Solve the following equations for X and Y :
`2X-Y=[(3,-3,0),(3,3,2)], 2Y+X=[(4,1,5),(-1,4,-4)]`

Explanation

To solve the equations for matrices \( X \) and \( Y \) given by:

1. \( 2X - Y = \begin{pmatrix} 3 & -3 & 0 \\ 3 & 3 & 2 \end{pmatrix} \)
2. \( 2Y + X = \begin{pmatrix} 4 & 1 & 5 \\ -1 & 4 & -4 \end{pmatrix} \)

we will follow these steps:

### Step 1: Rewrite the equations
We start with the two equations as given:

\[ 2X - Y = \begin{pmatrix} 3 & -3 & 0 \\ 3 & 3 & 2 \end{pmatrix} \tag{1} \]

\[ 2Y + X = \begin{pmatrix} 4 & 1 & 5 \\ -1 & 4 & -4 \end{pmatrix} \tag{2} \]

### Step 2: Express \( Y \) in terms of \( X \)
From equation (1), we can express \( Y \):

\[ Y = 2X - \begin{pmatrix} 3 & -3 & 0 \\ 3 & 3 & 2 \end{pmatrix} \]

### Step 3: Substitute \( Y \) into equation (2)
Now, substitute the expression for \( Y \) into equation (2):

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

### Step 4: Simplify the equation
Distributing the 2 in the equation gives:

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

Combining like terms:

\[ 5X - \begin{pmatrix} 6 & -6 & 0 \\ 6 & 6 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 1 & 5 \\ -1 & 4 & -4 \end{pmatrix} \]

### Step 5: Isolate \( X \)
Now, we isolate \( 5X \):

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

Calculating the right-hand side:

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

This simplifies to:

\[ 5X = \begin{pmatrix} 10 & -5 & 5 \\ 5 & 10 & 0 \end{pmatrix} \]

### Step 6: Solve for \( X \)
Now, divide both sides by 5:

\[ X = \begin{pmatrix} 10/5 & -5/5 & 5/5 \\ 5/5 & 10/5 & 0/5 \end{pmatrix} = \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 0 \end{pmatrix} \]

### Step 7: Substitute \( X \) back to find \( Y \)
Now that we have \( X \), we can substitute it back into the expression for \( Y \):

\[ Y = 2X - \begin{pmatrix} 3 & -3 & 0 \\ 3 & 3 & 2 \end{pmatrix} \]

Calculating \( 2X \):

\[ 2X = 2 \cdot \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 0 \end{pmatrix} = \begin{pmatrix} 4 & -2 & 2 \\ 2 & 4 & 0 \end{pmatrix} \]

Now substituting:

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

Calculating the right-hand side:

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

### Final Solution
Thus, the matrices \( X \) and \( Y \) are:

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