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 $$A \hspace{1 mm} \vec{x} = \left[ \begin{array}{c}5 \\ 3 \end{array}\right]$$. You can do that by the following method.

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

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:

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

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.