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.

Since it maps all the vectors into directions of single vector, hence it must be rank 1; in particular following solution will work 
$\left[\begin{array}{ccc}
\gamma a & \alpha a & \beta a\\
\gamma b & \alpha b & \beta b\\
\gamma c & \alpha c & \beta c
\end{array}\right]\left[\begin{array}{c}
x\\
y\\
z
\end{array}\right]=(\gamma x+\alpha y+\beta z)\left[\begin{array}{c}
a\\
b\\
c
\end{array}\right]
 $
Now you need to choose $\gamma\ ,\alpha \ \& \beta $ such that 
$\left[\begin{array}{ccc}
\gamma a & \alpha a & \beta a\\
\gamma b & \alpha b & \beta b\\
\gamma c & \alpha c & \beta c
\end{array}\right]\left[\begin{array}{c}
a\\
b\\
c
\end{array}\right]=\left[\begin{array}{c}
a\\
b\\
c
\end{array}\right]
 $
Which should be straight forward to find.
