Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having difficulty understanding what the following question is asking and was hoping someone could explain it to me.

$T = \begin{bmatrix}1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{bmatrix}$

Given a point on the line x = y = z, prove that the set of points in $\mathbb{R}^3$ that $T$ maps to that point make up the plane through that point that is perpendicular to the line $x = y = z$.

According to my instructor, I am suppose to solve for $Tx = \begin{bmatrix}a\\a\\a\end{bmatrix}$. Doing so I get

$\begin{bmatrix} 1/3 && 1/3 && 1/3 && a \\1/3 && 1/3 && 1/3 && a \\1/3 && 1/3 && 1/3 && a \end{bmatrix}$

Reducing:

$\begin{bmatrix} 1 && 1 && 1 && 3a \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \end{bmatrix}$

$ x_1 + x_2 + x_3 = 3a $

$ x_1 = 3a - x_2 - x_3$

So

$x = \begin{bmatrix} x_1\\ x_2\\x_3 \end{bmatrix} = x_1\begin{bmatrix} 3a \\ 0 \\ 0 \end{bmatrix} + x_2\begin{bmatrix} -1\\1\\0\end{bmatrix} + x_3\begin{bmatrix} -1\\0\\1 \end{bmatrix} $

Since I don't understand what the question is really asking, I'm not sure what to do with this answer.

share|improve this question
    
I really don't understand what your $\;x\;$ supposedly is or what does that mean... –  DonAntonio Nov 19 '13 at 17:17
    
I added the steps I took to arrive at my answer. I may have done it incorrectly. –  user109609 Nov 19 '13 at 17:28
    
Ah I understand my mistake now. Clears a lot of things up for me. Thank you. –  user109609 Nov 19 '13 at 17:35
    
Now I understand: you reduced not the matrix but the augmented matrix of the linear system...fine! I missed that part, so what you do is correct, yet I'm not sure that helps you with your quest. Perhaps some of my answer's ideas can help you. –  DonAntonio Nov 19 '13 at 17:35
    
No, it was fine reducing the augmented matrix! I thought you reduced, just like that, the original matrix, which would be fine if you were trying to find what its kernel is, say. It is fine what you did! –  DonAntonio Nov 19 '13 at 17:36

1 Answer 1

up vote 1 down vote accepted

Well, check that the set of points you denoted as $\;x\;$ indeed passes through the point on $\;x=y=z\;$ and is perpendicular to that line:

$$\begin{pmatrix}a\\a\\a\end{pmatrix}=\begin{pmatrix}1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\frac13\begin{pmatrix}x_1+x_2+x_3\\x_1+x_2+x_3\\x_1+x_2+x_3\end{pmatrix}\iff$$

$$x_1+x_2+x_3=3a$$

Now, check the vector $\;u=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\;$ defined by the above relation actually passes throught the point $\;A:=\begin{pmatrix}a\\a\\a\end{pmatrix}\,$ (very easy) ,and that the vector $\;A-u\;$ is perpendicular to the given line there (very easy, too).

share|improve this answer
    
Can you explain where the $A-u$ vector comes from? I don't understand why showing $A-u$ to be perpendicular to the line $x=y=z$ is helpful. –  Xoque55 Nov 25 '13 at 4:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.