Finding the determinant of {{1,2,3}, {4,5,6}, {7,8,9}} by inspection. $$
\det
\begin{bmatrix}
   1 & 2 & 3 \\
   4 & 5 & 6 \\
   7 & 8 & 9
\end{bmatrix}
=
0
$$
Is it possible to find the determinant of the matrix
$$
\begin{bmatrix}
   1 & 2 & 3 \\
   4 & 5 & 6 \\
   7 & 8 & 9
\end{bmatrix}
$$
by inspection? By finding the determinant of the matrix by "inspection" I mean to find it without putting it into cofactor form or using diagonals.
I have already tried interpreting the determinant geometrically as the parallelepiped formed by the the three column vectors in the matrix, but fail to see that they would result in a parallelepiped with a volume of 0. I have not tried manipulating the determinant by adding multiples of one row/column to another row/column as I feel that there should be a simpler solution.
 A: In this particular case the middle column is equal to the average of the first and 3rd, which automatically means, that there is a column of zero after a couple of equivalent transformations, which makes it zero.
A: The second row is the element by element  average of the other rows, thus a linear combination.  The determinant of a square matrix where any row is a linear combination of the others is zero.
A: You can use the fact that the third line is twice the second one minus the first one.
A: Alternatively, notice that
$$ \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}\begin{pmatrix}1\\-2\\1\end{pmatrix} = 0\begin{pmatrix}1\\-2\\1\end{pmatrix}, $$
so that $0$ is an eigenvalue of your matrix.
A: Your matrix is $\begin{pmatrix} 1&1&1\\4&4&4\\7&7&7\end{pmatrix}+\begin{pmatrix} 0&1&2\\0&1&2\\0&1&2\end{pmatrix}\quad$ So if you set $u=\begin{pmatrix} 1\\4\\7\end{pmatrix}$ and $v=\begin{pmatrix} 1\\1\\1\end{pmatrix}$
You can see that $A(e_1)=u+0v,\ A(e_2)=u+1v,\ A(e_3)=u+2v$
Meaning $\operatorname{rank}(A)$ is at most $2$ since the image is in $\operatorname{span}(u,v)$ of dimension $2$.
Thus $\det(A)=0$
A: The middle column is the average of the first and the third which makes the determinant to become zero. 
