Let \(F\) be a field, \(A\) a \(2 \times 3\) matrix over \(F\), and \((c_1, c_2, c_3)\) the vector in \(F^3\) defined by \[ c_1 = \begin{vmatrix} A_{12} & A_{13} \\ A_{22} & A_{23} \end{vmatrix}, \quad c_2 = \begin{vmatrix} A_{13} & A_{11} \\ A_{23} & A_{21} \end{vmatrix}, \quad c_3 = \begin{vmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{vmatrix}. \] Show that
\(\operatorname{rank}(A) = 2\) if and only if \((c_1, c_2, c_3) \neq 0\);
if \(A\) has rank 2, then \((c_1, c_2, c_3)\) is a basis for the solution space of the system of equations \(AX = 0\).
Solution
Write the rows of \(A\) as \[ r_1=(a_{11},a_{12},a_{13}),\qquad r_2=(a_{21},a_{22},a_{23}), \] and set \(c=(c_1,c_2,c_3)^t\) as in the statement.
For (a), the coordinates \(c_1,c_2,c_3\) are exactly the three \(2\times2\) minors of \(A\), up to the displayed cyclic ordering. A \(2\times3\) matrix has rank \(2\) exactly when at least one of its \(2\times2\) minors is nonzero. Therefore \[ \operatorname{rank}A=2 \iff (c_1,c_2,c_3)\ne0. \]
For (b), direct expansion gives \[ r_1c =a_{11}(a_{12}a_{23}-a_{13}a_{22}) +a_{12}(a_{13}a_{21}-a_{11}a_{23}) +a_{13}(a_{11}a_{22}-a_{12}a_{21})=0, \] and similarly \(r_2c=0\). Thus \(Ac=0\).
If \(\operatorname{rank}A=2\), rank-nullity gives \[ \dim\ker A=3-2=1. \] Part (a) shows \(c\ne0\), so the one-dimensional kernel is exactly \[ \ker A=\operatorname{span}\{c\}. \] Hence \((c_1,c_2,c_3)\) is a basis of the solution space of \(AX=0\).