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Image of plane (linear map)

  • Thread starter TsAmE
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  • #1
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Homework Statement



Let T: [itex]\mathbb{R}^3 \to \mathbb{R}^3[/itex] be the linear map represented by the matrix [itex]\begin{pmatrix} 4 & -1 & 0 \\ 6& 3 & -2\\ 12& 6 & -4\end{pmatrix}[/itex]

What is the image under T of the plane [itex]2x - 5y + 2z = -5[/itex]?

Homework Equations



None

The Attempt at a Solution



I made [itex]z = \mu[/itex] and [itex]y = \lambda[/itex] (since z and y are both excess variables) and so got the parametric equations of my plane to be:

[itex]x = \frac{-5}{2} + \frac{5}{2}\lambda - \mu[/itex]
[itex]y = \lambda[/itex]
[itex]z = \mu[/itex]

where [itex]\mu, \lambda\varepsilon \mathbb{R}[/itex] but the correct answer was:

[itex]\begin{pmatrix}x\\y \\z\end{pmatrix} = \begin{pmatrix}1\\3 \\4\end{pmatrix} + \lambda \begin{pmatrix}5\\2 \\0\end{pmatrix} + \mu\begin{pmatrix}1\\0 \\-1\end{pmatrix}[/itex]

Im not sure what I did wrong

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
33,070
4,771

Homework Statement



Let T: [itex]\mathbb{R}^3 \to \mathbb{R}^3[/itex] be the linear map represented by the matrix [itex]\begin{pmatrix} 4 & -1 & 0 \\ 6& 3 & -2\\ 12& 6 & -4\end{pmatrix}[/itex]

What is the image under T of the plane [itex]2x - 5y + 2z = -5[/itex]?

Homework Equations



None

The Attempt at a Solution



I made [itex]z = \mu[/itex] and [itex]y = \lambda[/itex] (since z and y are both excess variables) and so got the parametric equations of my plane to be:

[itex]x = \frac{-5}{2} + \frac{5}{2}\lambda - \mu[/itex]
[itex]y = \lambda[/itex]
[itex]z = \mu[/itex]

where [itex]\mu, \lambda\varepsilon \mathbb{R}[/itex] but the correct answer was:

[itex]\begin{pmatrix}x\\y \\z\end{pmatrix} = \begin{pmatrix}1\\3 \\4\end{pmatrix} + \lambda \begin{pmatrix}5\\2 \\0\end{pmatrix} + \mu\begin{pmatrix}1\\0 \\-1\end{pmatrix}[/itex]

Im not sure what I did wrong
The parametric equation of the plane and the answer to the problem are two different things. The image of your plane is the product of the given matrix and the vector with the parametric representation of the plane.
 
  • #3
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The parametric equation of the plane and the answer to the problem are two different things. The image of your plane is the product of the given matrix and the vector with the parametric representation of the plane.
I see, but where did I go wrong with the vector with the parametric representation of the plane, cause I cant see what I did wrong.
 
  • #4
33,070
4,771
I don't see anything wrong with that. I got about the same as what you got. It's possible that you and the book just have different representations of the same thing. You can check yours by substituting a few values of lambda and mu to see if you actually get points that are on the plane.
 
  • #5
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I tried checking the x values so I subbed x = 1 into [itex]x = \frac{-5}{2} + \frac{5}{2}\lambda - \mu[/itex] and got x = -1 and for [itex]\begin{pmatrix}x\\y \\z\end{pmatrix} = \begin{pmatrix}1\\3 \\4\end{pmatrix} + \lambda \begin{pmatrix}5\\2\\0\end{pmatrix} + \mu\begin{pmatrix}1\\0 \\-1\end{pmatrix}[/itex] I got x = 5. But these two values dont equal each other:confused:
 
  • #6
33,070
4,771
You have this parametrization of the plane:
[tex]\begin{pmatrix}x\\y \\z\end{pmatrix} = \begin{pmatrix}-5/2\\0 \\0\end{pmatrix} + \lambda \begin{pmatrix}5\\2\\0\end{pmatrix} + \mu\begin{pmatrix}1\\0 \\-1\end{pmatrix}[/tex]

Your book has this parametrization:
[tex]\begin{pmatrix}x\\y \\z\end{pmatrix} = \begin{pmatrix}1\\3 \\4\end{pmatrix} + \lambda \begin{pmatrix}5\\2\\0\end{pmatrix} + \mu\begin{pmatrix}1\\0 \\-1\end{pmatrix}[/tex]

Both are correct. You can check by substituting pairs of values for lambda and mu. For a given pair of values for lambda and mu, you'll get different points (x, y, z), but both points satisfy the equation of the plane, 2x - 5y + 2z = -5.
 
  • #7
132
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You have this parametrization of the plane:
[tex]\begin{pmatrix}x\\y \\z\end{pmatrix} = \begin{pmatrix}-5/2\\0 \\0\end{pmatrix} + \lambda \begin{pmatrix}5\\2\\0\end{pmatrix} + \mu\begin{pmatrix}1\\0 \\-1\end{pmatrix}[/tex]
Dont you mean I have this parametrization of the plane:
[tex]\begin{pmatrix}x\\y \\z\end{pmatrix} = \begin{pmatrix}\frac{-5}{2}\\0 \\0\end{pmatrix} + \lambda \begin{pmatrix}\frac{5}{2}\\1\\0\end{pmatrix} + \mu\begin{pmatrix}-1\\0 \\1\end{pmatrix}[/tex]

corresponding to:

[itex]x = \frac{-5}{2} + \frac{5}{2}\lambda - \mu[/itex]
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
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Yes, but of course those are parameterizations of the same plane with different values of [itex]\mu[/itex] giving the same point.
 
  • #9
33,070
4,771
Dont you mean I have this parametrization of the plane:
[tex]\begin{pmatrix}x\\y \\z\end{pmatrix} = \begin{pmatrix}\frac{-5}{2}\\0 \\0\end{pmatrix} + \lambda \begin{pmatrix}\frac{5}{2}\\1\\0\end{pmatrix} + \mu\begin{pmatrix}-1\\0 \\1\end{pmatrix}[/tex]

corresponding to:

[itex]x = \frac{-5}{2} + \frac{5}{2}\lambda - \mu[/itex]
What I wrote wasn't exactly the same as what you had, but it gives exactly the same plane. I debated saying that when I posted my reply, but didn't.

Any point in the plane whose equation is 2x -5y + 2z = -5 can be reached by a vector sum. That's what the parametric equations are saying. Since the plane doesn't go through the origin, the first vector goes from the origin to a point on the plane: the vector <-5/2, 0, 0> does this. From this point on the plane, any other point on the plane is a linear combination of two other vectors that I wrote as <5, 2, 0> and <1, 0, -1>.

So the vector from (0, 0, 0) to a point (x, y, z) on the plane is the vector sum <-5/2, 0, 0> + a<5, 2, 0> + b<1, 0, -1>.

The two vectors <5, 2, 0> and <1, 0, -1> are both perpendicular to the plane's normal, which happens to be <2, -5, 2>. You can verify this by checking that the dot product of each of these vectors with <2, -5, 2> is zero.

Any multiples of the vectors <5, 2, 0> and <1, 0, -1> will also generate all points in the plane. The vectors you had -- <5/2, 1, 0> and <-1, 0, 1> -- are just multiples of the vectors I chose.
 

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