Let \(A = \begin{bmatrix} 1 & -1 & 2\\ 1 & 1 & 1\end{bmatrix}\).
What must be the value of \(a\) so that
\(\begin{bmatrix} a \\ 1 \\ 2\end{bmatrix}\) is in \(N(A)\)?
The answer is “\(-3\)”.
Recall that \(\begin{bmatrix} a \\ 1 \\ 2\end{bmatrix}\) in \(N(A)\) means
\[\begin{bmatrix} 1 & -1 & 2\\ 1 & 1 & 1\end{bmatrix}
\begin{bmatrix} a \\ 1 \\ 2\end{bmatrix}=\begin{bmatrix} 0 \\ 0\end{bmatrix}.\]
Multiplying out the left-hand side gives
\[\begin{bmatrix} a+3\\ a+ 3\end{bmatrix}
=\begin{bmatrix} 0 \\ 0\end{bmatrix}.\]
Hence, \(a=-3\).