# Prove a matrix maps to a point that make up the plane perpendicular to a line.

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.

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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

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).
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