\( \left. \begin{array} { l } { x - 2 } \\ { x + 1 } \end{array} \right. \leq 0 \)

To solve the inequality \[ \begin{pmatrix} x - 2 \\ x + 1 \end{pmatrix} \leq 0, \] we need to analyze each component of the vector separately. This means we will solve the following two inequalities: 1. \( x - 2 \leq 0 \) 2. \( x + 1 \leq 0 \) ### Solving the first inequality: 1. \( x - 2 \leq 0 \) Adding 2 to both sides gives: \[ x \leq 2. \] ### Solving the second inequality: 2. \( x + 1 \leq 0 \) Subtracting 1 from both sides gives: \[ x \leq -1. \] ### Combining the results: Now we need to find the values of \( x \) that satisfy both inequalities. - From the first inequality, we have \( x \leq 2 \). - From the second inequality, we have \( x \leq -1 \). The more restrictive condition is \( x \leq -1 \). Therefore, the solution to the original vector inequality is: \[ x \leq -1. \] ### Conclusion: The solution set is: \[ (-\infty, -1]. \]