Finding $T((5,6))$ given $T((1,2))$ and $T((0,1))$ for linear transformation $T$

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The given question is:

Let $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a linear transformation such that $T((1,2))=(2,3)$ and $T((0,1))=(1,4)$.

What is $T((5,6))?$

The given solution's hint says

$$T((5,6))=5T((1,2)) - 4T((0,1))$$

My question is: Where does this $4$ come from?

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2 Answers

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Since $T$ is a linear transformation, you have $T(cx) = cT(x)$.

In particular:

$$5T((1,2)) - 4T((0,1)) = T(5(1,2)) - T(4(0,1)) = T((5,10)) - T((0,4))$$

We also know that $T(x) - T(y) = T(x - y)$, so the final expression is:

$$T((5,10)) - T((0,4)) = T((5-0, 10-4)) = T((5,6))$$

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If $T$ is a linear transformation, then $T(\alpha v + \beta w) = \alpha Tv + \beta Tw$

In your case $(5,6) = 5(1,2) -4(0,1)$

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