Image, Range, and Matrix of a Linear Transformation

[SiddharthThakur]

Question
Consider the linear transformation T(x1,x2,x3)= (2*x1 -2*x2- 4*x3 ,x1+2*x2+x3)
(a) Find the image of (3, -2, 2) under T.
(b) Does the vector (5, 3) belong to the range of T?
(c) Determine the matrix of the transformation.
(d) Is the transformation T onto? Justify your answer
(e) Is the transformation one-to one? Justify your answerPlease see the attachement.it will be more clear to you.

 

[Jameson]

Thread closed until issue of helping with a graded assignment addressed. Please send me a PM or reply to the other thread SiddharthThakur. This is not to embarrass you, it's just MHB policy. Let's talk about this so we can help you as much as possible.

 

[Jameson]

Question
Consider the linear transformation T(x1,x2,x3)= (2*x1 -2*x2- 4*x3 ,x1+2*x2+x3)
(a) Find the image of (3, -2, 2) under T.
(b) Does the vector (5, 3) belong to the range of T?
(c) Determine the matrix of the transformation.
(d) Is the transformation T onto? Justify your answer
(e) Is the transformation one-to one? Justify your answerPlease see the attachement.it will be more clear to you.

(a) The image of $(3, -2, 2)$ here is simply plugging in $x_1=3$, $x_2=-2$ and $x_3=2$ into the expression you are given.

(b) You want to solve \(\displaystyle A \hspace{1 mm} \vec{x} = \left[ \begin{array}{c}5 \\ 3 \end{array}\right]\). You can do that by the following method.

\(\displaystyle \left[ \begin{array}{ccc} 2 & -2 & 4\\ 1 & 2 & 3 \end{array}\right] \left[ \begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[ \begin{array}{c}5 \\ 3 \end{array}\right]\)

Change that to an augmented matrix and solve.

 

[Jameson]

Question
Consider the linear transformation T(x1,x2,x3)= (2*x1 -2*x2- 4*x3 ,x1+2*x2+x3)
(a) Find the image of (3, -2, 2) under T.
(b) Does the vector (5, 3) belong to the range of T?
(c) Determine the matrix of the transformation.
(d) Is the transformation T onto? Justify your answer
(e) Is the transformation one-to one? Justify your answer

Since the deadline for the assignment in question has passed, I will make some more comments on solving the problem so others can use the information in the future.

(а) $Т(3,-2,2)=(2(3)-2(-2)-4(2), 3+2(-2)+2)=(2,1)$

(b) To answer this we can start by answering part (c). The transformation matrix of $T$ is $\left[ \begin{array}{ccc} 2 & -2 & 4\\ 1 & 2 & 3 \end{array}\right]$ To find if the vector $[5,3]$ is in the range of $T$ we can solve the following matrix equation:

\(\displaystyle \left[ \begin{array}{ccc} 2 & -2 & -4\\ 1 & 2 & 3 \end{array}\right] \left[ \begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[ \begin{array}{c}5 \\ 3 \end{array}\right]\)

I will leave the computation to the reader, but if you row-reduce the augmented matrix from the above equation then you'll find that the the system is consistent, thus there exists a solution and the vector $[5,3]$ is in the range of $T$.

(c) Answered above.

(d) The transformation matrix is composed of 3 column vectors in $\mathbb{R}^2$ and after noticing that there are two of them which are linearly independent it follows that $T$ is in fact onto $\mathbb{R}^2$. This can also be shown by using the fact that there is a pivot position in every row.

(e) We have three column vectors in $\mathbb{R}^2$, so at least one of them must be a linear combination of the other two so $T$ isn't one-to-one. We can see this fact when row-reducing the augmented matrix for part (b). The solution contains a free variable which means that there is more than one solution, which doesn't fit the definition of being one-to-one. Two ways to check this in general is the columns of the transformation matrix must be linearly independent and when row-reducing the augmented matrix to solve for any $\vec{b}$, there can't be any free variables.

 

[blue_raver22]

(a) The image of (3, -2, 2) under T is (10, 5).
To find the image, we simply plug in the values of x1, x2, and x3 into the transformation equation:
T(3,-2,2) = (2*3 -2*(-2)- 4*2 ,3+2*(-2)+2) = (10, 5)

(b) To determine if (5, 3) belongs to the range of T, we need to find values of x1, x2, and x3 that would result in (5, 3) when plugged into the transformation equation:
T(x1,x2,x3) = (5, 3)
This would result in the following system of equations:
2*x1 - 2*x2 - 4*x3 = 5
x1 + 2*x2 + x3 = 3
Solving this system, we get x1 = 3, x2 = -2, and x3 = 1.
Therefore, (5, 3) belongs to the range of T.

(c) The matrix of the transformation can be found by writing the transformation equation in the form of a matrix:
T(x1,x2,x3) = (2 -2 -4 ,1 2 1) * (x1,x2,x3)
Therefore, the matrix of the transformation is:
[2 -2 -4]
[1 2 1]

(d) To determine if the transformation T is onto, we need to check if every element in the range of T has a corresponding element in the domain. In other words, we need to check if for every (x1, x2, x3) in the range of T, there exists a (x1', x2', x3') in the domain such that T(x1', x2', x3') = (x1, x2, x3).
In this case, since we have already shown that (5, 3) belongs to the range of T, we can choose (x1', x2', x3') = (3, -2, 2) and we get T(3, -2, 2) = (5, 3).
Therefore, T is onto.

(e) To determine if the transformation T is one-to-one, we need to check