The values of x, y, z and w such that
` [{:(x - y ,2z + w),(2x - y , 2x + w):}] = [{:(6,4),(12,15):}]` are

Options

A. x = 6 , y = 0 , z = 1/2 , w = 3
B. x = 2 , y = 7, z = 1 , w = 1
C. x = 7 , y = 2 , z = -1, w = 2
D. x = - 7, y = 2, , w = 1

Answer

To find the values of \( x, y, z, \) and \( w \) such that

\[ \begin{pmatrix} x - y & 2z + w \\ 2x - y & 2x + w \end{pmatrix} = \begin{pmatrix} 6 & 4 \\ 12 & 15 \end{pmatrix} \]

we will equate the corresponding elements of the matrices.

### Step 1: Set up the equations
From the equality of the matrices, we can set up the following equations:

1. \( x - y = 6 \) (Equation 1)
2. \( 2z + w = 4 \) (Equation 2)
3. \( 2x - y = 12 \) (Equation 3)
4. \( 2x + w = 15 \) (Equation 4)

### Step 2: Solve for \( y \) using Equation 1
From Equation 1, we can express \( y \) in terms of \( x \):

\[ y = x - 6 \]

### Step 3: Substitute \( y \) into Equation 3
Now, substitute \( y \) into Equation 3:

\[ 2x - (x - 6) = 12 \]

This simplifies to:

\[ 2x - x + 6 = 12 \]

\[ x + 6 = 12 \]

\[ x = 12 - 6 = 6 \]

### Step 4: Find \( y \)
Now that we have \( x \), we can find \( y \):

\[ y = 6 - 6 = 0 \]

### Step 5: Substitute \( x \) into Equation 4 to find \( w \)
Now substitute \( x \) into Equation 4 to find \( w \):

\[ 2(6) + w = 15 \]

\[ 12 + w = 15 \]

\[ w = 15 - 12 = 3 \]

### Step 6: Substitute \( w \) into Equation 2 to find \( z \)
Now substitute \( w \) into Equation 2 to find \( z \):

\[ 2z + 3 = 4 \]

\[ 2z = 4 - 3 \]

\[ 2z = 1 \]

\[ z = \frac{1}{2} \]

### Final Values
Thus, we have the values:

\[ x = 6, \quad y = 0, \quad z = \frac{1}{2}, \quad w = 3 \]

### Summary
The values of \( x, y, z, \) and \( w \) are:

- \( x = 6 \)
- \( y = 0 \)
- \( z = \frac{1}{2} \)
- \( w = 3 \)