Question

question 9
let ( t : mathbb{r}^2 \to mathbb{r}^2 ) be the linear transformation defined by ( t\begin{bmatrix} x \\ y end{bmatrix} = \begin{bmatrix} x + y \\ y end{bmatrix} ). find ( t^{-1}\begin{bmatrix} 2 \\ 1 end{bmatrix} ).
(a) ( \begin{bmatrix} 1 \\ 1 end{bmatrix} )
(b) ( \begin{bmatrix} 3 \\ 1 end{bmatrix} )
(c) ( \begin{bmatrix} 2 \\ 1 end{bmatrix} )
(d) ( \begin{bmatrix} 1 \\ 2 end{bmatrix} )
(e) none of the above.

Explanation:

Step1: Define \( T^{-1}\)

Let \( T^{-1}

$$\begin{bmatrix}a\\b\end{bmatrix}$$

=

$$\begin{bmatrix}x\\y\end{bmatrix}$$

\). By definition of inverse transformation, \( T

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=

$$\begin{bmatrix}a\\b\end{bmatrix}$$

\). Given \( T

$$\begin{bmatrix}x\\y\end{bmatrix}$$

=

$$\begin{bmatrix}x + y\\y\end{bmatrix}$$

\), so we have the system:

$$
LATEXBLOCK6
$$

Step2: Solve for \(x\) and \(y\)

We are given \(a = 2\) and \(b = 1\) (since we need \(T^{-1}

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

\)). From the second equation \(y=b = 1\). Substitute \(y = 1\) into the first equation \(x+y=a\), we get \(x+1 = 2\), so \(x=2 - 1=1\).
Thus, \(T^{-1}

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

=

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

\)

Answer:

A. \(

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

\)