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The linear transformation $A:\mathbb{R}^2\to \mathbb{R}^2$ is given by the images of basis vectors:
$A((1,1))=(2,1)$ and $A((1,0))=(0,3)$.

  1. Find a matrix of linear transformation $A$ in the basis $(1,1), (1,0)$.

  2. Find $A((3,2))$.

  3. Find vector $x=(x_1,x_2)$ such that the matrix $\begin{pmatrix}-6 &-6\\ 3 &4\end{pmatrix}$ is matrix of the linear transformation $A$ in the basis $x$, $(0,3)$.

Please help me about this.

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    I think that 1. is [2 1; 0 3]. Right?2011-12-17
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    2. (3,2)= a(1,1)+b(1,0) = (a+b,a) . From this I have a=2, b=1. (3,2)=2(1,1)+(1,0) A((3,2)) = A(2(1,1)) + A((1,0)) = 2A((1,1))+A((1,0)) = 2(2,1) + (0,3) = (4,2)+(0,3) = (4,5)2011-12-17
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    is this good? Correct me if not.2011-12-17
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    The image of the basis vectors, are the columns of the transformation matrix, so you can trivially do 1.2011-12-17
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    For (1) you already know how to do it as Paxinum has told you above. For (2), I think when you ask to find $A((3,2))$ it is technically not correct. This is because your matrix $A$ has columns in a basis different from the standard basis (I am assuming that $(3,2) = 3e_1 + 2e_2$, $\{e_1,e_2\}$ the standard basis of $\mathbb{R}^2$.2011-12-17
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    Therefore, you need to ask what is $(3,2)$ in the basis you have specified above? You will need to solve a system of two equations in two unknowns to get this.2011-12-17
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    I did it. I wrote it in comment above. can u check it, please?2011-12-17
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    And I think that Plaxinum is not right.2011-12-17
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    Are you sure the second column in the matrix in part 3. is correct? (Something is wrong in my deleted answer if it is.)2011-12-17
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    yes, it's [-6 -6; 3 4]2011-12-17
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    I got (x1,x2)=(-1,3)2011-12-17
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    @Gorcia You are correct. For part 1., things depend on what you mean by "matrix of linear transformation" and if you are using the given basis in both the domain and range (which is what I assume in my answer).2011-12-17

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