Matrices

The condition is that the points all lie on some line $ax+by+c=0,$ where $a,b$ are not both zero. In this case, $\begin{bmatrix} x_1&y_1&1 \\ x_2&y_2&1 \\ x_3&y_3&1 \end{bmatrix} \begin{bmatrix} a\\b\\c \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix},$ so the matrix has a nontrivial nullspace. By rank-nullity , the matrix has rank at most $2.$

On the other hand, the matrix $\begin{bmatrix} 1&0&1 \\ 2&0&1 \\ 3&0&1 \end{bmatrix}$ has rank $2,$ so the answer is $\fbox{2}.$

If the above three points (x1, y1), (x2, y2), and (x3, y3) are collinear, then they each lie on some line y = mx + b. We can then rewrite these points as:

(x1, m x1 + b); (x2, m x2 + b); (x3, m*x3 + b) (i).

Let us now express the columns of the above matrix as the vectors:

v1 = [x1 x2 x3]^T; v2 = [m x1 + b m x2 + b m*x3 + b]^T; v3 = [1 1 1]^T (ii)

It becomes apparent that v2 = m v1 + b v3 => v2 is linearly dependent on v1 and v3. Hence the maximum rank of the matrix is 2.