i) If \( x+y=\begin{array}{ll}5 & 2 \\ 0 & 9\end{array} \) and \( x-y=\begin{array}{cc}3 & 6 \\ 0 & -1\end{array} \) then calculate \( x \) and \( y \).

To solve for \( x \) and \( y \) given the equations: \[ x + y = \begin{pmatrix} 5 & 2 \\ 0 & 9 \end{pmatrix} \] \[ x - y = \begin{pmatrix} 3 & 6 \\ 0 & -1 \end{pmatrix} \] we can use the method of elimination. We can add the two equations to eliminate \( y \): \[ (x + y) + (x - y) = \begin{pmatrix} 5 & 2 \\ 0 & 9 \end{pmatrix} + \begin{pmatrix} 3 & 6 \\ 0 & -1 \end{pmatrix} \] Calculating the right-hand side: \[ \begin{pmatrix} 5 + 3 & 2 + 6 \\ 0 + 0 & 9 - 1 \end{pmatrix} = \begin{pmatrix} 8 & 8 \\ 0 & 8 \end{pmatrix} \] Thus, we have: \[ 2x = \begin{pmatrix} 8 & 8 \\ 0 & 8 \end{pmatrix} \] Now, we can solve for \( x \) by dividing both sides by 2: \[ x = \begin{pmatrix} 4 & 4 \\ 0 & 4 \end{pmatrix} \] Next, we can substitute \( x \) back into one of the original equations to find \( y \). We will use the equation \( x + y = \begin{pmatrix} 5 & 2 \\ 0 & 9 \end{pmatrix} \): \[ \begin{pmatrix} 4 & 4 \\ 0 & 4 \end{pmatrix} + y = \begin{pmatrix} 5 & 2 \\ 0 & 9 \end{pmatrix} \] To isolate \( y \), we subtract \( x \) from both sides: \[ y = \begin{pmatrix} 5 & 2 \\ 0 & 9 \end{pmatrix} - \begin{pmatrix} 4 & 4 \\ 0 & 4 \end{pmatrix} \] Calculating the right-hand side: \[ y = \begin{pmatrix} 5 - 4 & 2 - 4 \\ 0 - 0 & 9 - 4 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 0 & 5 \end{pmatrix} \] Thus, we have: \[ x = \begin{pmatrix} 4 & 4 \\ 0 & 4 \end{pmatrix}, \quad y = \begin{pmatrix} 1 & -2 \\ 0 & 5 \end{pmatrix} \] In conclusion, the values of \( x \) and \( y \) are: \[ \boxed{x = \begin{pmatrix} 4 & 4 \\ 0 & 4 \end{pmatrix}, \quad y = \begin{pmatrix} 1 & -2 \\ 0 & 5 \end{pmatrix}} \]