Find the values of `'x','y','z'` if the matrix : `A=[(0,2y,z),(x,y,-z),(x,-y,z)]` satisfies the equation `A'A=I`

Answer

To solve the problem of finding the values of \(x\), \(y\), and \(z\) such that the matrix

\[ A = \begin{pmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{pmatrix} \]

satisfies the equation \(A^T A = I\) (where \(I\) is the identity matrix), we will follow these steps:

### Step 1: Compute the Transpose of Matrix \(A\)

The transpose of matrix \(A\), denoted as \(A^T\), is obtained by swapping the rows and columns:

\[ A^T = \begin{pmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{pmatrix} \]

### Step 2: Compute \(A^T A\)

Now we will multiply \(A^T\) by \(A\):

\[ A^T A = \begin{pmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{pmatrix} \begin{pmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{pmatrix} \]

Calculating the entries of \(A^T A\):

1. **First Row:**
- \( (0 \cdot 0 + x \cdot x + x \cdot x) = 2x^2 \)
- \( (0 \cdot 2y + x \cdot y + x \cdot -y) = 0 \)
- \( (0 \cdot z + x \cdot -z + x \cdot z) = 0 \)

2. **Second Row:**
- \( (2y \cdot 0 + y \cdot x + -y \cdot x) = 0 \)
- \( (2y \cdot 2y + y \cdot y + -y \cdot -y) = 6y^2 \)
- \( (2y \cdot z + y \cdot -z + -y \cdot z) = 0 \)

3. **Third Row:**
- \( (z \cdot 0 + -z \cdot x + z \cdot x) = 0 \)
- \( (z \cdot 2y + -z \cdot y + z \cdot -y) = 0 \)
- \( (z \cdot z + -z \cdot -z + z \cdot z) = 3z^2 \)

Putting this together, we get:

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

### Step 3: Set \(A^T A\) Equal to Identity Matrix \(I\)

We know that \(A^T A = I\), where

\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Thus, we can set up the following equations:

1. \(2x^2 = 1\)
2. \(6y^2 = 1\)
3. \(3z^2 = 1\)

### Step 4: Solve for \(x\), \(y\), and \(z\)

1. From \(2x^2 = 1\):
\[ x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \]

2. From \(6y^2 = 1\):
\[ y^2 = \frac{1}{6} \implies y = \pm \frac{1}{\sqrt{6}} \]

3. From \(3z^2 = 1\):
\[ z^2 = \frac{1}{3} \implies z = \pm \frac{1}{\sqrt{3}} \]

### Final Values

Thus, the values of \(x\), \(y\), and \(z\) are:

\[ x = \pm \frac{1}{\sqrt{2}}, \quad y = \pm \frac{1}{\sqrt{6}}, \quad z = \pm \frac{1}{\sqrt{3}} \]