(i) if `A+B=[{:(3,4),(-1,0):}]and A-B =[{:(1,2),(5,6):}],` then find the matrix A and B.
(ii) `if X+Y=[{:(2,1),(1,2):}]and 2X-Y=[{:(1,2),(2,1):}]` then find X and Y .

Explanation

To solve the given problems step by step, we will break them down into manageable parts.

### Part (i)

Given:
1. \( A + B = \begin{pmatrix} 3 & 4 \\ -1 & 0 \end{pmatrix} \) (Equation 1)
2. \( A - B = \begin{pmatrix} 1 & 2 \\ 5 & 6 \end{pmatrix} \) (Equation 2)

**Step 1: Add Equation 1 and Equation 2.**

\[ (A + B) + (A - B) = \begin{pmatrix} 3 & 4 \\ -1 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 2 \\ 5 & 6 \end{pmatrix} \]

This simplifies to:

\[ 2A = \begin{pmatrix} 3 + 1 & 4 + 2 \\ -1 + 5 & 0 + 6 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 4 & 6 \end{pmatrix} \]

**Step 2: Solve for A.**

\[ A = \frac{1}{2} \begin{pmatrix} 4 & 6 \\ 4 & 6 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix} \]

**Step 3: Substitute A back into Equation 1 to find B.**

Using Equation 1:

\[ A + B = \begin{pmatrix} 3 & 4 \\ -1 & 0 \end{pmatrix} \]

Substituting \( A \):

\[ \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix} + B = \begin{pmatrix} 3 & 4 \\ -1 & 0 \end{pmatrix} \]

**Step 4: Solve for B.**

\[ B = \begin{pmatrix} 3 & 4 \\ -1 & 0 \end{pmatrix} - \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -3 & -3 \end{pmatrix} \]

Thus, the matrices are:
\[ A = \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 1 \\ -3 & -3 \end{pmatrix} \]

### Part (ii)

Given:
1. \( X + Y = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \) (Equation 1)
2. \( 2X - Y = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \) (Equation 2)

**Step 1: Add Equation 1 and Equation 2.**

\[ (X + Y) + (2X - Y) = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} + \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \]

This simplifies to:

\[ 3X = \begin{pmatrix} 2 + 1 & 1 + 2 \\ 1 + 2 & 2 + 1 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix} \]

**Step 2: Solve for X.**

\[ X = \frac{1}{3} \begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]

**Step 3: Substitute X back into Equation 1 to find Y.**

Using Equation 1:

\[ X + Y = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \]

Substituting \( X \):

\[ \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} + Y = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \]

**Step 4: Solve for Y.**

\[ Y = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Thus, the matrices are:
\[ X = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \quad Y = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]