Better way to come up with a singular 3x3 matrix than trial and error

Question

I am given a task to give an example of a singular matrix, for a 2x2 is very simple to try out some combinations of numbers to eventually get a determinant=0. But I find it a time consuming to go by trial and error for a 3x3 matrix.

Criteria:

1. No zero entries
2. No rows are multiple of other rows

What I know:

1. Formula to find determinant of 3x3 matrix.
2. Zero determinant means Singular matrix.

Answer 1

How about the zero matrix?

How about the matrix $$\begin{pmatrix} 1&0&0\\0&1&0\\0&0&0\end{pmatrix}?$$

Or any matrix with a row of zeros.

More generally, a matrix will be singular if its rows or columns are linearly dependent. So choose any two rows you want, and then choose some multiples of each and add them together to get the third row. This will give a less obvious singular matrix.

For example, suppose I choose $(1, 2, 3)$ and $(1, -2, 2)$ as my first two rows, and then subtract them for the third. This gives

$$\begin{pmatrix}1&2&3 \\ 1&-2&3 \\ 0 & 4 & 1 \end{pmatrix},$$

which is perhaps less obviously singular.

Answer 2

Here is a method that, in principle, can give you any rank-$2$ invertible matrix. Select any two matrices for which each row is a multiple of the first (i.e. rank-$1$ matrices), and add them together.

For example: consider the matrices $$ \pmatrix{1 & 2 & 3\\2&4&6\\4&8&12}, \quad \pmatrix{1&1&1\\2&2&2\\1&1&1}. $$ Adding these together yields $$ \pmatrix{2&3&4\\4&6&8\\5&9&13}. $$ This matrix has rank $2$.

The only thing that can go wrong with this method is that we "accidentally" end up with a matrix for which condition 2 fails to hold. In order to make sure that this does not happen, it suffices to make sure that the rows of the first matrix are not multiples of the rows of the second and the columns of the first matrix are not multiples of the columns of the second.

Answer 3

The simplest way to devise matrices with certain properties is in general by $PDP^{-1}$ or $PJP^{-1}$ where $D$ is diagonal or $J$ the Jordan form, and $P$ any invertible matrix.

In your case you want $M$ to be singular, so take a $D$ or $J$ with a $0$ or more in the diagonal (i.e. a zero eigenvalue).

Then multiply by your prefered $P$ (hint, the simplest one is $P=I$).

It might be a bit uneasy to come with some $P,P^{-1}$ couples with integer values, that would give you an example $M$ a bit more "interesting" than just a diagonal one. but for $3\times3$ or $2\times2$ it is possible to devise some manually.

But when we just want a singular matrix, notice there is an easier way, by ensuring the matrix is not full rank. An obvious way to reduce the rank is to choose two columns at random and then calculate the third one to be a linear combination of the first two.

E.g. $\begin{pmatrix}2 & -5 & .\\3 & 1 & .\\7 & 4 & .\end{pmatrix}\overset{C1-4C_2}{\longrightarrow}\begin{pmatrix}2 & -5 & 22\\3 & 1 & -1\\7 & 4 & -9\end{pmatrix}$