# $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $T\begin{bmatrix} 1 \\1 \end{bmatrix}=3$ and $T\begin{bmatrix} -1 \\2 \end{bmatrix}=6$ [closed]

Seeking some help with the following midterm review question:

Give a linear transformation $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $T\begin{bmatrix} 1 \\1 \end{bmatrix}=3$ and $T\begin{bmatrix} -1 \\2 \end{bmatrix}=6$ and find a matrix $A$ such that $T=T_A$. So $T$ is left multiplication by $A$.

## closed as off-topic by user137731, Claude Leibovici, user26857, Shailesh, NamasteOct 30 '16 at 11:40

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• I think the only was this can work is to pick a 1 by 2 matrix $[a ..b]$. If you perform the multiplications, you get a system of two linear equations with 2 unknowns from which $a$ and $b$ can be found. Otherwise I wouldn't know... – imranfat Oct 30 '16 at 2:26
• @ levap thank you for your edit. Any idea how to proceed with thus problem? – jh123 Oct 30 '16 at 3:09

## 1 Answer

You can substitute $A = \begin{bmatrix} a & b\end{bmatrix}$ and solve the resulting system of equations. Alternatively:

$$A \begin{bmatrix} 1&-1\\1&2\end{bmatrix} = \begin{bmatrix}3&6\end{bmatrix}$$

$$A = \begin{bmatrix}3&6\end{bmatrix}\begin{bmatrix} 1&-1\\1&2\end{bmatrix}^{-1}$$

• is this the complete answer? – jh123 Oct 30 '16 at 3:08
• No, but it should help you figure out the answer. – Michael Biro Oct 30 '16 at 3:09
• okay I will try to proceed fromhere – jh123 Oct 30 '16 at 3:10
• @ Michael Biro so all I would need to do is solve the system of equations? – jh123 Oct 30 '16 at 3:13
• Yes, that should work. – Michael Biro Oct 30 '16 at 3:21