Question #236244

Consider the linear transformations :

f:R^2→R^2 g:R^2→R^2

f (x,y)→(2x-y,3x+y) g(x,y)→(4x-2y,2x+y)


a) Determine the matrix of linear transformation f relative to the basis

{e1=(1,2),e2=(-1,1)}

b)dertimine (fog)(xy)

c) find g^-1


1
Expert's answer
2021-09-20T16:52:19-0400

f(x,y)=(2xy, 3x+y),   g(x,y)=(4x2y, 2x+y)f(x,y)=(2x-y,\ 3x+y),\ \ \ g(x,y)=(4x-2y, \ 2x+y)


a)Let AA be the matrix of ff with respect to the standard basis,

BB be the matrix of ff with respect to the basis e1=(1,2), e2=(1,1)e_1=(1,2), \ e_2=(-1,1) , and PP be the transition matrix.

A=(2131)A=\begin{pmatrix} 2 & -1 \\ 3 & 1 \end{pmatrix}

P=(1121)P=\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} , P1=13(1121)=(1/31/32/31/3)P^{-1}=\frac{1}{3} \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}=\begin{pmatrix} 1/3 & 1/3 \\ -2/3 & 1/3 \end{pmatrix}

B=P1AP=(1/31/32/31/3)(2131)(1121)=(1/31/32/31/3)(0352)=(5/35/35/34/3)B=P^{-1}AP =\begin{pmatrix} 1/3 & 1/3 \\ -2/3 & 1/3 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} =\begin{pmatrix} 1/3 & 1/3 \\ -2/3 & 1/3 \end{pmatrix} \begin{pmatrix} 0 & -3 \\ 5 & -2 \end{pmatrix}=\begin{pmatrix} 5/3 & -5/3 \\ 5/3 & 4/3 \end{pmatrix}


b) (fg)(x,y)=f(4x2y, 2x+y)=(2(4x2y)(2x+y), 3(4x2y)+(2x+y))=(6x5y,14x5y)(f\circ g)(x,y) = f(4x-2y,\ 2x+y)=\big(2(4x-2y)-(2x+y),\ 3(4x-2y)+(2x+y)\big)= (6x-5y, 14x-5y)


c) g(x,y)=(4x2y, 2x+y)=(z,w)g(x,y)=(4x-2y, \ 2x+y)=(z,w)

Let GG be the matrix of ff : G=(4221)G=\begin{pmatrix} 4 & -2 \\ 2 & 1 \end{pmatrix}

Then G1=18(1224)=(1/81/41/41/2)G^{-1}=\frac {1}{8} \begin{pmatrix} 1 & 2 \\ -2 & 4 \end{pmatrix}=\begin{pmatrix} 1/8 & 1/4 \\ -1/4 & 1/2 \end{pmatrix}


g1(z,w)=(z8+w4, z4+w2)g^{-1}(z,w)=(\frac{z}{8}+\frac{w}{4}, \ -\frac{z}{4}+\frac{w}{2})


Answer:

a) B=(5/35/35/34/3)B= \begin{pmatrix} 5/3 & -5/3 \\ 5/3 & 4/3 \end{pmatrix}

b) (fg)(x,y)=(6x5y,14x5y)(f\circ g)(x,y) = (6x-5y, 14x-5y)

c) g1(z,w)=(z8+w4, z4+w2)g^{-1}(z,w)=(\frac{z}{8}+\frac{w}{4}, \ -\frac{z}{4}+\frac{w}{2})


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