Find `A=[(0,2y,z),(x,y,-z),(x,-y,z)]` satisfies `A^(T) = A^(-1)`

Options

A. `pm1/sqrt2`
B. `pm1/sqrt3`
C. `pm1/sqrt6`
D. None of these

Answer

To solve the problem, we need to find the values of \(x\), \(y\), and \(z\) such that the matrix \(A\) satisfies the condition \(A^T = A^{-1}\).

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

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

### Step 2: Use the condition \(A^T = A^{-1}\)
Since \(A^T = A^{-1}\), we can multiply \(A\) by \(A^T\) and set it equal to the identity matrix \(I\):
\[ A \cdot A^T = I \]

### Step 3: Calculate \(A \cdot A^T\)
Now we will multiply \(A\) and \(A^T\):
\[ A \cdot A^T = \begin{pmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{pmatrix} \cdot \begin{pmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{pmatrix} \]

Calculating the elements of the resulting matrix:

1. First row, first column:
\[ 0 \cdot 0 + 2y \cdot 2y + z \cdot z = 4y^2 + z^2 \]

2. First row, second column:
\[ 0 \cdot x + 2y \cdot y + z \cdot (-z) = 2y^2 - z^2 \]

3. First row, third column:
\[ 0 \cdot x + 2y \cdot (-y) + z \cdot z = -2y^2 + z^2 \]

4. Second row, first column:
\[ x \cdot 0 + y \cdot 2y + (-z) \cdot z = 2y^2 - z^2 \]

5. Second row, second column:
\[ x \cdot x + y \cdot y + (-z) \cdot (-z) = x^2 + y^2 + z^2 \]

6. Second row, third column:
\[ x \cdot x + y \cdot (-y) + (-z) \cdot z = x^2 - y^2 + z^2 \]

7. Third row, first column:
\[ x \cdot 0 + (-y) \cdot 2y + z \cdot z = -2y^2 + z^2 \]

8. Third row, second column:
\[ x \cdot x + (-y) \cdot y + z \cdot (-z) = x^2 - y^2 - z^2 \]

9. Third row, third column:
\[ x \cdot x + (-y) \cdot (-y) + z \cdot z = x^2 + y^2 + z^2 \]

Putting it all together, we have:
\[ A \cdot A^T = \begin{pmatrix} 4y^2 + z^2 & 2y^2 - z^2 & -2y^2 + z^2 \\ 2y^2 - z^2 & x^2 + y^2 + z^2 & x^2 - y^2 + z^2 \\ -2y^2 + z^2 & x^2 - y^2 - z^2 & x^2 + y^2 + z^2 \end{pmatrix} \]

### Step 4: Set \(A \cdot A^T\) equal to the identity matrix
The identity matrix \(I\) is:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Set up equations
From the equality \(A \cdot A^T = I\), we can set up the following equations:
1. \(4y^2 + z^2 = 1\) (Equation 1)
2. \(x^2 + y^2 + z^2 = 1\) (Equation 2)
3. \(2y^2 - z^2 = 0\) (Equation 3)
4. \(x^2 - y^2 - z^2 = 0\) (Equation 4)

### Step 6: Solve the equations
From Equation 3:
\[ z^2 = 2y^2 \]

Substituting \(z^2\) into Equation 1:
\[ 4y^2 + 2y^2 = 1 \implies 6y^2 = 1 \implies y^2 = \frac{1}{6} \implies y = \pm \frac{1}{\sqrt{6}} \]

Now substitute \(y^2\) back into the expression for \(z^2\):
\[ z^2 = 2 \cdot \frac{1}{6} = \frac{1}{3} \implies z = \pm \frac{1}{\sqrt{3}} \]

Now substitute \(y^2\) and \(z^2\) into Equation 2 to find \(x^2\):
\[ x^2 + \frac{1}{6} + \frac{1}{3} = 1 \implies x^2 + \frac{1}{6} + \frac{2}{6} = 1 \implies x^2 + \frac{3}{6} = 1 \implies x^2 = \frac{3}{6} = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \]

### Final Values
Thus, the values are:
\[ x = \pm \frac{1}{\sqrt{2}}, \quad y = \pm \frac{1}{\sqrt{6}}, \quad z = \pm \frac{1}{\sqrt{3}} \]