2. Let $$ \left.A=\left[\begin{array}{cc}1 & -3 \ 3 & 5 \ -1 & 7\end{array} ight], \vec{u}=\left\lvert\, \begin{array}{c}2 \ -1\end{array} ight. ight], \vec{b}=\left|\begin{array}{c}3 \ 2 \ -5\end{array} ight|, \vec{c}=\left|\begin{array}{l}3 \ 2 \ 5\end{array} ight| $$ and defined transformation $$ T: R^{2} ightarrow R^{3} $$ by $$ T(\vec{x})=A(\vec{x}) $$. a. Find $$ T(\vec{u}) $$. b. Find an $$ \vec{x} $$ in $$ R^{2} $$ whose image under $$ T $$ is $$ \vec{b} $$ c. Is more than one $$ \vec{x} $$ whose image under $$ T $$ is $$ \vec{b} $$ ? d. Determine if $$ \vec{c} $$ is in the range of the transformation $$ T $$

Question

2. Let $$ \left.A=\left[\begin{array}{cc}1 & -3 \ 3 & 5 \ -1 & 7\end{array}ight], \vec{u}=\left\lvert\, \begin{array}{c}2 \ -1\end{array}ight.ight], \vec{b}=\left|\begin{array}{c}3 \ 2 \ -5\end{array}ight|, \vec{c}=\left|\begin{array}{l}3 \ 2 \ 5\end{array}ight| $$ and defined transformation $$ T: R^{2} ightarrow R^{3} $$ by $$ T(\vec{x})=A(\vec{x}) $$. a. Find $$ T(\vec{u}) $$. b. Find an $$ \vec{x} $$ in $$ R^{2} $$ whose image under $$ T $$ is $$ \vec{b} $$ c. Is more than one $$ \vec{x} $$ whose image under $$ T $$ is $$ \vec{b} $$ ? d. Determine if $$ \vec{c} $$ is in the range of the transformation $$ T $$

UpStudy AI · Accepted Answer

To solve the problem, we will go through each part step by step.

### Part a: Find $$ T(\vec{u}) $$

Given the transformation $$ T(\vec{x}) = A\vec{x} $$ and the matrix $$ A $$:

$$
A = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}
$$

and the vector $$ \vec{u} = \begin{bmatrix} 2 \ -1 \end{bmatrix} $$.

We compute $$ T(\vec{u}) $$:

$$
T(\vec{u}) = A\vec{u} = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix} \begin{bmatrix} 2 \ -1 \end{bmatrix}
$$

Calculating the matrix multiplication:

1. First row: $$ 1 \cdot 2 + (-3) \cdot (-1) = 2 + 3 = 5 $$
2. Second row: $$ 3 \cdot 2 + 5 \cdot (-1) = 6 - 5 = 1 $$
3. Third row: $$ -1 \cdot 2 + 7 \cdot (-1) = -2 - 7 = -9 $$

Thus,

$$
T(\vec{u}) = \begin{bmatrix} 5 \ 1 \ -9 \end{bmatrix}
$$

### Part b: Find an $$ \vec{x} $$ in $$ R^{2} $$ whose image under $$ T $$ is $$ \vec{b} $$

We need to find $$ \vec{x} $$ such that $$ T(\vec{x}) = \vec{b} $$, where $$ \vec{b} = \begin{bmatrix} 3 \ 2 \ -5 \end{bmatrix} $$.

This leads to the equation:

$$
A\vec{x} = \begin{bmatrix} 3 \ 2 \ -5 \end{bmatrix}
$$

Let $$ \vec{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} $$. Then we have the system of equations:

1. $$ x_1 - 3x_2 = 3 $$
2. $$ 3x_1 + 5x_2 = 2 $$
3. $$ -x_1 + 7x_2 = -5 $$

We can solve this system. From the first equation, we can express $$ x_1 $$:

$$
x_1 = 3 + 3x_2
$$

Substituting $$ x_1 $$ into the second equation:

$$
3(3 + 3x_2) + 5x_2 = 2
$$
$$
9 + 9x_2 + 5x_2 = 2
$$
$$
14x_2 = 2 - 9
$$
$$
14x_2 = -7 \implies x_2 = -\frac{1}{2}
$$

Now substituting $$ x_2 $$ back to find $$ x_1 $$:

$$
x_1 = 3 + 3\left(-\frac{1}{2}\right) = 3 - \frac{3}{2} = \frac{3}{2}
$$

Thus, one solution is:

$$
\vec{x} = \begin{bmatrix} \frac{3}{2} \ -\frac{1}{2} \end{bmatrix}
$$

### Part c: Is there more than one $$ \vec{x} $$ whose image under $$ T $$ is $$ \vec{b} $$?

To determine if there are multiple solutions, we can analyze the system of equations. The matrix $$ A $$ has 3 rows and 2 columns, which means it can have at most 2 linearly independent rows.

If the rank of the augmented matrix $$ [A | \vec{b}] $$ is equal to the rank of $$ A $$, then there are infinitely many solutions. If the rank of $$ [A | \vec{b}] $$ is greater than the rank of $$ A $$, then there is no solution.

We can row-reduce the matrix $$ A $$ and the augmented matrix $$ [A | \vec{b}] $$ to check this.

After performing row reduction, we find that the rank of $$ A $$ is 2, and the rank of $$ [A | \vec{b}] $$ is also 2. Therefore, there are infinitely many solutions, as the system is consistent and underdetermined.

### Part d: Determine if $$ \vec{c} $$ is in the range of the transformation $$ T $$

We need to check if $$ \vec{c} = \begin{bmatrix} 3 \ 2 \ 5 \end{bmatrix} $$ is in the range of $$ T $$. This means we need to solve:

$$
A\vec{x} = \begin{bmatrix} 3 \ 2 \ 5 \end{bmatrix}
$$

Setting up the system of equations:

1. $$ x_1 - 3x_2 = 3 $$
2. $$ 3x_1 + 5x_2 = 2 $$
3. $$ -x_1 + 7x_2 = 5 $$

We can solve this system similarly.

From the first equation, we have:

$$
x_1 = 3 + 3x_2
$$

Substituting into the second equation:

$$
3(3 + 3x_2) + 5x_2 = 2
$$
$$
9 + 9x_2 + 5x_2 = 2
$$
$$
14x_2 = 2 - 9
$$
$$
14x_2 = -7 \implies x_2 = -\frac{1}{2}
$$

Substituting back to find $$ x_1 $$:

$$
x_1 = 3 + 3\left(-\frac{1}{2}\right) = 3 - \frac{3}{2} = \frac{3}{2}
$$

Now substituting $$ x_1 $$ and $$ x_2 $$ into the third equation:

$$
-x_1 + 7x_2 = 5
$$
$$
-\frac{3}{2} + 7\left(-\frac{1}{2}\right) = 5
$$
$$
-\frac{3}{2} - \frac{7}{2} = 5
$$
$$
-\frac{10}{2} = 5 \implies -5 = 5
$$

This is a contradiction, which means that there is no solution for $$ A\vec{x} = \vec{c} $$. Therefore, $$ \vec{c} $$ is **not** in the range of the transformation $$ T $$.

### Summary of Results:
- a. $$ T(\vec{u}) = \begin{bmatrix} 5 \ 1 \ -9 \end{bmatrix} $$
- b. $$ \vec{x} = \begin{bmatrix} \frac{3}{2} \ -\frac{1}{2} \end{bmatrix} $$
- c. Yes, there are infinitely many solutions for $$ T(\vec{x}) = \vec{b} $$.
- d. $$ \vec{c} $$ is not in the range of the transformation $$ T $$.
a. $$ T(\vec{u}) = \begin{bmatrix} 5 \ 1 \ -9 \end{bmatrix} $$ b. $$ \vec{x} = \begin{bmatrix} \frac{3}{2} \ -\frac{1}{2} \end{bmatrix} $$ c. Yes, there are infinitely many solutions for $$ T(\vec{x}) = \vec{b} $$. d. $$ \vec{c} $$ is not in the range of the transformation $$ T $$.

Let and defined transformation

by .
a. Find .
b. Find an in whose image under is
c. Is more than one whose image under is ?
d. Determine if is in the range of the transformation

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