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Consider the linear map from $R^3 \rightarrow R^3$ which takes $\vec{e_1}$ to $\vec{a_1}=\begin{bmatrix} 1\\0\\-1\end{bmatrix}$, takes $\vec{e_2}$ to $\vec{a_2}=\begin{bmatrix}0\\1\\3\end{bmatrix}$, and takes $\vec{e_3}$ to $\vec{a_3}=\begin{bmatrix}2\\5\\1\end{bmatrix}$. What is the matrix of this linear map?

Here is how I am approaching this question:

From what is given I know that:

$$ \vec{e_1} \times \vec{x} = \vec{a_1}\\ \vec{e_2} \times \vec{x} = \vec{a_2}\\ \vec{e_3} \times \vec{x} = \vec{a_3}\\ $$

Then I will end up with 3 equations per each (total of 9 equations). But I do now know where should I go from there? Should I use row echelon form to find the result? How should I deal with 9 equations?

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Suppose that the matrix of this linear map is: $$ M = \begin{bmatrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \\ \end{bmatrix} $$ Then since $M\vec{e_1} = \vec{a_1}$, we know that: $$ \begin{bmatrix} m_{11} \\ m_{21} \\ m_{31} \end{bmatrix} = \begin{bmatrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} $$

Continuing in this manner, we conclude that: $$ M = [\vec{a_1} ~|~ \vec{a_2} ~|~ \vec{a_3}] = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 5 \\ -1 & 3 & 1 \\ \end{bmatrix} $$

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