Let \(T:\mathbb{R}^3\rightarrow \mathbb{R}\) be given by
\(T\left(\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix}\right)=
2x_1-x_2+x_3.\)
The vector \(\begin{bmatrix} a\\1\\-3\end{bmatrix}\) is
in \(\ker(T)\). What must be the value of \(a\)?
The answer is 2.
Let \(v\) denote the vector \(\begin{bmatrix} a\\1\\-3\end{bmatrix}\).
We need \(T(v) = 0\). Hence, \(2a - 1 + (-3) = 0\), implying that
\( 2a = 4\). This gives \(a = 2\).