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 \(Y\), we have the following equations:

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

### Step 1: Multiply the first equation by 2
We start by multiplying the entire first equation by 2 to facilitate elimination:

\[ 4X - 2Y = \begin{pmatrix} 6 & -6 & 0 \\ 6 & 6 & 4 \end{pmatrix} \]

### Step 2: Rewrite the second equation
The second equation remains as it is:

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

### Step 3: Add the two equations
Now, we add the modified first equation and the second equation:

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

This simplifies to:

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

Calculating the right-hand side gives:

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

### Step 4: Solve for \(X\)
Now, divide both sides by 5 to find \(X\):

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

### Step 5: Substitute \(X\) back into the second equation
Now we substitute \(X\) back into the second equation to find \(Y\):

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

### Step 6: Rearrange to solve for \(Y\)
Rearranging gives:

\[ 2Y = \begin{pmatrix} 4 & 1 & 5 \\ -1 & 4 & -4 \end{pmatrix} - \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 0 \end{pmatrix} \]

Calculating the right-hand side:

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

### Step 7: Solve for \(Y\)
Now, divide both sides by 2:

\[ Y = \frac{1}{2} \begin{pmatrix} 2 & 2 & 4 \\ -2 & 2 & -4 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 2 \\ -1 & 1 & -2 \end{pmatrix} \]

### Final Result
Thus, the solutions for the matrices \(X\) and \(Y\) are:

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