In a $2\times2$ matrix, it is quite easy to see if the vectors lie on a plane or not. By vector, I mean the columns of the matrix. I usually determine if the numbers are of a certain multiple. From there, I can judge that the columns of the matrix lie on a plane and that means its determinant is zero and therefore not invertible. So for instance:
$$
\left |\begin{bmatrix}
2 & 4\\
4 & 8
\end{bmatrix} \right | = 0
$$
Both vectors $\begin{bmatrix}
2\\ 4
\end{bmatrix}$ and $\begin{bmatrix}
4\\ 8
\end{bmatrix}$ lie on the same plane because the second vector is just 2 times of the first vector.
But in a $3\times3$ or $N\times N$ matrix, how can I get to know if a matrix is invertible or not by observing it without going through to calculate its determinant? For example in a matrix say $\begin{bmatrix}
1 & 4 & 7\\
2 & 5 & 8\\
3 & 6 & 9
\end{bmatrix}$, the numbers are not of a common multiple of each other. But still, the determinant of this matrix gives zero, which means the matrix is not invertible. But I wouldn't have known this without going through the trouble to calculate its determinant value. Since the numbers are not of a multiple, I am also not sure if the columns of the matrix lie on the same plane or not.
Thanks for any help.
Best Answer
In small cases, if you don't want to calculate the determinant, you can try to observe if one column can be written as a linear combination of the others, not just if it is a multiple of some other column. In your $3\times 3$ case, you have $C3=2C2-C1$, which shows the columns are a linearly dependent set, and so the matrix is not invertible. In the $2\times 2$ case, writing one column as a linear combination of the other is the same as seeing it is a multiple of the other, but this is not the strategy for more than two vectors.
Added: Geometrically, the other vector does not generally lie in the same direction as one of the other vectors, unless it is a multiple of just one vector, or if the other two vectors themselves are multiples of each other. If the two other vectors are a linearly independent pair, then in this case they span a two dimensional subspace isomorphic to the plane, which just looks like a plane passing through the origin at some angle. Since the third vector is in the span of the other two vectors, it will lie in that "plane" they span. As far as $n$-dimensional space, I have I hard time picturing it geometrically when $n\geq 4$. The general idea applies though that the vector will lie in the subspace spanned by the other vectors, but it doesn't necessarily have to be a multiple of any one vector.