Let
\[ A = \left[ \begin{array}{c c c c} 3 & - 6 & 2 & - 1 \\ - 2 & 4 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 1 & - 2 & 1 & 0 \end{array} \right]. \]
For which \((y_{1}, y_{2}, y_{3}, y_{4})\) does the system of equations AX = Y have a solution?
Solution
The system \(AX=Y\) is solvable exactly for those \(Y=(y_1,y_2,y_3,y_4)^t\) satisfying \[ 2y_1+3y_2-7y_3=0, \qquad 3y_1+y_2-7y_4=0. \]
1. Every vector in the column space of \(A\) satisfies the two displayedrelations.
The vectors \[ u=(-2,-3,7,0)^t, \qquad v=(-3,-1,0,7)^t \] satisfy \[ u^tA=0, \qquad v^tA=0. \] Hence if \(Y=AX\), then \[ u^tY=u^tAX=0, \qquad v^tY=v^tAX=0, \] which are exactly the stated equations.
2. The two relations are also sufficient.
Row reduction shows \(\operatorname{rank}A=2\). Therefore the left nullspace \[ \ker(A^t) \] has dimension \(4-2=2\). The two vectors \(u,v\) in 1 are linearly independent, so they form a basis of this left nullspace.
The column space is the annihilator of the left nullspace. Consequently a vector \(Y\) lies in the column space of \(A\) exactly when \(u^tY=v^tY=0\), namely exactly when the two displayed relations hold.