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

Options

A. x=2,y=1,z=3
B. x=3,y=1,z=2
C. x=1,y=2,z=3
D. x=1,y=3,z=2

Answer

To solve the equation given by the matrices, we will equate the corresponding elements of the two matrices and derive equations from them. Let's go through the solution step by step.

### Step 1: Write down the equations from the matrix equality
We have the matrices:
\[ \begin{pmatrix} x+y & y-z \\ z-2x & y-x \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix} \]

From this equality, we can equate the corresponding elements:
1. \( x + y = 3 \) (Equation 1)
2. \( y - z = -1 \) (Equation 2)
3. \( z - 2x = 1 \) (Equation 3)
4. \( y - x = 1 \) (Equation 4)

### Step 2: Solve for \( y \) in terms of \( x \)
From Equation 4:
\[ y - x = 1 \implies y = x + 1 \quad \text{(Equation 5)} \]

### Step 3: Substitute \( y \) into Equation 1
Substituting Equation 5 into Equation 1:
\[ x + (x + 1) = 3 \]
This simplifies to:
\[ 2x + 1 = 3 \]
Subtracting 1 from both sides:
\[ 2x = 2 \]
Dividing by 2:
\[ x = 1 \quad \text{(Value of \( x \))} \]

### Step 4: Substitute \( x \) back to find \( y \)
Using the value of \( x \) in Equation 5:
\[ y = 1 + 1 = 2 \quad \text{(Value of \( y \))} \]

### Step 5: Substitute \( y \) into Equation 2 to find \( z \)
Now, substitute \( y = 2 \) into Equation 2:
\[ 2 - z = -1 \]
Adding \( z \) to both sides:
\[ 2 = z - 1 \]
Adding 1 to both sides:
\[ z = 3 \quad \text{(Value of \( z \))} \]

### Final Values
Thus, we have found:
- \( x = 1 \)
- \( y = 2 \)
- \( z = 3 \)

### Conclusion
The values of \( x, y, z \) are \( x = 1, y = 2, z = 3 \).