If `{:[(x,1),(-1,-y)]:}+{:[(y,1),(3,x)]:}={:[(1,2),(2,1)]:}`, then

Options

A. x=-1, y=0
B. x=1, y=0
C. x=0, y=1
D. x=1, y=1

Answer

To solve the equation given by the matrices, we start by writing down the matrices involved:

1. The first matrix is \( \begin{pmatrix} x & 1 \\ -1 & -y \end{pmatrix} \).
2. The second matrix is \( \begin{pmatrix} y & 1 \\ 3 & x \end{pmatrix} \).
3. The resulting matrix is \( \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \).

We can express the equation as follows:

\[ \begin{pmatrix} x & 1 \\ -1 & -y \end{pmatrix} + \begin{pmatrix} y & 1 \\ 3 & x \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \]

### Step 1: Set up the equations from the matrix addition

From the addition of the matrices, we can equate the corresponding elements:

1. From the (1,1) position: \( x + y = 1 \) (Equation 1)
2. From the (1,2) position: \( 1 + 1 = 2 \) (This does not provide any new information)
3. From the (2,1) position: \( -1 + 3 = 2 \) (This does not provide any new information)
4. From the (2,2) position: \( -y + x = 1 \) (Equation 2)

### Step 2: Solve the equations

Now we have two equations to work with:

1. \( x + y = 1 \) (Equation 1)
2. \( -y + x = 1 \) (Equation 2)

We can rearrange Equation 1 to express \( y \) in terms of \( x \):

\[ y = 1 - x \]

Now, substitute this expression for \( y \) into Equation 2:

\[ -x + x = 1 \]

This simplifies to:

\[ x - (1 - x) = 1 \]

### Step 3: Simplify and solve for \( x \)

Combining the terms gives:

\[ x - 1 + x = 1 \]

This simplifies to:

\[ 2x - 1 = 1 \]

Adding 1 to both sides:

\[ 2x = 2 \]

Dividing by 2:

\[ x = 1 \]

### Step 4: Substitute \( x \) back to find \( y \)

Now that we have \( x \), we can substitute back into Equation 1 to find \( y \):

\[ 1 + y = 1 \]

Subtracting 1 from both sides gives:

\[ y = 0 \]

### Conclusion

Thus, the solution to the equations is:

\[ x = 1, \quad y = 0 \]