Let `A=[(x,2y),(-1,y)], x , y in R`
If AA'=`[(1,0),(0,alpha)] (alpha in R)` , then `alpha+y^2` is equal to :

Options

A. `sqrt2-1`
B. 1
C. 2
D. `sqrt2`

Answer

To solve the problem, we need to find the value of \( \alpha + y^2 \) given the matrix \( A \) and the condition \( AA' = \begin{pmatrix} 1 & 0 \\ 0 & \alpha \end{pmatrix} \).

### Step-by-step Solution:

1. **Define the Matrix \( A \)**:
\[ A = \begin{pmatrix} x & 2y \\ -1 & y \end{pmatrix} \]

2. **Calculate the Transpose of \( A \)**:
\[ A' = \begin{pmatrix} x & -1 \\ 2y & y \end{pmatrix} \]

3. **Multiply \( A \) with \( A' \)**:
\[ AA' = \begin{pmatrix} x & 2y \\ -1 & y \end{pmatrix} \begin{pmatrix} x & -1 \\ 2y & y \end{pmatrix} \]
The product is calculated as follows:
- First row, first column:
\[ x^2 + 4y^2 \]
- First row, second column:
\[ -x + 2y^2 \]
- Second row, first column:
\[ -x + 2y^2 \]
- Second row, second column:
\[ 1 + y^2 \]
Thus, we have:
\[ AA' = \begin{pmatrix} x^2 + 4y^2 & -x + 2y^2 \\ -x + 2y^2 & 1 + y^2 \end{pmatrix} \]

4. **Set \( AA' \) equal to the given matrix**:
\[ \begin{pmatrix} x^2 + 4y^2 & -x + 2y^2 \\ -x + 2y^2 & 1 + y^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & \alpha \end{pmatrix} \]

5. **Equate the corresponding elements**:
- From the first element:
\[ x^2 + 4y^2 = 1 \quad (1) \]
- From the second element:
\[ -x + 2y^2 = 0 \quad (2) \]
- From the third element:
\[ 1 + y^2 = \alpha \quad (3) \]

6. **Solve the equations**:
From equation (2):
\[ -x + 2y^2 = 0 \implies x = 2y^2 \]
Substitute \( x = 2y^2 \) into equation (1):
\[ (2y^2)^2 + 4y^2 = 1 \implies 4y^4 + 4y^2 = 1 \]
Rearranging gives:
\[ 4y^4 + 4y^2 - 1 = 0 \]
Let \( z = y^2 \):
\[ 4z^2 + 4z - 1 = 0 \]
Using the quadratic formula:
\[ z = \frac{-4 \pm \sqrt{(4)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{-4 \pm \sqrt{16 + 16}}{8} = \frac{-4 \pm 4\sqrt{2}}{8} = \frac{-1 \pm \sqrt{2}}{2} \]
Since \( z = y^2 \) must be non-negative, we take:
\[ y^2 = \frac{-1 + \sqrt{2}}{2} \]

7. **Substitute \( y^2 \) back to find \( \alpha \)**:
Using equation (3):
\[ \alpha = 1 + y^2 = 1 + \frac{-1 + \sqrt{2}}{2} = \frac{2 - 1 + \sqrt{2}}{2} = \frac{1 + \sqrt{2}}{2} \]

8. **Find \( \alpha + y^2 \)**:
\[ \alpha + y^2 = \frac{1 + \sqrt{2}}{2} + \frac{-1 + \sqrt{2}}{2} = \frac{1 + \sqrt{2} - 1 + \sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2} \]

### Final Answer:
\[ \alpha + y^2 = \sqrt{2} \]