\fAns : 1 [ ] v = 2 , w= 1 , u= 2 1 2 3 Let a , b R such that av+bw=o >a 2 +b 1 = 0 1 2 0 [ > 2 a+ b = 0 a+2 b 0 >2 a+b=0 a+2 b=0 Solve these two equation we get a=0 , b=0 Thus { v , w } is linearly independent Span { v , w }=av +bw ,a ,b R a 2 +b 1 1 2 [ 2 a+b a+2 b {[ ] x : x =2 a+b , y=a+ 2b y {[ ] } x : x , y R . y R2 Hence Span { v , w }=R2 } 2 There fore = { v , w } isbasis for R . [ ] Further Let u= 2 =a 2 + b 1 3 1 2 >2 a+b=2a+ 2b=3 Solve these two equation for ab we get a= 7 8 , b= 3 3 Thus [ u ] = ( a , b )= (73 , 83 ) 2 (a) [ ] A= 2 1 1 1 Here| A|=21=1= A is invertible {[ ] [ ]} > = 2 , 1 isbasis for R2 1 1 Let arbitary x= x 1 R2 thenthere exist a1 , a2 R such that x2 [ ] x 1 =a 2 +a 1 = [ x ] = a1 2 1 1 x2 a2 [ ] Now x 1 =a 2 + a2 1 1 1 x2 [ ] [ ][ ] > x 1 = 2 1 a1 1 1 a2 x2 > x= A [ x ] 1 Now Since A is invertible so A exist So A1 x= A1 ( A [ x ] ) > A1 x=( A1 A ) [ x ] 1 > A x=I [ x ] . > [ x ] = A1 x Where I is 2 2 order identity (b) Generalize form Let P=[ C 1 C 2 ... ... ..... C n ] be the invertible C i are column of P > = {C1 ,C 2 , ... ... ..... , Cn } form basis for R n then x1 x2 2 for any x= ... R there exist ai R ,i =1,2,... , n such that ... xn x=a1 C1 +a 2 C 2+ ... ... ... .+an Cn > x=[ C 1 C 2 ... ... ..... C n ] a1 a2 ... ... an > x=P [ x ] 1 Since P invertible implies P exist P1 x=P1 ( P [ x ] ) > P1 x =( P1 P ) [ x ] 1 > P x =I [ x ] . > [ x ] =P1 x Where I is n n identity Ans : 1 [ ] v = 2 , w= 1 , u= 2 1 2 3 Let a , b R such that av+bw=o >a 2 +b 1 = 0 1 2 0 [ > 2 a+ b = 0 a+2 b 0 >2 a+b=0 a+2 b=0 Solve these two equation we get a=0 , b=0 Thus { v , w } is linearly independent Span { v , w }=av +bw ,a ,b R a 2 +b 1 1 2 [ 2 a+b a+2 b {[ ] x : x =2 a+b , y=a+ 2b y {[ ] } x : x , y R . y R2 Hence Span { v , w }=R2 } 2 There fore = { v , w } isbasis for R . [ ] Further Let u= 2 =a 2 + b 1 3 1 2 >2 a+b=2a+ 2b=3 Solve these two equation for ab we get a= 7 8 , b= 3 3 Thus [ u ] = ( a , b )= (73 , 83 ) 2 (a) [ ] A= 2 1 1 1 Here| A|=21=1= A is invertible {[ ] [ ]} > = 2 , 1 isbasis for R2 1 1 Let arbitary x= x 1 R2 thenthere exist a1 , a2 R such that x2 [ ] x 1 =a 2 +a 1 = [ x ] = a1 2 1 1 x2 a2 [ ] Now x 1 =a 2 + a2 1 1 1 x2 [ ] [ ][ ] > x 1 = 2 1 a1 1 1 a2 x2 > x= A [ x ] 1 Now Since A is invertible so A exist So A1 x= A1 ( A [ x ] ) > A1 x=( A1 A ) [ x ] 1 > A x=I [ x ] . > [ x ] = A1 x Where I is 2 2 order identity (b) Generalize form Let P=[ C 1 C 2 ... ... ..... C n ] be the invertible C i are column of P > = {C1 ,C 2 , ... ... ..... , Cn } form basis for R n then x1 x2 2 for any x= ... R there exist ai R ,i =1,2,... , n such that ... xn x=a1 C1 +a 2 C 2+ ... ... ... .+an Cn > x=[ C 1 C 2 ... ... ..... C n ] a1 a2 ... ... an > x=P [ x ] 1 Since P invertible implies P exist P1 x=P1 ( P [ x ] ) > P1 x =( P1 P ) [ x ] 1 > P x =I [ x ] . > [ x ] =P1 x Where I is n n identity