Given vectors p = 2 −2 and q = 3 4 , solve for vector r in the equation (1)/(2)p − q + r = 0 0 .

The given equation is $\frac{1}{2}\mathbf{p} - \mathbf{q} + \mathbf{r} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.
Substitute $\mathbf{p} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$:

$$\frac{1}{2}\begin{pmatrix} 2 \\ -2 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \end{pmatrix} + \mathbf{r} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

$$\begin{pmatrix} \frac{1}{2} \times 2 \\ \frac{1}{2} \times (-2) \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \end{pmatrix} + \mathbf{r} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ $$\begin{pmatrix} 1 \\ -1 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \end{pmatrix} + \mathbf{r} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

$$\begin{pmatrix} 1 - 3 \\ -1 - 4 \end{pmatrix} + \mathbf{r} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ $$\begin{pmatrix} -2 \\ -5 \end{pmatrix} + \mathbf{r} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

Subtract $\begin{pmatrix} -2 \\ -5 \end{pmatrix}$ from both sides:

$$\mathbf{r} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} - \begin{pmatrix} -2 \\ -5 \end{pmatrix}$$ $$\mathbf{r} = \begin{pmatrix} 0 - (-2) \\ 0 - (-5) \end{pmatrix}$$ $$\mathbf{r} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$$

The value of $\mathbf{r}$ is: