Compute $T([-1, 4, 2])$ given the effect of three vectors on a $T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{2}$ linear transformation

Question

I have this that $T:\mathbb{R}^{3}\Rightarrow \mathbb{R}^{2}$ and these results of the linear transformation:

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

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

$T\left ( \begin{bmatrix} 0\\ -1\\ 1 \end{bmatrix}\right )=\begin{bmatrix}3\\-5\end{bmatrix}$

And I need to find $T([-1, 4, 2])$.

This is my solution:

Write a vector $\vec{b}$ in terms of the three transformed $\mathbb{R}^{3}$ vectors:

$\vec{b}=\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}=k_{1}\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}+k_{2}\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}+k_{3}\begin{bmatrix} 0\\ -1\\ 1 \end{bmatrix}$

Then,

\begin{bmatrix} 1 & 0 & 0 & b_{1}\\ 0 & 1 & -1 & b_{2}\\ 0 & 0 & 1 & b_{3} \end{bmatrix}

Reducing the matrix:

\begin{bmatrix} 1 & 0 & 0 & b_{1}\\ 0 & 1 & 0 & b_{2} + b_{3}\\ 0 & 0 & 1 & b_{3} \end{bmatrix}

So any $\vec{b}$ with the linear transformation can be written as:

$T\left ( \begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix}\right )=b_{1}T\left ( \begin{bmatrix} 1\\0\\0 \end{bmatrix} \right )+(b_{2}+b_{3})T\left ( \begin{bmatrix} 0\\1\\0 \end{bmatrix} \right )+b_{3}T\left ( \begin{bmatrix} 0\\-1\\1 \end{bmatrix} \right )$

With this result I can calculate $T([-1, 4, 2])$:

$T\left ( \begin{bmatrix} -1\\ 4\\ 2 \end{bmatrix}\right )=-1T\left ( \begin{bmatrix} 1\\0\\0 \end{bmatrix} \right )+(4+2)T\left ( \begin{bmatrix} 0\\1\\0 \end{bmatrix} \right )+2T\left ( \begin{bmatrix} 0\\-1\\1 \end{bmatrix} \right )=\begin{bmatrix}33\\-17\end{bmatrix}$

Is my result correct?

Answer 1

Your result is correct. You can find it easily noting that: $$ [-1,4,2]=-1[1,0,0]+6[0,1,0]+2[0,-1,1] $$ so, by linearity: $$ T[-1,4,2]=-[-3,1]+6[4,-1]+2[3,-5] $$