Find the null space of the matrix
\[ \Mtrx{A} = \left[ \begin{array}{rrr} 3 & 1 & 7 \\ -2 & 1 & 4 \\ \end{array} \right] \]We are looking for the solutions of the vector equation
\[ \left[ \begin{array}{rrr} 3 & 1 & 7 \\ -2 & 1 & 4 \\ \end{array} \right]\cdot \left[ \begin{array}{r} x \\ y \\ z \end{array} \right]\ =\ \left[ \begin{array}{cc} 0 \\ 0 \end{array} \right] \]We see that the solutions of this vector equation correspond to the simultaneous solutions of the system of two homogeneous linear equations
\[ \begin{array}{rcrcrcl} 3x & + & y & + & 7z & = & 0 \\ -2x& + & y & + & 4z & = & 0 \end{array} \]The augmented coefficient matrix of this system of linear equations is
\[ \begin{array}{rrr|r} 3 & 1 & 7 & 0 \\ -2& 1 & 4 & 0 \end{array} \]The RREF of this matrix is
\[ \begin{array}{rrr|r} 1 & 0 & \tfrac{3}{5} & 0 \\ 0 & 1 & \tfrac{26}{5}& 0 \end{array} \]This means that \((x,y,z)\) solves the system exactly when, for arbitrary \(z\) in \(\RNr\),
| \(x\) | \(=\) | \(-\tfrac{3}{5}\) |
| \(y\) | \(=\) | \(-\tfrac{26}{5}\) |
In other words,
| \((x,y,z)\) | \(=\) | \(z\cdot (-\tfrac{3}{5},-\tfrac{26}{5},1)\) |
Therefore, the null space of \(\Mtrx{A}\) consists of all vectors of the form \(z\cdot (-\tfrac{3}{5},-\tfrac{26}{5},1)\), with \(z\) in \(\RNr\) arbitrary. This is the line in \(\RNrSpc{3}\) through the origin in the direction of the vector \((-3,-26,5)\).