Linear Transformation of a vector from $R^3$ to $R^2$

Question

Consider the basis $S =\{v_1,v_2,v_3\} $ for $R^3$ where $v_1=(1,1,1), v_2=(1,1,0) ,v_3=(1,0,0) $. T is a linear transformation from $R^3$ to $R^2$ such that $T (v_1)=(1,0), T(v_2)= (2,-1) , T(v_3)= (4,3) $. Then $T(2,-3,5)$ is- ?

I am familiar with the concept of linear transformation and I was thinking of first finding the matrix of transformation. But the question gives no information about the basis of $R^2$ but does for $R^3$. However I tried with the basis $(1,0)$ and $(0,1) $ but my answer doesn't match with any of the options.

Answer 1

Since$$(2,-3,5)=5v_1-8v_2+5v_3,$$then $T(2,-3,5)=\cdots$?

Answer 2

As shown in José’s answer, you want to find $(2,-3,5)$ as a linear combination of $v_1, v_2$, and $v_3$. That is, you want to find a solution to

$$\begin{bmatrix} | & | & | \\ v_1 & v_2 & v_3 \\ | & | & | \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix} 2\\-3\\5\end{bmatrix}.$$

Let $A$ be the matrix above with the $v_i$ as its columns. Since the $v_i$ form a basis, that means that $A$ must be invertible, and thus the solution is given by $x = A^{-1}(2,-3,5)^T$. Fortunately, in this case the inverse is fairly easy to find.

Now that you have your linear combination, you can proceed with José’s hint.