S o l u t i o n : L e t u s c o n s i d e r b a s i s f o r R 2 g i v e n b y : S = { [ − 2 1 ] , [ 1 3 ] } a n d l e t u s c o n s i d e r l i n e a r o p e r a t o r T : R 2 → R 3 f o r w h i c h : T ( v 1 ) = [ − 1 4 0 ] a n d T ( v 2 ) = [ 0 − 5 9 ] L e t u s f i r s t f i n d f o r m u l a f o r T ( x 1 , x 2 ) : ( x 1 , x 2 ) = a v 1 + b v 2 = a [ − 2 1 ] + b [ 1 3 ] [ x 1 x 2 ] = [ − 2 ( a ) 1 ( a ) ] + [ 1 ( b ) 3 ( b ) ] [ x 1 x 2 ] = [ − 2 a a ] + [ b 3 b ] [ x 1 x 2 ] = [ − 2 a + b a + 3 b ] ⇒ x 1 = − 2 a + b a n d x 2 = a + 3 b , S o l v i n g t h e s e e q u a t i o n s f o r a a n d b , w e g e t a = 1 7 ( − 3 x 1 + x 2 ) a n d b = 1 7 ( x 1 + 2 x 2 ) w h i c h g i v e s u s : ( x 1 , x 2 ) = a v 1 + b v 2 = 1 7 ( − 3 x 1 + x 2 ) v 1 + 1 7 ( x 1 + 2 x 2 ) v 2 T h e r e f o r e , w e h a v e : T ( x 1 , x 2 ) = T ( a v 1 + b v 2 ) = a T ( v 1 ) + b T ( v 2 ) = 1 7 ( − 3 x 1 + x 2 ) T ( v 1 ) + 1 7 ( x 1 + 2 x 2 ) T ( v 2 ) = 1 7 ( − 3 x 1 + x 2 ) [ − 1 4 0 ] + 1 7 ( x 1 + 2 x 2 ) [ 0 − 5 9 ] = 1 7 { ( − 3 x 1 + x 2 ) [ − 1 4 0 ] + ( x 1 + 2 x 2 ) [ 0 − 5 9 ] } = 1 7 { [ 3 x 1 − x 2 − 12 x 1 + 4 x 2 0 ] + [ 0 − 5 x 1 − 10 x 2 9 x 1 + 18 x 2 ] } = 1 7 [ 3 x 1 − x 2 − 17 x 1 − 6 x 2 9 x 1 + 18 x 2 ] W h i c h g i v e s : T ( 4 , − 5 ) = 1 7 [ 3 ( 4 ) − ( − 5 ) − 17 ( 4 ) − 6 ( − 5 ) 9 ( 4 ) + 18 ( − 5 ) ] = 1 7 [ 12 + 5 − 68 + 30 36 − 90 ] = 1 7 [ 17 − 38 − 54 ] = [ 17 7 − 38 7 − 54 7 ] = [ 2.428571 − 5.428571 − 7.71428571 ] Solution~:~Let~ us ~consider~ basis~ for ~ R^2 ~ given ~by:
\\S= \{ \begin{bmatrix}-2 \\1 \end{bmatrix},\begin{bmatrix}1 \\3 \end{bmatrix} \} ~and ~let ~ us ~consider ~ linear ~ operator~ T:R^2 \rightarrow R^3~ for ~which :
\\T(v_1)= \begin{bmatrix}-1 \\4 \\0 \end{bmatrix}~ and~ T(v_2)=\begin{bmatrix}0 \\-5 \\9 \end{bmatrix}
\\Let ~ us ~ first ~ find ~ formula ~ for ~ T(x_1,x_2):
\\(x_1,x_2)=av_1+bv_2
\\~~~~~~~~~~~~~~=a\begin{bmatrix}-2 \\1 \end{bmatrix}+b\begin{bmatrix}1 \\3 \end{bmatrix}
\\\begin{bmatrix}x_1 \\x_2 \end{bmatrix}=\begin{bmatrix}-2(a) \\1(a) \end{bmatrix}+\begin{bmatrix}1(b) \\3(b) \end{bmatrix}
\\\begin{bmatrix}x_1 \\x_2 \end{bmatrix}=\begin{bmatrix}-2a \\a \end{bmatrix}+\begin{bmatrix}b \\3b \end{bmatrix}
\\\begin{bmatrix}x_1 \\x_2 \end{bmatrix}=\begin{bmatrix}-2a+b \\a+3b \end{bmatrix}
\\\Rightarrow~x_1=-2a+b~ and ~x_2 =a+3b, Solving ~these ~ equations ~ for ~a ~and ~b, we ~get
\\a=\frac{1}{7}(-3x_1+x_2) ~and ~b=\frac{1}{7}(x_1+2x_2)~ which ~gives ~us:
\\(x_1,x_2)=av_1+bv_2=\frac{1}{7}(-3x_1+x_2)v_1+\frac{1}{7}(x_1+2x_2)v_2
\\Therefore, we ~have:
\\T(x_1,x_2)=T(av_1+bv_2)=aT(v_1)+bT(v_2)
\\~~~~~~~~~~~~~~~~~=\frac{1}{7}(-3x_1+x_2)T(v_1)+\frac{1}{7}(x_1+2x_2)T(v_2)
\\~~~~~~~~~~~~~~~~~=\frac{1}{7}(-3x_1+x_2)\begin{bmatrix}-1 \\4 \\0 \end{bmatrix}+\frac{1}{7}(x_1+2x_2)\begin{bmatrix}0 \\-5 \\9 \end{bmatrix}
\\~~~~~~~~~~~~~~~~~=\frac{1}{7}~\{(-3x_1+x_2)\begin{bmatrix}-1 \\4 \\0 \end{bmatrix}+(x_1+2x_2)\begin{bmatrix}0 \\-5 \\9 \end{bmatrix}\}
\\~~~~~~~~~~~~~~~~~=\frac{1}{7}\{\begin{bmatrix} 3x_1-x_2 \\-12x_1+4x_2 \\0 \end{bmatrix}+\begin{bmatrix}0 \\-5x_1-10x_2 \\9x_1+18x_2 \end{bmatrix}\}
\\~~~~~~~~~~~~~~~~~=\frac{1}{7}\begin{bmatrix} 3x_1-x_2 \\-17x_1-6x_2 \\9x_1+18x_2 \end{bmatrix}
\\Which~gives:
\\T(4,-5)=\frac{1}{7}\begin{bmatrix} 3(4)-(-5) \\-17(4)-6(-5) \\9(4)+18(-5) \end{bmatrix}
\\~~~~~~~~~~~~~~~~~=\frac{1}{7}\begin{bmatrix} 12+5 \\-68+30 \\36-90 \end{bmatrix}
\\~~~~~~~~~~~~~~~~~=\frac{1}{7}\begin{bmatrix} 17 \\-38 \\-54 \end{bmatrix}
\\~~~~~~~~~~~~~~~~~=\begin{bmatrix} \frac{17}{7} \\-\frac{38}{7} \\-\frac{54}{7} \end{bmatrix}
\\~~~~~~~~~~~~~~~~~=\begin{bmatrix} 2.428571 \\-5.428571 \\-7.71428571 \end{bmatrix} S o l u t i o n : L e t u s co n s i d er ba s i s f or R 2 g i v e n b y : S = { [ − 2 1 ] , [ 1 3 ] } an d l e t u s co n s i d er l in e a r o p er a t or T : R 2 → R 3 f or w hi c h : T ( v 1 ) = ⎣ ⎡ − 1 4 0 ⎦ ⎤ an d T ( v 2 ) = ⎣ ⎡ 0 − 5 9 ⎦ ⎤ L e t u s f i rs t f in d f or m u l a f or T ( x 1 , x 2 ) : ( x 1 , x 2 ) = a v 1 + b v 2 = a [ − 2 1 ] + b [ 1 3 ] [ x 1 x 2 ] = [ − 2 ( a ) 1 ( a ) ] + [ 1 ( b ) 3 ( b ) ] [ x 1 x 2 ] = [ − 2 a a ] + [ b 3 b ] [ x 1 x 2 ] = [ − 2 a + b a + 3 b ] ⇒ x 1 = − 2 a + b an d x 2 = a + 3 b , S o l v in g t h ese e q u a t i o n s f or a an d b , w e g e t a = 7 1 ( − 3 x 1 + x 2 ) an d b = 7 1 ( x 1 + 2 x 2 ) w hi c h g i v es u s : ( x 1 , x 2 ) = a v 1 + b v 2 = 7 1 ( − 3 x 1 + x 2 ) v 1 + 7 1 ( x 1 + 2 x 2 ) v 2 T h ere f ore , w e ha v e : T ( x 1 , x 2 ) = T ( a v 1 + b v 2 ) = a T ( v 1 ) + b T ( v 2 ) = 7 1 ( − 3 x 1 + x 2 ) T ( v 1 ) + 7 1 ( x 1 + 2 x 2 ) T ( v 2 ) = 7 1 ( − 3 x 1 + x 2 ) ⎣ ⎡ − 1 4 0 ⎦ ⎤ + 7 1 ( x 1 + 2 x 2 ) ⎣ ⎡ 0 − 5 9 ⎦ ⎤ = 7 1 {( − 3 x 1 + x 2 ) ⎣ ⎡ − 1 4 0 ⎦ ⎤ + ( x 1 + 2 x 2 ) ⎣ ⎡ 0 − 5 9 ⎦ ⎤ } = 7 1 { ⎣ ⎡ 3 x 1 − x 2 − 12 x 1 + 4 x 2 0 ⎦ ⎤ + ⎣ ⎡ 0 − 5 x 1 − 10 x 2 9 x 1 + 18 x 2 ⎦ ⎤ } = 7 1 ⎣ ⎡ 3 x 1 − x 2 − 17 x 1 − 6 x 2 9 x 1 + 18 x 2 ⎦ ⎤ Whi c h g i v es : T ( 4 , − 5 ) = 7 1 ⎣ ⎡ 3 ( 4 ) − ( − 5 ) − 17 ( 4 ) − 6 ( − 5 ) 9 ( 4 ) + 18 ( − 5 ) ⎦ ⎤ = 7 1 ⎣ ⎡ 12 + 5 − 68 + 30 36 − 90 ⎦ ⎤ = 7 1 ⎣ ⎡ 17 − 38 − 54 ⎦ ⎤ = ⎣ ⎡ 7 17 − 7 38 − 7 54 ⎦ ⎤ = ⎣ ⎡ 2.428571 − 5.428571 − 7.71428571 ⎦ ⎤
#332078 on Oct 2022
Comments