MHB Solving Linear Transformations in R2: Step by Step Guide

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To solve the linear transformation problem in R2, the transformation phi is defined by its action on the basis vectors, with phi((1,0)) = (4,5) and phi((0,1)) = (9,11). The image of any vector (x,y) under phi can be expressed as phi(x,y) = (4x + 9y, 5x + 11y). To find the matrix representation of phi with respect to the basis B formed by vectors w1 = (3,5) and w2 = (10,17), one must compute phi(w1) and phi(w2) and express these results as linear combinations of the basis vectors. This leads to a system of equations to determine the coefficients α1, α2, β1, and β2 for the matrix M_B^B(φ). The final matrix representation can then be constructed using these coefficients.
Fernando Revilla
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Could anyone give me a step by step method on how to solve these type of questions please? Very confused and need to know for my exams.

Let V denote the real vector space R2 and phi: V -> V be a real linear transformation such that phi((1,0)) = (4,5) and phi((0,1)) = (9,11). Express the imagine phi((x,y)) of (x,y) in terms of x and y.

Assume that w1 = (3,5) and w2 = (10,17) form an ordered basis B for V. Working from the definition determine the matrix M(subscript and superscript B) (phi) of phi with respect to the basis B.

I have given a link to the topic there so the OP can see my response.
 
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Using that $\phi$ is a linear map:
$$\phi(x,y)=\phi[x(1,0)+y(0,1)]=x\phi (1,0)+y\phi(0,1)\\=x(4,5)+y(9,11)=(4x+9y,5x+11y)$$
On the other hand:
$$\phi(w_1)=\phi (3,5)=(4\cdot 3+9\cdot 5,\;5\cdot 3+11\cdot 5)=(57,70)$$
Now, write $\phi(w_1)=(57,70)=\alpha_1w_1+\alpha_2 w_2$ and you'll easily find $\alpha_1$ and $\alpha_2$ by means of a simple system. In a similar way, you'll get $\phi(w_2)=\beta_1w_1+\beta_2 w_2$. Then,
$$M_B^B= \begin{bmatrix}{\alpha_1}&{\beta_1}\\{\alpha_2}&{ \beta_2}\end{bmatrix}=\ldots$$
 
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