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Question: For set of vectors {$x_1,x_2$}, $x_1=(1,3)^T, x_2=(4,6)^T$ are in $R^2$. Find the matrix of linear transformation $T:R^2\rightarrow R^3$ such that $Tx_1=(-2,2,-7)^T$ and $Tx_2=(-2,-4,-10)^T$

My attempt: So I'm guessing there exists a matrix (a transformation matrix) and it must be of order (2,3) for it to give (-2,2,-7)$^T$ when multiplied by (1,3)$^T$ :

$\left[\begin{array}{cc} t_{11} & t_{12}\\ t_{21} & t_{22}\\ t_{31} & t_{32}\end{array} \right]$ $\left[ \begin{array}{c} 1\\ 3\end{array} \right]$ = $\left[ \begin{array}{cc} -2\\ 2\\ -7\end{array} \right]$

But how am I supposed to find out all those t$_{ij}$ variables? Also, using same on second set of vectors give me inconsistent equations:

$\begin{array}{cc} t_{11} + 3t_{12} = -2 & 4t_{11} + 6t_{12} = -2\\ t_{21} + 3t_{22} = 2 & 4t_{21} + 6t_{22} = -4\\ t_{31} + 3t_{32} = -7 & 4t_{31} + 6t_{32} = -10\end{array}$

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up vote 1 down vote accepted

It might be easier if you use a different vector. Notice that $x_2-2x_1=(2,0)^T$. Since you know that $T(x_2-2x_1)=T(x_2)-2T(x_1)$, you can set up a simpler set of equations. Then try to get a vector $(0,a)^T$, for some $a$.

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