Find the value of y, if the matrix `A = ((0,2y,z),(x,y,-z),(x,-y,z))` obeys the law `A^(T).A = I`.

Answer

To find the value of \( y \) from the matrix \( A = \begin{pmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{pmatrix} \) such that \( A^T A = I \), we will follow these steps:

### Step 1: Compute the Transpose of Matrix \( A \)
The transpose of matrix \( A \), denoted \( A^T \), is obtained by swapping rows with columns:
\[ A^T = \begin{pmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{pmatrix} \]

**Hint:** Remember that the transpose of a matrix is formed by flipping it over its diagonal.

### Step 2: Compute the Product \( A^T A \)
Now, we need to compute the product \( A^T A \):
\[ A^T A = \begin{pmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{pmatrix} \begin{pmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{pmatrix} \]

We will calculate each element of the resulting matrix.

### Step 3: Calculate Each Element of \( A^T A \)
1. **Element (1,1)**:
\[ 0 \cdot 0 + x \cdot x + x \cdot x = 2x^2 \]

2. **Element (1,2)**:
\[ 0 \cdot 2y + x \cdot y + x \cdot (-y) = 0 \]

3. **Element (1,3)**:
\[ 0 \cdot z + x \cdot (-z) + x \cdot z = 0 \]

4. **Element (2,1)**:
\[ 2y \cdot 0 + y \cdot x + (-y) \cdot x = 0 \]

5. **Element (2,2)**:
\[ 2y \cdot 2y + y \cdot y + (-y) \cdot (-y) = 4y^2 + y^2 + y^2 = 6y^2 \]

6. **Element (2,3)**:
\[ 2y \cdot z + y \cdot (-z) + (-y) \cdot z = 2yz - yz - yz = 0 \]

7. **Element (3,1)**:
\[ z \cdot 0 + (-z) \cdot x + z \cdot x = 0 \]

8. **Element (3,2)**:
\[ z \cdot 2y + (-z) \cdot y + z \cdot (-y) = 2yz - yz - yz = 0 \]

9. **Element (3,3)**:
\[ z \cdot z + (-z) \cdot (-z) + z \cdot z = z^2 + z^2 + z^2 = 3z^2 \]

Thus, we have:
\[ A^T A = \begin{pmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{pmatrix} \]

### Step 4: Set \( A^T A \) Equal to the Identity Matrix
Since \( A^T A = I \), we equate:
\[ \begin{pmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

This gives us the following equations:
1. \( 2x^2 = 1 \)
2. \( 6y^2 = 1 \)
3. \( 3z^2 = 1 \)

### Step 5: Solve for \( y \)
From the equation \( 6y^2 = 1 \):
\[ y^2 = \frac{1}{6} \implies y = \pm \frac{1}{\sqrt{6}} \]

### Final Result
Thus, the value of \( y \) is:
\[ y = \pm \frac{1}{\sqrt{6}} \]