Let `A=[(0, 2y,z),(x,y,-z),(x,-y,z)]` such that `A^(T)A=I`, then the value of `x^(2)+y^(2)+z^(2)` is

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 + z^2 \) given the matrix \( A \) such that \( A^T A = I \).

### Step-by-Step Solution:

1. **Define the Matrix \( A \)**:
\[ A = \begin{pmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{pmatrix} \]

2. **Calculate the Transpose of \( A \)**:
The transpose \( A^T \) is obtained by swapping rows and columns:
\[ A^T = \begin{pmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{pmatrix} \]

3. **Multiply \( A^T \) and \( A \)**:
We need to compute \( A^T A \):
\[ A^T A = \begin{pmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{pmatrix} \begin{pmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{pmatrix} \]

We will calculate each entry of the resulting matrix.

4. **Calculate Each Entry of \( A^T A \)**:
- **First Row**:
- First column: \( 0 \cdot 0 + x \cdot x + x \cdot x = 2x^2 \)
- Second column: \( 0 \cdot 2y + x \cdot y + x \cdot (-y) = 0 \)
- Third column: \( 0 \cdot z + x \cdot (-z) + x \cdot z = 0 \)

- **Second Row**:
- First column: \( 2y \cdot 0 + y \cdot x + (-y) \cdot x = 0 \)
- Second column: \( 2y \cdot 2y + y \cdot y + (-y) \cdot (-y) = 4y^2 + y^2 + y^2 = 6y^2 \)
- Third column: \( 2y \cdot z + y \cdot (-z) + (-y) \cdot z = 2yz - yz - yz = 0 \)

- **Third Row**:
- First column: \( z \cdot 0 + (-z) \cdot x + z \cdot x = 0 \)
- Second column: \( z \cdot 2y + (-z) \cdot y + z \cdot (-y) = 2yz - yz - yz = 0 \)
- Third column: \( z \cdot z + (-z) \cdot (-z) + z \cdot z = z^2 + z^2 + z^2 = 3z^2 \)

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

6. **Set Equal to 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 equations:
- \( 2x^2 = 1 \) → \( x^2 = \frac{1}{2} \)
- \( 6y^2 = 1 \) → \( y^2 = \frac{1}{6} \)
- \( 3z^2 = 1 \) → \( z^2 = \frac{1}{3} \)

7. **Calculate \( x^2 + y^2 + z^2 \)**:
\[ x^2 + y^2 + z^2 = \frac{1}{2} + \frac{1}{6} + \frac{1}{3} \]

To add these fractions, find a common denominator (which is 6):
\[ = \frac{3}{6} + \frac{1}{6} + \frac{2}{6} = \frac{6}{6} = 1 \]

### Final Answer:
\[ x^2 + y^2 + z^2 = 1 \]