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1. Define [tex] T: \mathbb{R}^2 \rightarrow \mathbb{R}^2 [/tex] by
[tex]\[ T\left(\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}\right)=\begin{bmatrix}
2x_1 - 3x_2 \\
-x_1 + x_2
\end{bmatrix}. \][/tex]

Find each of the following:

a) [tex] T\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\right) [/tex]

b) [tex] T\left(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right) [/tex]

c) [tex] T\left(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\right) [/tex]

d) [tex] T\left(\begin{bmatrix} -1 \\ 0 \end{bmatrix}\right) [/tex]

Sagot :

GinnyAnsweridn
To find the image of a vector under the transformation [tex]\( T \)[/tex], we need to apply the transformation [tex]\( T \)[/tex] to each given vector according to the formula:

[tex]\[ T\left(\left[\begin{array}{c}x_1 \\ x_2\end{array}\right]\right) = \left[\begin{array}{c}2 x_1 - 3 x_2 \\ -x_1 + x_2\end{array}\right] \][/tex]

Let's calculate each specified transformation step-by-step:

### Part (a)
For [tex]\( T\left(\left[\begin{array}{c}0 \\ 0\end{array}\right]\right) \)[/tex]:
[tex]\[ x_1 = 0, \quad x_2 = 0 \][/tex]

Using the transformation formula:
[tex]\[ T\left(\left[\begin{array}{c}0 \\ 0\end{array}\right]\right) = \left[\begin{array}{c}2(0) - 3(0) \\ -0 + 0\end{array}\right] = \left[\begin{array}{c}0 \\ 0\end{array}\right] \][/tex]

So, [tex]\( T\left(\left[\begin{array}{c}0 \\ 0\end{array}\right]\right) = \left[\begin{array}{c}0 \\ 0\end{array}\right] \)[/tex].

### Part (b)
For [tex]\( T\left(\left[\begin{array}{c}1 \\ 1\end{array}\right]\right) \)[/tex]:
[tex]\[ x_1 = 1, \quad x_2 = 1 \][/tex]

Using the transformation formula:
[tex]\[ T\left(\left[\begin{array}{c}1 \\ 1\end{array}\right]\right) = \left[\begin{array}{c}2(1) - 3(1) \\ -1 + 1\end{array}\right] = \left[\begin{array}{c}2 - 3 \\ -1 + 1\end{array}\right] = \left[\begin{array}{c}-1 \\ 0\end{array}\right] \][/tex]

So, [tex]\( T\left(\left[\begin{array}{c}1 \\ 1\end{array}\right]\right) = \left[\begin{array}{c}-1 \\ 0\end{array}\right] \)[/tex].

### Part (c)
For [tex]\( T\left(\left[\begin{array}{c}2 \\ 1\end{array}\right]\right) \)[/tex]:
[tex]\[ x_1 = 2, \quad x_2 = 1 \][/tex]

Using the transformation formula:
[tex]\[ T\left(\left[\begin{array}{c}2 \\ 1\end{array}\right]\right) = \left[\begin{array}{c}2(2) - 3(1) \\ -2 + 1\end{array}\right] = \left[\begin{array}{c}4 - 3 \\ -2 + 1\end{array}\right] = \left[\begin{array}{c}1 \\ -1\end{array}\right] \][/tex]

So, [tex]\( T\left(\left[\begin{array}{c}2 \\ 1\end{array}\right]\right) = \left[\begin{array}{c}1 \\ -1\end{array}\right] \)[/tex].

### Part (d)
For [tex]\( T\left(\left[\begin{array}{c}-1 \\ 0\end{array}\right]\right) \)[/tex]:
[tex]\[ x_1 = -1, \quad x_2 = 0 \][/tex]

Using the transformation formula:
[tex]\[ T\left(\left[\begin{array}{c}-1 \\ 0\end{array}\right]\right) = \left[\begin{array}{c}2(-1) - 3(0) \\ -(-1) + 0\end{array}\right] = \left[\begin{array}{c}-2 - 0 \\ 1 + 0\end{array}\right] = \left[\begin{array}{c}-2 \\ 1\end{array}\right] \][/tex]

So, [tex]\( T\left(\left[\begin{array}{c}-1 \\ 0\end{array}\right]\right) = \left[\begin{array}{c}-2 \\ 1\end{array}\right] \)[/tex].

### Summary
The results for each part are:

(a) [tex]\( T\left(\left[\begin{array}{c}0 \\ 0\end{array}\right]\right) = \left[\begin{array}{c}0 \\ 0\end{array}\right] \)[/tex]
(b) [tex]\( T\left(\left[\begin{array}{c}1 \\ 1\end{array}\right]\right) = \left[\begin{array}{c}-1 \\ 0\end{array}\right] \)[/tex]
(c) [tex]\( T\left(\left[\begin{array}{c}2 \\ 1\end{array}\right]\right) = \left[\begin{array}{c}1 \\ -1\end{array}\right] \)[/tex]
(d) [tex]\( T\left(\left[\begin{array}{c}-1 \\ 0\end{array}\right]\right) = \left[\begin{array}{c}-2 \\ 1\end{array}\right] \)[/tex]
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