S o l u t i o n : ( a ) L e t S = ( 1 , 1 ) , ( 1 , 0 ) = v 1 , v 2 w h e r e v 1 = ( 1 , 1 ) , v 2 = ( 1 , 0 ) Solution: (a)~Let ~S={(1,1),(1,0)}=v_{1},v_{2}~~~where ~v_{1}=(1,1),~v_{2}=(1,0) S o l u t i o n : ( a ) L e t S = ( 1 , 1 ) , ( 1 , 0 ) = v 1 , v 2 w h ere v 1 = ( 1 , 1 ) , v 2 = ( 1 , 0 )
w e h a v e , S i s b a s i s f o r R 2 ( i ) S i s l i n e a r l y i n d e p e n d e n t . ( i i ) S i s s p a n n i n g s e t o f R 2 we~have,~S ~ is ~basis ~for~R^2
(i)S ~is~linearly~independent.(ii)S ~is~spanning~set~of~R^2 w e ha v e , S i s ba s i s f or R 2 ( i ) S i s l in e a r l y in d e p e n d e n t . ( ii ) S i s s p annin g se t o f R 2 ( i ) T o c h e c k : S i s l i n e a r l y i n d e p e n d e n t . (i)To ~check:S~is~linearly~independent. ( i ) T o c h ec k : S i s l in e a r l y in d e p e n d e n t .
C o n s i d e r , l i n e a r c o m b i n a t i o n k 1 v 1 + k 2 v 2 = 0 w h e r e k 1 , k 2 ∈ R 2 Consider,~ linear~ combination ~k_{1}v_{1}+k_{2}v_{2}=0~where~k_{1},k_{2}\in R^2 C o n s i d er , l in e a r co mbina t i o n k 1 v 1 + k 2 v 2 = 0 w h ere k 1 , k 2 ∈ R 2
∴ k 1 ( 1 , 1 ) + k 2 ( 1 , 0 ) = 0 \ \therefore k_{1}(1,1)+k_{2}(1,0)=0 ∴ k 1 ( 1 , 1 ) + k 2 ( 1 , 0 ) = 0
⇒ ( k 1 , k 1 ) + ( k 2 , 0 ) = 0 \Rightarrow (k_{1},k_{1})+(k_{2},0)=0 ⇒ ( k 1 , k 1 ) + ( k 2 , 0 ) = 0
⇒ ( k 1 + k 2 , k 1 ) = 0 \Rightarrow ~(k_{1}+k_{2},k_{1})=0 ⇒ ( k 1 + k 2 , k 1 ) = 0
∴ w e g e t t w o h o m o g e n e o u s e q u a t i o n s a s \therefore ~we~ get~ two~ homogeneous ~equations ~as ∴ w e g e t tw o h o m o g e n eo u s e q u a t i o n s a s
k 1 + k 2 = 0........................ ( 1 ) a n d k 1 = 0.......................... ( 2 ) p u t k 1 = 0 i n e q u a t i o n ( 1 ) , w e g e t k 2 = 0 H e n c e w e g e t k 1 = k 2 = 0 ∴ S i s l i n e a r l y i n d e p e n d e n t S e t . k_{1}+k_{2}=0 ........................(1)\\ and~ k_{1}=0.......................... (2)\\ put~ k_{1}=0~in~equation~(1),we~get~ k_{2}=0\\Hence~we~get~ k_{1}= k_{2}=0 \\ \therefore ~S ~is~ linearly~ independent Set. k 1 + k 2 = 0........................ ( 1 ) an d k 1 = 0.......................... ( 2 ) p u t k 1 = 0 in e q u a t i o n ( 1 ) , w e g e t k 2 = 0 He n ce w e g e t k 1 = k 2 = 0 ∴ S i s l in e a r l y in d e p e n d e n tS e t .
( i i ) T o c h e c k : S i s s p a n n i n g s e t o f R 2 . L e t V = ( a , b , c ) ∈ R 2 b e a n y e l e m e n t a n d v 1 = ( 1 , 1 ) a n d v 1 = ( 1 , 0 ) . L e t i f p o s s i b l e V = k 1 v 1 + k 2 v 2 w h e r e k 1 , k 2 ∈ R 2 ∴ ( a , b ) = k 1 ( 1 , 1 ) + k 2 ( 1 , 0 ) ( a , b ) = ( k 1 , k 1 ) + ( k 2 , 0 ) ( a , b ) = ( k 1 + k 2 , k 1 ) k 1 + k 2 = a a n d k 1 = b (ii) To~ check~:~S~ is~spanning~set~of~R^2.~\\ ~~~~~Let~V=(a,b,c)\in R^2~be~any~element~and~v_{1}=(1,1)~and ~v_{1}=(1,0).\\~~~~~Let~if~possible~V=k_{1}v_{1}+k_{2}v_{2}~where~k_{1},k_{2}\in R^2 ~\\ ~\ \therefore (a,b)=k_{1}(1,1)+k_{2}(1,0)~\\(a,b)=(k_{1},k_{1})+(k_{2},0)~\\(a,b)=(k_{1}+k_{2},k_{1})~\\k_{1}+k_{2}=a~and~k_{1}=b ( ii ) T o c h ec k : S i s s p annin g se t o f R 2 . L e t V = ( a , b , c ) ∈ R 2 b e an y e l e m e n t an d v 1 = ( 1 , 1 ) an d v 1 = ( 1 , 0 ) . L e t i f p oss ib l e V = k 1 v 1 + k 2 v 2 w h ere k 1 , k 2 ∈ R 2 ∴ ( a , b ) = k 1 ( 1 , 1 ) + k 2 ( 1 , 0 ) ( a , b ) = ( k 1 , k 1 ) + ( k 2 , 0 ) ( a , b ) = ( k 1 + k 2 , k 1 ) k 1 + k 2 = a an d k 1 = b ( i i ) T o c h e c k : S s p a n n i n g s e t o f R 2 . L e t V = ( a , b ) ∈ R 2 b e a n y e l e n e n t a n d v 1 = ( 1 , 1 ) , v 2 = ( 1 , 0 ) . L e t i f p o s s i b l e V = k 1 v 1 + k 2 v 2 w h e r e k 1 , k 2 ∈ R 2 ∴ ( a , b ) = k 1 v 1 + k 2 v 2 = k 1 ( 1 , 1 ) + k 2 ( 1 , 0 ) = ( k 1 + k 2 , k 1 ) k 1 + k 2 = a a n d k 1 = b (ii)~ To~check~:~S~spanning~set~of~R^2.\\~~~~~~~Let~V=(a,b)\in R^2~be~any~elenent~and~v_{1}=(1,1),v_{2}=(1,0).\\~~~~~~~Let~if~possible~V=k_{1}v_{1}+k_{2}v_{2}~where~k_{1},k_{2}\in R^2\\ \therefore (a,b)=k_{1}v_{1}+k_{2}v_{2}=k_{1}(1,1)+k_{2}(1,0)=(k_{1}+k_{2},k_{1})\\k_{1}+k_{2}=a~and~k_{1}=b ( ii ) T o c h ec k : S s p annin g se t o f R 2 . L e t V = ( a , b ) ∈ R 2 b e an y e l e n e n t an d v 1 = ( 1 , 1 ) , v 2 = ( 1 , 0 ) . L e t i f p oss ib l e V = k 1 v 1 + k 2 v 2 w h ere k 1 , k 2 ∈ R 2 ∴ ( a , b ) = k 1 v 1 + k 2 v 2 = k 1 ( 1 , 1 ) + k 2 ( 1 , 0 ) = ( k 1 + k 2 , k 1 ) k 1 + k 2 = a an d k 1 = b
H e r e , N u m b e r o f e q u a t i o n s = N u m b e r o f v a r i a b l e s ∴ A = [ 1 1 1 0 ] ∣ A ∣ = − 1 ≠ 0 ∴ A i s n o n − s i n g u l a r . A i s i n v e r t i b l e . ∴ S y s t e m i s c o n s i s t e n t . ∴ L i n e a r s y s t e m A X = B h a s s o l u t i o n . ∴ E v e r y e l e m e n t o f R 2 c a n b e e x p r e s s e d a s a l i n e a r c o m b i n a t i o n o f v 1 a n d v 2 . S s p a n s R 2 . ∴ A l l t w o c o n d i t i o n s h o l d . ∴ S i s b a s i s f o r R 2 Here, ~~~Number ~of~ equations~=~Number~ of~ variables\\ \therefore A=\begin{bmatrix}
1 & 1\\1 & 0 \end{bmatrix}\\ ~|A|=-1\neq0\\ ~\therefore A ~is~ non-singular.\\ ~A ~is~ invertible.\\ \therefore System~ is~ consistent.\\ \therefore Linear~ system~ AX=B~has~solution.\\ \therefore Every~~element~of~R^2~can~be~expressed~as~a~linear~combination~of~v_{1}~and~v_{2}.\\ S~spans~R^2.\\ \therefore~ All~ two~ conditions~ hold.\\ \therefore S~is~basis~for~R^2 Here , N u mb er o f e q u a t i o n s = N u mb er o f v a r iab l es ∴ A = [ 1 1 1 0 ] ∣ A ∣ = − 1 = 0 ∴ A i s n o n − s in gu l a r . A i s in v er t ib l e . ∴ S ys t e m i s co n s i s t e n t . ∴ L in e a r sys t e m A X = B ha s so l u t i o n . ∴ E v ery e l e m e n t o f R 2 c an b e e x p resse d a s a l in e a r co mbina t i o n o f v 1 an d v 2 . S s p an s R 2 . ∴ A ll tw o co n d i t i o n s h o l d . ∴ S i s ba s i s f or R 2
( b ) L e t P = ( 1 , 1 , 0 ) , ( 1 , 0 , 1 ) , ( 0 , 1 , 1 ) = u 1 , u 2 , u 3 w h e r e u 1 = ( 1 , 1 , 1 ) , u 2 = ( 1 , 0 , 1 ) , u 3 = ( 0 , 1 , 1 ) (b)~Let ~P={(1,1,0),(1,0,1),(0,1,1)}=u_{1},u_{2},u_{3}~~~where ~u_{1}=(1,1,1),~u_{2}=(1,0,1),u_{3}=(0,1,1) ( b ) L e t P = ( 1 , 1 , 0 ) , ( 1 , 0 , 1 ) , ( 0 , 1 , 1 ) = u 1 , u 2 , u 3 w h ere u 1 = ( 1 , 1 , 1 ) , u 2 = ( 1 , 0 , 1 ) , u 3 = ( 0 , 1 , 1 )
w e h a v e , P i s b a s i s f o r R 3 i f ( i ) P i s l i n e a r l y i n d e p e n d e n t . ( i i ) P i s s p a n n i n g s e t o f R 3 we~have,P ~is ~basis ~for~R^3~if
(i)P ~is~linearly~independent.(ii)P ~is~spanning~set~of~R^3 w e ha v e , P i s ba s i s f or R 3 i f ( i ) P i s l in e a r l y in d e p e n d e n t . ( ii ) P i s s p annin g se t o f R 3
( i ) T o c h e c k : P i s l i n e a r l y i n d e p e n d e n t . (i)To ~check:P~is~linearly~independent. ( i ) T o c h ec k : P i s l in e a r l y in d e p e n d e n t .
C o n s i d e r , l i n e a r c o m b i n a t i o n k 1 v 1 + k 2 v 2 + k 3 v 3 = 0 w h e r e k 1 , k 2 , k 3 ∈ R 3 Consider,~ linear~ combination ~k_{1}v_{1}+k_{2}v_{2}+k_{3}v_{3}=0~where~k_{1},k_{2},k_{3}\in R^3 C o n s i d er , l in e a r co mbina t i o n k 1 v 1 + k 2 v 2 + k 3 v 3 = 0 w h ere k 1 , k 2 , k 3 ∈ R 3
∴ k 1 ( 1 , 1 , 0 ) + k 2 ( 1 , 0 , 1 ) + k 3 ( 0 , 1 , 1 ) = 0 \ \therefore k_{1}(1,1,0)+k_{2}(1,0,1)+k_{3}(0,1,1)=0 ∴ k 1 ( 1 , 1 , 0 ) + k 2 ( 1 , 0 , 1 ) + k 3 ( 0 , 1 , 1 ) = 0
⇒ ( k 1 , k 1 , 0 , ) + ( k 2 , 0 , k 2 ) + ( 0 , k 3 , k 3 ) = 0 ⇒ k 1 + k 2 = 0 k 1 + k 3 = 0 k 2 + k 3 = 0 H e r e , N u m b e r o f e q u a t i o n s = N u m b e r o f v a r i a b l e s ∴ A = [ 1 1 0 1 0 1 0 1 1 ] ∣ A ∣ = − 2 ≠ 0 ∴ A i s n o n − s i n g u l a r . A i s i n v e r t i b l e . ∴ S y s t e m i s c o n s i s t e n t . ∴ L i n e a r s y s t e m A X = 0 h a s u n i q u e t r i v i a l s o l u t i o n . k 1 = k 2 = k 3 = 0 ∴ P i s l i n e a r l y i n d e p e n d e n t s e t \Rightarrow (k_{1},k_{1},0,)+(k_{2},0,k_{2})+(0,k_{3},k_{3})=0 \\ \Rightarrow k_{1}+k_{2}=0 \\~~~~~k_{1}+k_{3}=0 \\~~~~~k_{2}+k_{3}=0 \\Here, ~~~Number ~of~ equations~=~Number~ of~ variables\\ \therefore A=\begin{bmatrix}1 & 1&0\\1 & 0 &1\\0&1&1\end{bmatrix}\\ ~|A|=-2\neq0\\ ~\therefore A ~is~ non-singular.\\ ~A ~is~ invertible.\\ \therefore System~ is~ consistent.\\ \therefore Linear~ system~ AX=0~has~unique~trivial~solution.\\ ~~k_{1}=k_{2}=k_{3}=0\\ \therefore ~P~ is ~linearly~ independent ~set ⇒ ( k 1 , k 1 , 0 , ) + ( k 2 , 0 , k 2 ) + ( 0 , k 3 , k 3 ) = 0 ⇒ k 1 + k 2 = 0 k 1 + k 3 = 0 k 2 + k 3 = 0 Here , N u mb er o f e q u a t i o n s = N u mb er o f v a r iab l es ∴ A = ⎣ ⎡ 1 1 0 1 0 1 0 1 1 ⎦ ⎤ ∣ A ∣ = − 2 = 0 ∴ A i s n o n − s in gu l a r . A i s in v er t ib l e . ∴ S ys t e m i s co n s i s t e n t . ∴ L in e a r sys t e m A X = 0 ha s u ni q u e t r i v ia l so l u t i o n . k 1 = k 2 = k 3 = 0 ∴ P i s l in e a r l y in d e p e n d e n t se t
( i i ) T o c h e c k : P i s s p a n n i n g s e t o f R 3 . L e t V = ( a , b , c ) ∈ R 3 b e a n y e l e n e n t a n d u 1 = ( 1 , 1 , 0 ) , u 2 = ( 1 , 0 , 1 ) , u 3 = ( 0 , 1 , 1 ) . L e t i f p o s s i b l e V = k 1 v 1 + k 2 v 2 + k 3 v 3 w h e r e k 1 , k 2 , k 3 ∈ R 3 ∴ ( a , b , c ) = k 1 v 1 + k 2 v 2 + k 3 v 3 = k 1 ( 1 , 1 , 0 ) + k 2 ( 1 , 0 , 1 ) + k 3 ( 0 , 1 , 1 ) ∴ k 1 + k 2 = a k 1 + k 3 = b a n d k 2 + k 3 = c w h i c h i s n o n − h o m o g e n o u s s y s t e m . (ii)~ To~check~:~P~is~spanning~set~of~R^3.\\~~~~~~~Let~V=(a,b,c)\in R^3~be~any~elenent~and~u_{1}=(1,1,0),u_{2}=(1,0,1),u_{3}=(0,1,1).\\~~~~~~~Let~if~possible~V=k_{1}v_{1}+k_{2}v_{2}+k_{3}v_{3}~where~k_{1},k_{2},k_{3}\in R^3\\ \therefore (a,b,c)=k_{1}v_{1}+k_{2}v_{2}+k_{3}v_{3}~=k_{1}(1,1,0)+k_{2}(1,0,1)+k_{3}(0,1,1)\\ \therefore k_{1}+k_{2}=a~\\~~~~ k_{1}+k_{3}=b~~and\\~~~~k_{2}+k_{3}=c~\\which ~is~non-homogenous~system. ( ii ) T o c h ec k : P i s s p annin g se t o f R 3 . L e t V = ( a , b , c ) ∈ R 3 b e an y e l e n e n t an d u 1 = ( 1 , 1 , 0 ) , u 2 = ( 1 , 0 , 1 ) , u 3 = ( 0 , 1 , 1 ) . L e t i f p oss ib l e V = k 1 v 1 + k 2 v 2 + k 3 v 3 w h ere k 1 , k 2 , k 3 ∈ R 3 ∴ ( a , b , c ) = k 1 v 1 + k 2 v 2 + k 3 v 3 = k 1 ( 1 , 1 , 0 ) + k 2 ( 1 , 0 , 1 ) + k 3 ( 0 , 1 , 1 ) ∴ k 1 + k 2 = a k 1 + k 3 = b an d k 2 + k 3 = c w hi c h i s n o n − h o m o g e n o u s sys t e m .
H e r e , N u m b e r o f e q u a t i o n s = N u m b e r o f v a r i a b l e s ∴ L e t A = [ 1 1 0 1 0 1 0 1 1 ] ∣ A ∣ = − 2 ≠ 0 ∴ A i s n o n − s i n g u l a r . A i s i n v e r t i b l e . ∴ S y s t e m i s c o n s i s t e n t . ∴ L i n e a r s y s t e m A X = B h a s s o l u t i o n . ∴ E v e r y e l e m e n t i n R 3 c a n b e e x p r e s s e d a s a l i n e a r c o m b i n a t i o n o f u 1 , u 2 , u 3 ∴ P s p a n s R 3 ∴ A l l t w o c o n d i t i o n s s a t i s f i e d . ∴ P i s b a s i s f o r R 3 \\Here, ~~~Number ~of~ equations~=~Number~ of~ variables\\ \therefore~Let~ A=\begin{bmatrix}1 & 1&0\\1 & 0 &1\\0&1&1\end{bmatrix}\\ ~|A|=-2\neq0\\ ~\therefore A ~is~ non-singular.\\ ~A ~is~ invertible.\\ \therefore System~ is~ consistent.\\ \therefore Linear~ system~ AX=B~has~solution.\\ \therefore Every~element~in~R^3~can~be~expressed~as~a~linear~combination~of~u_{1},u_{2},u_{3}\\ \therefore ~P~spans~R^3\\ \therefore All~two~conditions~satisfied.\\ \\ \therefore P~is~basis~for~R^3 Here , N u mb er o f e q u a t i o n s = N u mb er o f v a r iab l es ∴ L e t A = ⎣ ⎡ 1 1 0 1 0 1 0 1 1 ⎦ ⎤ ∣ A ∣ = − 2 = 0 ∴ A i s n o n − s in gu l a r . A i s in v er t ib l e . ∴ S ys t e m i s co n s i s t e n t . ∴ L in e a r sys t e m A X = B ha s so l u t i o n . ∴ E v ery e l e m e n t in R 3 c an b e e x p resse d a s a l in e a r co mbina t i o n o f u 1 , u 2 , u 3 ∴ P s p an s R 3 ∴ A ll tw o co n d i t i o n s s a t i s f i e d . ∴ P i s ba s i s f or R 3
Now , To find the matrix of T w.r.t. these ordered basis.
L e t B = u 1 , u 2 = [ 1 1 ] , [ 1 0 ] a n d B ′ = v 1 , v 2 , v 3 = [ 1 1 0 ] , [ 1 0 1 ] , [ 0 1 1 ] Let ~B={u_1,u_2}=\begin{bmatrix}1 \\1 \end{bmatrix},\begin{bmatrix}1 \\0 \end{bmatrix} ~and ~B'=v_1,v_2,v_3=\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix},\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix} L e t B = u 1 , u 2 = [ 1 1 ] , [ 1 0 ] an d B ′ = v 1 , v 2 , v 3 = ⎣ ⎡ 1 1 0 ⎦ ⎤ , ⎣ ⎡ 1 0 1 ⎦ ⎤ , ⎣ ⎡ 0 1 1 ⎦ ⎤
T : R 2 → R 3 d e f i n e d b y T ( [ x 1 x 2 ] ) = ( [ x 1 + x 2 x 1 x 1 − x 2 ] ) T [ 1 1 ] = [ 1 + 1 1 1 − 1 ] = [ 2 1 0 ] T [ 1 0 ] = [ 1 + 0 1 1 − 0 ] = [ 1 1 1 ] T:R^2 \rightarrow R^3~ defined~ by ~T(\begin{bmatrix}x_1 \\x_2 \end{bmatrix}) =(\begin{bmatrix}x_1+x_2 \\x_1\\x_1-x_2 \end{bmatrix})\\ T\begin{bmatrix}1 \\1 \end{bmatrix} =\begin{bmatrix}1+1 \\1\\1-1 \end{bmatrix}=\begin{bmatrix}2 \\1\\0 \end{bmatrix}\\T\begin{bmatrix}1 \\0 \end{bmatrix} =\begin{bmatrix}1+0 \\1\\1-0 \end{bmatrix}=\begin{bmatrix}1 \\1\\1 \end{bmatrix} T : R 2 → R 3 d e f in e d b y T ( [ x 1 x 2 ] ) = ( ⎣ ⎡ x 1 + x 2 x 1 x 1 − x 2 ⎦ ⎤ ) T [ 1 1 ] = ⎣ ⎡ 1 + 1 1 1 − 1 ⎦ ⎤ = ⎣ ⎡ 2 1 0 ⎦ ⎤ T [ 1 0 ] = ⎣ ⎡ 1 + 0 1 1 − 0 ⎦ ⎤ = ⎣ ⎡ 1 1 1 ⎦ ⎤
To find the matrix of T w.r.t. bases B and B' construct matrix [ v 1 , v 2 , v 3 ∣ T ( u 1 ) , T ( u 2 ) ] [v_1,v_2,v_3|T(u_1),T(u_2)] [ v 1 , v 2 , v 3 ∣ T ( u 1 ) , T ( u 2 )] and transformed to reduced row echelon form as
[ 1 1 0 ∣ 2 1 1 0 1 ∣ 1 1 0 1 1 ∣ 0 1 ] A p p l y i n g R 2 − R 1 → R 2 [ 1 1 0 ∣ 2 1 0 − 1 1 ∣ − 1 0 0 1 1 ∣ 0 1 ] A p p l y i n g − R 2 → R 2 [ 1 1 0 ∣ 2 1 0 1 − 1 ∣ 1 0 0 1 1 ∣ 0 1 ] \begin{bmatrix}1&1&0 ~| ~2&1\\1 & 0 & 1|~1&1\\0&1&1|~0&~1\end{bmatrix}\\Applying~R_2-R_1\rightarrow~R_2\\\begin{bmatrix}1&1&0 |~~~~~2&1\\0 & -1 &1|-1&0\\0&1&1|0&1\end{bmatrix}
\\Applying~-R_2\rightarrow~R_2\\\begin{bmatrix}1&1&~~~0| 2&1\\0&1& -1|1&0\\0&1&~~~~1|~0&~1\end{bmatrix} ⎣ ⎡ 1 1 0 1 0 1 0 ∣ 2 1∣ 1 1∣ 0 1 1 1 ⎦ ⎤ A ppl y in g R 2 − R 1 → R 2 ⎣ ⎡ 1 0 0 1 − 1 1 0∣ 2 1∣ − 1 1∣0 1 0 1 ⎦ ⎤ A ppl y in g − R 2 → R 2 ⎣ ⎡ 1 0 0 1 1 1 0∣2 − 1∣1 1∣ 0 1 0 1 ⎦ ⎤
A p p l y i n g R 3 − R 2 → R 3 [ 1 1 0 ∣ 2 1 0 1 − 1 ∣ 1 0 0 0 2 ∣ − 1 1 ] \\Applying~R_3-R_2\rightarrow~R_3\\\begin{bmatrix}1&1&0| 2&1\\0&1& -1|1&0\\~0&~0&~2|-1&1\end{bmatrix} A ppl y in g R 3 − R 2 → R 3 ⎣ ⎡ 1 0 0 1 1 0 0∣2 − 1∣1 2∣ − 1 1 0 1 ⎦ ⎤
A p p l y i n g 1 2 R 3 → R 3 [ 1 1 0 ∣ 2 1 0 1 − 1 ∣ 1 0 0 0 1 ∣ − 1 2 1 2 ] \\Applying~\frac{1}{2}R_3\rightarrow~R_3\\\begin{bmatrix}1&1&~~~0| 2&1\\0&1& -1|1&0\\0&0&~~~~1|~-\frac{1}{2}&~\frac{1}{2}\end{bmatrix} A ppl y in g 2 1 R 3 → R 3 ⎣ ⎡ 1 0 0 1 1 0 0∣2 − 1∣1 1∣ − 2 1 1 0 2 1 ⎦ ⎤
A p p l y i n g R 2 + R 3 → R 2 [ 1 1 0 ∣ 2 1 0 1 0 ∣ 1 2 1 2 0 0 1 ∣ − 1 2 1 2 ] \\Applying~R_2+R_3\rightarrow~R_2\\\begin{bmatrix}1&1&~~~0| 2&1\\0&1& 0|\frac{1}{2}&\frac{1}{2}\\0&0&~~~~1|~-\frac{1}{2}&~\frac{1}{2}\end{bmatrix} A ppl y in g R 2 + R 3 → R 2 ⎣ ⎡ 1 0 0 1 1 0 0∣2 0∣ 2 1 1∣ − 2 1 1 2 1 2 1 ⎦ ⎤
A p p l y i n g R 1 − R 2 → R 1 [ 1 0 0 ∣ 3 2 1 2 0 1 0 ∣ 1 2 1 2 0 0 1 ∣ − 1 2 1 2 ] = [ I 3 ∣ A ] \\Applying~R_1-R_2\rightarrow~R_1\\\begin{bmatrix}1&0&~~~0| \frac{3}{2}&\frac{1}{2}\\0&1& 0|\frac{1}{2}&\frac{1}{2}\\0&0&~~~~1|~-\frac{1}{2}&~\frac{1}{2}\end{bmatrix}=[I_3|A] A ppl y in g R 1 − R 2 → R 1 ⎣ ⎡ 1 0 0 0 1 0 0∣ 2 3 0∣ 2 1 1∣ − 2 1 2 1 2 1 2 1 ⎦ ⎤ = [ I 3 ∣ A ]
∴ \therefore ∴ The matrix of T w.r.t. Bases B and B' is
A = [ T ] B , B ′ = [ 3 2 1 2 1 2 1 2 − 1 2 1 2 ] A=[T]_{B,B'}=\begin{bmatrix}\frac{3}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}\\-\frac{1}{2}&\frac{1}{2}\end{bmatrix} A = [ T ] B , B ′ = ⎣ ⎡ 2 3 2 1 − 2 1 2 1 2 1 2 1 ⎦ ⎤
Comments