1+ 17 x3 =0x 2+ 7 x3 =0. Setting x3 =k then wehave > x 1=17 k x 2=7 k x 3=k Thus genral solution is (17 k ,7 k , k ) where k R 1 3 4 1 2 3 Now Take raw of [ ]that is (1,3,4 )( 1,2,3 ) ( 1,3,4 ) . (17 k , 7 k , k ) =17 k + 21 k4 k =0 ( 1,2,3 ) . (17 k , 7 k , k )=17 k +14 k +3 k =0 1 3 4 1 2 3 Thus genral solution is (17 k ,7 k , k ) is orthogonal the raw of [] (b) (i) Let x =( x 1 , x 2 , x 3 ) be the three unknowndefine a . x =0 b . x =0 > (3 , 2,1 ) . ( x 1 , x 2 , x 3 ) =0 ( 0,2,2 ) . ( x 1 , x 2 , x 3 )=0 >3 x1 +2 x 2x 3=0 0. x12 x 22 x 3=0 [ ] [ ] x 3 2 1 1 = 0 ... ... ... . (1) 0 2 2 x 2 0 x3 > [ > x 3 0 3 1 =0 0 2 2 x2 0 x3 [ 3 x 13 x 3=02 x 22 x3 =0 setting x 3=k then x 1=k x 2=k solution is ( x 1 , x 2 , x 3 ) =(k ,k , k ) . 3 2 1 0 2 2 Thus genral solution (k ,k , k ) is orthogonal the raw of [ ] >Thus genral solution (k ,k , k ) is orthogonal the a b > solutionspace of ( 1 ) is orthogonal the a b ( ii ) Solution space=k (1 ,1,1 ) where k R Span { (1,1,1 ) } Straight line passing ( 0,0,0 ) (1 ,1,1 ) ( iii ) >3 x1 +2 x 2x 3=0 0. x12 x 22 x 3=0 [ ] [ ] x 3 2 1 1 = 0 ... ... ... . (1) 0 2 2 x 2 0 x3 > [ > x 3 0 3 1 =0 0 2 2 x2 0 x3 [ 3 x 13 x 3=02 x 22 x3 =0 setting x 3=k then x 1=k x 2=k Genral solution is ( x1 , x2 , x3 ) =(k ,k , k ) . . 5 (a) P=( 1,1,2 ) , Q=( 0,3,4 )R=( 6,1,8 ) Let u=QP=(1,4,2 ) v =RP=( 5,2,6 ) [ i j k u v = 1 4 2 5 2 6 ( 244 ) i (610 ) j+ (220 ) k 20 i+16 j22 k=(20,16,22) . 1 Area of triangle PQR is= |u v| 2 1 400+256+ 484 2 1 1 1140= 2 285= 285. 2 2 (b) u= (1 , 2,4 ) , v =( 3,4,2 ) w =(1 ,2,5 ) u . ( v w )=Det ([ 1 2 4 3 4 2 1 2 5 ) 1 ( 20+ 4 ) 2 ( 152 ) +4 ( 6+4 ) 2426+40 10 . (c ) u . ( v w )=3 Since u . ( v w )=v . ( w u )= w . ( u v ) . (i) v . ( u w )=v . ( w u )=3 ( ii ) ( u w ) . v =v . ( u w )=3 ( iii ) v . ( w w )=0. Ans : 1 (a) Let V =Set of all real values functions x ( , ) We have prove that V isVector Space field real line R : the Addition Axiom: ( 0 ) Closure property if f ( x ) , g ( x ) V then f ( x ) +g (x)V , f (x ), g( x ) V ( i ) Associative property If f ( x ), g(x ) ,h ( x)V then f ( x ) + ( g ( x )+ h ( x ) ) =f ( x ) +g ( x ) +h ( x ) ( f ( x ) + g ( x ) ) + h ( x )=( f ( x )+ g ( x ) ) +h ( x ) > f ( x ) + ( g ( x )+ h ( x ) ) =( f ( x ) + g ( x ) ) + h ( x ) , f (x ), g( x ), h(x ) V Thus Associative property hold ( ii ) Commutaive property if f ( x ) , g ( x ) V then f ( x ) +g ( x )=g ( x ) + f ( x ) , f , g V Thus Commutative property hold ( iii ) existence of identity Since zero functionO ( x ) V such that for all x R we have O ( x )=0 f ( x )+ O ( x )=f ( x ) +0=0+ f ( x ) > f ( x ) +O ( x ) =O ( x ) + f ( x )=f ( x ) , f ( x ) V Thus identity of V exist Ois identity for V ( iv ) Existence of inverse For each f ( x ) V we havef ( x ) V such that f ( x ) + (f ( x ) )=f ( x )f ( x )=f ( x )+ f ( x )=(f + f ) ( x ) =O ( x ) > f ( x ) + (f ( x ) )=f ( x ) + f ( x )=O ( x ) Thus inverse element exist for each f ( x ) V f ( x ) is inverse for f ( x ) Multiplication Axiom : ( 0 ) Closure property if k R then cf ( x ) V , k R ( i ) Compatibility of Scaler multiplication with field multiplication if , a , b R an d f ( x ) V then a ( bf ( x )) =abf ( x )=( ab ) f ( x ) >a ( bf ( x ) )= ( ab ) f ( x ) , f ( x ) V ( ii ) Identity element of scaler multiplication since 1 R1 f ( x ) =f ( x )= 1 f ( x )=f ( x ) , f ( x ) V ( iii ) Distributive of scaler multiplication w . r .t . vector addition if a Rf ( x ) , g ( x ) V then a { f ( x ) + g ( x ) }=af ( x )+ bg ( x ) , a Rf ( x ) , g ( x ) V ( iv ) Distributive of scaler multiplication w . r . t . field addition if a , b Rf ( x ) V then ( a+b ) f ( x ) =af ( x ) +bf (x) a , b Rf V Thus V satisfied all condition of vector space hence V is vector space proved (b) Let V =Set of all vector u =(u 1 , u2 , ... ... .u n , ... ... ... .) We have prove that V isVector Space field real line R : the Addition Axiom: ( 0 ) Closure property if u , v V then u + v =( u1 , u2 ,... ... . un , ... ... ... . ) + ( v 1 , v 2 , ... ... . v n , ... ... .... ) (u1 + v n ,u2 + v 2 , ... ... . un +v n , ... ... ... .)V , u , v V ( i ) Associative property If u , v , w V then u + ( v + w ) =( u1 ,u 2 , ... ... . un ,... ) + {( v 1 , v 2 , ... ... . v n ,.. )+ ( w 1 , w 2 , ... w n , .. ) } ( u1 ,u 2 , ... ... . un , ... ) + ( v 1+ w1 , v 2+ w2 , ... ... . v n +w n , .. ) ( u1 + v1 + w1 ,u2 + v 2+ w2 , ... ... . ,un + v n +w n , .. ) ( u1 + v n , u2 +v 2 , ... ... .u n+ v n , ... ... ... . ) + ( w 1 , w 2 , ... w n ,.. ) { u +v }+ w. > u + ( v + w )={ u + v }+ w , , u , v , w V Thus Associative property hold ( ii ) Commutaive property if u , v V then u + v =( u1 , u2 ,... ... . un , ... ... ... . ) + ( v 1 , v 2 , ... ... . v n , ... ... .... ) ( u1 +v 1 , u2+ v n , ... ... .u n+ v n , ... ... ... . ) ( v 1 +u1 , v 2 +u2 , ... ... , v n +un , .... ) v + u =u +v =v + u , u , v V Thus Commutative property hold ( iii ) existence of identity Since zero vector 0 V such that we have 0 + u =( 0,0, ....,0, ... .. ) + ( u1 ,u 2 , ... ... . un , ... ... ... . ) = ( 0+u 1 , 0+u2 , ... ... . 0+un , ... ... ... . ) ( u1 +0 , u2 +0 , ... ... .u n+ 0 , ... ... ... . ) ( u1 ,u 2 , ... ... . un , ... ... .... )=u . u + 0 =( u1 ,u 2 , ... ... . un , ... ... ... . ) + ( 0,0, ....,0, ... .. ) ( u1 +0 , u2 +0 , ... ... .u n+ 0 , ... ... ... . ) ( u1 ,u 2 , ... ... . un , ... ... .... )=u . > 0 +u =u + 0 =u , u V Thus identity of V exist 0 isidentity for V ( iv ) Existence of inverse For each u V we haveu V such that u + (u )=u u=( u1 ,u2 , ... ... .u n , ... ... ... . ) ( u1 ,u 2 , ... ... . un , ... ... ... . ) ( u1u 1 , u2u2 , ... . ,u nun , ... ... ) ( 0,0, ....,0, ... .. )=0 . u + u=( u1 , u2 , ... ... . un , ... ... ... . ) + ( u1 ,u 2 , ... ... . un , ... ... ... . ) (u1+ u1 ,u2 +u2 , ... ... ,un +u n , ... . ) ( 0,0, ....,0, ... .. )=0 . > u + (u ) =u + u =0 Thus inverse element exist for each u V u isinverse for u Multiplication Axiom : ( 0 ) Closure property if k R then k u V , k R ( i ) Compatibility of Scaler multiplication with field multiplication if , a , b Ru V thena ( b u ) =a ( b u1 , b u2 , ... ... , b un , ... ... ... . ) ( a b u1 , a b u2 ,... ... ,a b u n , ... ... ... . ) ( ab) ( u1 , u2 ,... ... . un , ... ... ... . )=( ab ) u >a ( b u )=( ab ) u , u V ( ii ) Identity element of scaler multiplication since 1 R1 u =( 1 u1 , 1 u2 ,... ... . , 1u n , ... ... ... . ) =( u1 ,u 2 , ... ... . un , ... ... ... . ) >1 u=u , u V ( iii ) Distributive of scaler multiplication w . r .t . vector addition if a Ru , v V then a { u + v }=a ( u1 + v 1 , u2 +v n , ... ... . un +v n , ... ... ... . ) ( a ( u 1+ v 1 ) , a ( u 2+ v 2 ) , ... ... .a ( u n+ v n ) , ... ... ... . ) ( a u1 +a v 1 , a u2 +a v n , ... ... . a un +a v n , ... ... ... . ) ( a u1 , a u2 , ... ... . , a un , ... ... ... . ) + ( a v1 , a v 2 , ... ... . a v n , ... ... ... . ) a u +a v . a { u + v }=a u +a v , a Ru , v V ( iv ) Distributive of scaler multiplication w . r . t . field addition if a , b Ru V then ( a+ b ) u=( ( a+b ) u1 ,(a+ b) u2 , ... ... . ,(a+b)un , ... ... ... . ) ( a u1 +b u1 , a u2+ b u2 , ... .... , a un +b u n , ... ... ... . ) ( a u1 , a u2 , ... ... . , a un , ... ... ... . ) + ( b u1 , b u2 ,... ... . , b un , ... ... ... . ) a u +b u > ( a+ b ) u=a u + b u , a , b Ru V Thus V satisfied all condition of vector space hence V is vector space proved , 2 (a) (i) 97 x15 x 2=2 ( 2+ x + 4 x 2 )+ 1 ( 1x+3 x 2 )2 ( 3+2 x +5 x2 ) ( ii ) 6+11 x +6 x 2=4 ( 2+ x+ 4 x 2 )5 ( 1x +3 x 2 ) +1 ( 3+2 x +5 x 2 ) ( iii ) 0=0 ( 2+ x+ 4 x 2 ) +0 ( 1x +3 x 2 )+ 0 ( 3+2 x+5 x 2 ) ( iv ) 7+8 x +9 x 2=0 ( 2+ x +4 x 2 )2 ( 1x+3 x 2 ) +3 ( 3+2 x+5 x 2 ) . (b) [ 2 3 1 : 2 0 14 0: 3 01 6 Define= 1 1 0 5 2 :7 0113 3 2 1:3 0 1 4 [ [ [ [ [[ 0 5 1 :4 0116 R 1 R12 R2 1 1 0:3 0 1 6 R4 R 4 3 R 2 0 5 2:7 0 113 05 1 :6 0214 0 5 1 :4 0116 R 3 R 3R1 1 1 0:3 0 1 6 R4 R 4 R 1 0 0 3:3 0 2 3 0 0 2:2 01 2 0 5 1 :4 0116 1 1 0:3 0 1 6 [ R 3 R3R 4 ] 0 0 1:1 0 3 1 0 0 2:2 01 2 0 5 1 :4 0116 1 1 0:3 0 1 6 [ R 4 R 42 R3 ] 0 0 1:1 0 3 1 0 0 0 :0 0 5 0 ( i ) , ( ii ) , (iv ) span { v1 , v2 , v3 } Only ( iii ) is not span { v1 , v2 , v3} 3 (a) (i) [ 1 2 34 Define a = 0 1 01 1 3 33 [ [ 1 2 34 0 1 01 ,[ R3 R3R 1] 0 1 01 1 2 34 0 1 01 , , [ R3 R3R 2 ] 0 0 00 Thus we have1 ( 1,2,3,4 )1 ( 0,1,0,1 )+1 ( 1,3,3,3 )=0 Hence { v1 , v 2 , v 3 } is linearly dependent ( ii ) ( 1,2,3,4 ) =1 ( 0,1,0,1 ) +1 ( 1,3,3,3 ) . 1 ( 0,1,0,1 )=1 ( 1,2,3,4 ) +1 ( 1,3,3,3 ) ( 1,3,3,3 )=1 ( 1,2,3,4 ) +1 ( 0,1,0,1 ) (b) Suppose that if { v 1 , v2, v3 } is linearly independent thenif a v 1 +b v 2+ c v 3=0 >a=0, b=0, c=0 Now if a1 v 1+ a3 v 3=0 >a 1 v1 =a3 v3 If a1 0a3 0 . Then v 1 v 3 are hence { v 1 , v2 , v3 } is linearly dependent We reach contradiction Thus a1=0a 2=0 > { v1 , v3 } linearly independent Similarly if b2 v 2=0 then b 2=0 since v2 0 Hence { v 2 } islinearly independent . 4 (a) (i) Define A= [ a11 a12 then a21 a22 A a =k a b b > [ [ ] [ ] a11 a12 a =k a a21 a22 b b >a 11 a+a12 b=ka a21 a+a22 b=kb LHS=RHS whena 11=k=a 22 a12=a 21=0 [ ] A= k 0 . 0 k 5 (a) [ ] Let A= a b then c d this send a vector which is on xaxis is xaxis [ ][ ] [ ] a b 0=0 c d 0 0 [ ][ ] [ ] a b 3 = 3 = 3 a=33 c=0= a=1c=0 c d 0 0 Because we make a b 2 = 0 [angle triangle] c d 1 1 [ ][ ] [ ] >2 a+b=02 c+ d=1 Thus we have a=1 , c=02 a+b=02c +d =1 >a=1 , b=2 , c=0 , d=1 [ A= 1 2 0 1 (b) [ ] Let A= a b then question c d [ ][ ] [ ] a b 0=0 c d 0 0 [ ][ ] [ ] a b 1 = 2 = a=2c=0 c d 0 0 [ ][ ] [ ] a b 0.5 = 1 = a +b= 2c + d=1 c d 1 1 2 2 Thus we have a=2c= 0a 2c + b= +d=1 2 2 >a=2 , b=1 , c=0 , d =1 [ ] A= 2 1 0 1 The send the triangle ( 0,0 ) , ( 1,0 ) , ( 0.5,1 ) the triangle see graph . 5.4 Introduction to Linear Transformations: Basic Matrix Transformations in R2 and 1&3 To introduce the idea of a linear transformation, consider, for example, an object located at point {2, 1] of the siscoordinate system. Assume that, under a transformation, this object is moved to the point {3, 6]. Now, observe that (i i)(i)=()- (5-1) As you can see, the displacement {transformation} is described by the matrix 3 2 A=(1 4). [5.2) We will call A a mah'ix nnsfonndon or mam"): operator. In this section we will study some basic types of matrix transformations, which can be interpreted geometrically in two and threedimensional spaces. These transformations are important in computer graphics, physics and other areas of applied mathematics. In the next section we analyze one of these applications in more detail. 4 (b) Let us suppose that reflection xaxis is B then [ ] B x = x B 0 = 0 = 0 ... ... ... ... ... ....(1) 0 0 y y y [ ] > B= 10 01 Now define a rotation whenany coordinate rotate degree anticlockwise [ [ [ A= Cos Sin Sin Cos A x = xCos A 0 = ySin 0 xSin y yCos Let L be the reflection theline y =tan x +k then. [ [ LA x = xCos L A 0 = ySin 0 xSin y yCos > LA x =A x LA 0 = A 0 0 0 y y A1 LA= x A1 LA= 0 ... ... ... ( 2 ) 0 y Compare ( 1 )( 2 ) A1 LA=B [ [ ] [ [ > L= AB A1= Cos Sin 10 Cos Sin = cos 2 sin2 Sin Cos 01 Sin Cos sin 2 cos 2 Now Sinceline y=tan x +k makes angle degree with xaxis so put = then 4 [ ] L= 0 1 . 1 0