Let \(t =
\begin{bmatrix}
2 \\ 0 \\ -1
\end{bmatrix}+
\begin{bmatrix}
0 \\ 1 \\ 4
\end{bmatrix}-
2~\begin{bmatrix}
1 \\ 0 \\ -1
\end{bmatrix}.\)
Is \(t_3 \leq 4\)?
The answer is “false”.
To see this, we first simplify \(t\):
\[ t =
\begin{bmatrix}
2 \\ 0 \\ -1
\end{bmatrix}+
\begin{bmatrix}
0 \\ 1 \\ 4
\end{bmatrix}-
2~\begin{bmatrix}
1 \\ 0 \\ -1
\end{bmatrix}
= \begin{bmatrix} 2 \\ 1 \\3 \end{bmatrix}
-2~\begin{bmatrix}
1 \\ 0 \\ -1
\end{bmatrix}
= \begin{bmatrix} 2 \\ 1 \\3 \end{bmatrix}
- \begin{bmatrix} 2 \\ 0 \\-2 \end{bmatrix}
= \begin{bmatrix} 0 \\ 1 \\ 5 \end{bmatrix}.
\]
Since \(t_3\) denotes the third entry of
\(t\) which has value \(5\), we see that
\(t_3\gt 4\).