Find matrices 'X' and 'Y' if :
`2X+3Y=[(2,3),(4,0)]` and `3X+2Y=[(2,-2),(-1,5)]`

Answer

To find the matrices \( X \) and \( Y \) given the equations:

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

We will solve these equations step by step.

### Step 1: Multiply the equations to eliminate one variable

Let's multiply Equation 1 by 3 and Equation 2 by 2 to make the coefficients of \( X \) the same.

\[ 3(2X + 3Y) = 3\begin{pmatrix} 2 & 3 \\ 4 & 0 \end{pmatrix} \implies 6X + 9Y = \begin{pmatrix} 6 & 9 \\ 12 & 0 \end{pmatrix} \]

\[ 2(3X + 2Y) = 2\begin{pmatrix} 2 & -2 \\ -1 & 5 \end{pmatrix} \implies 6X + 4Y = \begin{pmatrix} 4 & -4 \\ -2 & 10 \end{pmatrix} \]

### Step 2: Subtract the two equations

Now we will subtract the second modified equation from the first modified equation:

\[ (6X + 9Y) - (6X + 4Y) = \begin{pmatrix} 6 & 9 \\ 12 & 0 \end{pmatrix} - \begin{pmatrix} 4 & -4 \\ -2 & 10 \end{pmatrix} \]

This simplifies to:

\[ 5Y = \begin{pmatrix} 6 - 4 & 9 + 4 \\ 12 + 2 & 0 - 10 \end{pmatrix} = \begin{pmatrix} 2 & 13 \\ 14 & -10 \end{pmatrix} \]

### Step 3: Solve for \( Y \)

Now, we can solve for \( Y \) by dividing both sides by 5:

\[ Y = \frac{1}{5} \begin{pmatrix} 2 & 13 \\ 14 & -10 \end{pmatrix} = \begin{pmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{pmatrix} \]

### Step 4: Substitute \( Y \) back to find \( X \)

Now we substitute \( Y \) back into Equation 1 to find \( X \):

\[ 2X + 3\begin{pmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 4 & 0 \end{pmatrix} \]

Calculating \( 3Y \):

\[ 3Y = \begin{pmatrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{pmatrix} \]

Now we substitute:

\[ 2X + \begin{pmatrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 4 & 0 \end{pmatrix} \]

### Step 5: Solve for \( 2X \)

Subtract \( 3Y \) from both sides:

\[ 2X = \begin{pmatrix} 2 & 3 \\ 4 & 0 \end{pmatrix} - \begin{pmatrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{pmatrix} \]

Calculating the right side:

\[ = \begin{pmatrix} 2 - \frac{6}{5} & 3 - \frac{39}{5} \\ 4 - \frac{42}{5} & 0 + 6 \end{pmatrix} = \begin{pmatrix} \frac{10}{5} - \frac{6}{5} & \frac{15}{5} - \frac{39}{5} \\ \frac{20}{5} - \frac{42}{5} & 6 \end{pmatrix} \]

This simplifies to:

\[ = \begin{pmatrix} \frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6 \end{pmatrix} \]

### Step 6: Solve for \( X \)

Now divide by 2:

\[ X = \frac{1}{2} \begin{pmatrix} \frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6 \end{pmatrix} = \begin{pmatrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{pmatrix} \]

### Final Answer

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

\[ X = \begin{pmatrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{pmatrix}, \quad Y = \begin{pmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{pmatrix} \]