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.

  • $\begingroup$ You can perform a certain amount of row-reduction, which will show you if the matrix is invertible or not without having to compute the determinant. Elementary row (or elementary column) operations don't change the rank, so you can perform them until you can spot "by eye" if the matrix has full rank or not. $\endgroup$ Commented Jun 23, 2011 at 19:32
  • 2
    $\begingroup$ Note also that in the general $N\times N$ case, the matrix is invertible if and only if the rows (equivalently, columns) do not lie in the same $N-1$-dimensional hyperplane (rather than only in the same plane). For $2\times 2$ you are checking to see if they are in the same line; for $3\times 3$, in the same plane; for $4\times 4$, in the same $3$-dimensional hyperplane, etc. $\endgroup$ Commented Jun 23, 2011 at 19:35
  • $\begingroup$ There are some ad hoc "spot by eye" -tricks. Such as with your example matrix: the difference between the two first columns is $(3,3,3)^T$. You also see that the same vector is repeated as the difference between the second and the third columns. This is a tell-tale sign of a liner dependency relation. This is (one of the ways) how teachers quickly produce examples of singular matrices :-) $\endgroup$ Commented Jun 23, 2011 at 19:39
  • $\begingroup$ What should I spot for when doing the reduction? I would eliminate the entries one by one, targeting to turn it into an identity matrix, right? $\endgroup$
    – xenon
    Commented Jun 23, 2011 at 19:40
  • $\begingroup$ @xEnOn: Just start simplifying the matrix into a triangular matrix; if at some point your pivot is to the right of the main diagonal, you're done (not invertible). Eventually you'll come down either to considering a $2\times 2$ block in the bottom right, or spot sooner that the matrix is not invertible for some "obvious" reason. You don't need to go down all the way to the identity in any case. $\endgroup$ Commented Jun 23, 2011 at 19:53

1 Answer 1


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.

  • $\begingroup$ Thanks! Geometrically, how does linear dependent look like? Are they still on the same direction vector in a 3D space or N-dimension space? $\endgroup$
    – xenon
    Commented Jun 23, 2011 at 19:39
  • $\begingroup$ Thanks for the explanation! It is very helpful. :) $\endgroup$
    – xenon
    Commented Jun 23, 2011 at 20:20

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