Question

question 5
consider a linear transformation ( t : mathbb{r}^2 \to mathbb{r}^2 ) such that
( t \begin{bmatrix} 1 \\ 1 end{bmatrix} = \begin{bmatrix} -1 \\ 1 end{bmatrix} ) and ( t \begin{bmatrix} 2 \\ 1 end{bmatrix} = \begin{bmatrix} 1 \\ 1 end{bmatrix} ).
then ( t \begin{bmatrix} 0 \\ 1 end{bmatrix} ) is equal to:
(a) ( \begin{bmatrix} 1 \\ 1 end{bmatrix} ).
(b) ( \begin{bmatrix} -1 \\ 3 end{bmatrix} ).
(c) ( \begin{bmatrix} -3 \\ 1 end{bmatrix} ).
(d) ( \begin{bmatrix} 3 \\ 2 end{bmatrix} ).
(e) none of the above.

Explanation:

Step1: Express \(

$$\begin{bmatrix}0\\1\end{bmatrix}$$

\) as a linear combination
Let \(

$$\begin{bmatrix}0\\1\end{bmatrix}$$

=a

$$\begin{bmatrix}1\\1\end{bmatrix}$$

+b

$$\begin{bmatrix}2\\1\end{bmatrix}$$

\). Then we have the system of equations:
\(a + 2b=0\) (from the first component) and \(a + b = 1\) (from the second component).
Subtract the second equation from the first: \((a + 2b)-(a + b)=0 - 1\), so \(b=- 1\).
Substitute \(b = - 1\) into \(a + b=1\), we get \(a-1 = 1\), so \(a = 2\).
Thus, \(

$$\begin{bmatrix}0\\1\end{bmatrix}$$

=2

$$\begin{bmatrix}1\\1\end{bmatrix}$$

-1

$$\begin{bmatrix}2\\1\end{bmatrix}$$

\).

Step2: Use linearity of \(T\)

Since \(T\) is a linear transformation, \(T(c\mathbf{u}+d\mathbf{v})=cT(\mathbf{u})+dT(\mathbf{v})\) for scalars \(c,d\) and vectors \(\mathbf{u},\mathbf{v}\).
So \(T

$$\begin{bmatrix}0\\1\end{bmatrix}$$

=2T

$$\begin{bmatrix}1\\1\end{bmatrix}$$

-1\times T

$$\begin{bmatrix}2\\1\end{bmatrix}$$

\).
We know \(T

$$\begin{bmatrix}1\\1\end{bmatrix}$$

=

$$\begin{bmatrix}-1\\1\end{bmatrix}$$

\) and \(T

$$\begin{bmatrix}2\\1\end{bmatrix}$$

=

$$\begin{bmatrix}1\\1\end{bmatrix}$$

\).
Then \(2

$$\begin{bmatrix}-1\\1\end{bmatrix}$$

-

$$\begin{bmatrix}1\\1\end{bmatrix}$$

=

$$\begin{bmatrix}-2 - 1\\2-1\end{bmatrix}$$

=

$$\begin{bmatrix}-3\\1\end{bmatrix}$$

\).

Answer:

\(\boldsymbol{\text{(C) }

$$\begin{bmatrix}-3\\1\end{bmatrix}$$

}\)