If `[(x-y,2x-y),(2x+z,3z+w)]=[(-1,0),(5,13)]`

Options

A. z=3,w=4
B. z=4,w=3
C. z=1,w=2
D. z=2,w=-1

Explanation

To solve the problem, we start with the given matrix equation:

\[ \begin{pmatrix} x - y & 2x - y \\ 2x + z & 3z + w \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 5 & 13 \end{pmatrix} \]

From this equation, we can equate the corresponding elements of the matrices to form a system of equations:

1. \( x - y = -1 \) (Equation 1)
2. \( 2x - y = 0 \) (Equation 2)
3. \( 2x + z = 5 \) (Equation 3)
4. \( 3z + w = 13 \) (Equation 4)

### Step 1: Solve Equations 1 and 2

From Equation 1:
\[ x - y = -1 \implies y = x + 1 \]

Now, substitute \( y \) in Equation 2:
\[ 2x - (x + 1) = 0 \]
\[ 2x - x - 1 = 0 \implies x - 1 = 0 \implies x = 1 \]

### Step 2: Find \( y \)

Using the value of \( x \) in Equation 1:
\[ y = x + 1 = 1 + 1 = 2 \]

### Step 3: Find \( z \)

Now we use the value of \( x \) in Equation 3:
\[ 2(1) + z = 5 \implies 2 + z = 5 \implies z = 5 - 2 = 3 \]

### Step 4: Find \( w \)

Now we substitute \( z \) into Equation 4:
\[ 3(3) + w = 13 \implies 9 + w = 13 \implies w = 13 - 9 = 4 \]

### Final Values

Thus, we find:
- \( x = 1 \)
- \( y = 2 \)
- \( z = 3 \)
- \( w = 4 \)

### Conclusion

The values of \( z \) and \( w \) are:
\[ z = 3, \quad w = 4 \]