Finding a $3 \times 3$ Matrix that maps points in $\mathbb{R}^3$ onto the a given line Give a $3 \times 3$ matrix that maps all points in $\mathbb{R}^3$ onto the line $[x,y,z] = t[a,b,c]$ and does not move the points that are on that line. Prove your matrix has these properties.
Can someone verify if I am doing this correctly?
I first find a matrix that takes the standard basis to a basis that has $\begin{bmatrix}a \\ b \\ c \end{bmatrix}$ in it:
$\begin{bmatrix} a & 0 & 0 \\ b & 1 & 0 \\ c & 0 & 1 \end{bmatrix} = A$
now I choose a matrix that projects $\mathbb{R^3}$ onto the given line:
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = B$
Now I need to invert $A$ to go back to the standard basis and so $A^{-1} B A$ will project all of $\mathbb{R}^3$ onto the given line.
Multiplying that out: 
$\begin{bmatrix}1/a & 0 & 0\\-b/a & 1 & 0 \\-c/a & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} a & 0 & 0 \\ b & 1 & 0 \\ c & 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 1 & 0 & 0\\-b & 0 & 0 \\-c & 0 & 0 \end{bmatrix}$
To show that this maps points in $\mathbb{R}^3$ to $[x, y, z] = t[a, b, c]$
$\begin{bmatrix} 1 & 0 & 0\\-b & 0 & 0 \\-c & 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y\\z\end{bmatrix} = \begin{bmatrix}x \\ -bx \\-cx \end{bmatrix} = x\begin{bmatrix}1\\-b\\-c\end{bmatrix}$ 
I apologize if some of my explanations don't make sense. I am trying to solve this the way my tutor showed me but I may have misunderstood some of his explanations.

You didn't invert correctly.  You need a $1$ in the $(2,2)$ and $(3,3)$ positions.  Also, if you multiply reverse the order (multiply by $A$ on left and $A^{-1}$ on right) you get $(x,bx/a,cx/a)$.  This defines the same line $(ax,bx,cx)$.
