Let $T : \mathbb{R}^3 \to \mathbb{R}^3$ be a linear trasformation. Find $T(x)$

Question

Let $T : \mathbb{R}^3 \to \mathbb{R}^3$ be a linear trasformation with

$T \left(\begin{bmatrix} 1 \\ -2 \\ -1 \\ \end{bmatrix}\right) = \begin{bmatrix} 1 \\ -1 \\ 2 \\ \end{bmatrix} $ and $T \left(\begin{bmatrix} -2 \\ 1 \\ 3 \\ \end{bmatrix}\right) = \begin{bmatrix} 0 \\ -2 \\ 4 \\ \end{bmatrix} $ find $T \left(\begin{bmatrix} -1 \\ -4 \\ 3 \\ \end{bmatrix}\right) = $

Can someone help me out with this its not making sense to me

Answer 1

If we could find scalars $\alpha$ and $\beta$ such that $$ \begin{bmatrix}-1\\-4\\3\end{bmatrix}=\alpha\begin{bmatrix}1\\-2\\-1\end{bmatrix}+\beta\begin{bmatrix}-2\\1\\3\end{bmatrix} $$ then the linearity of $T$ would imply that \begin{align*} T\left(\begin{bmatrix}-1\\-4\\3\end{bmatrix}\right) &=\alpha T\left(\begin{bmatrix}1\\-2\\-1\end{bmatrix}\right)+\beta T\left(\begin{bmatrix}-2\\1\\3\end{bmatrix}\right) \\ &= \alpha\begin{bmatrix}1\\-1\\2\end{bmatrix}+\beta\begin{bmatrix}0\\-2\\4\end{bmatrix} \end{align*} Can you find such $\alpha$ and $\beta$?

Answer 2

Hint:

Note that

$$\begin{pmatrix}-1\\-4\\\;\;3\end{pmatrix}=3\begin{pmatrix}\;\;1\\-2\\-1\end{pmatrix}+2\begin{pmatrix}-2\\\;\;1\\\;\;3\end{pmatrix}$$

Answer 3

We write the vector $$x=\begin{bmatrix} -1 \\ -4 \\ 3 \\ \end{bmatrix}$$ as a linear combination of the two vectors $$u=\begin{bmatrix} 1 \\ -2 \\ -1 \\ \end{bmatrix}\quad;\quad v=\begin{bmatrix} -2 \\ 1 \\ 3 \\ \end{bmatrix}$$ so we look for $\alpha,\beta\in\Bbb R$ such that $$x=\alpha u+\beta v$$ hence we find using the components $$-1=\alpha-2\beta\quad;\quad-4=-2\alpha+\beta$$ so we find $\alpha=3$ and $\beta=2$ and we verify that the third equality of components is valid with these values. Now the answer to the question is easy using the linearity of $T$.