Consider the matrix `A=[(x, 2y,z),(2y,z,x),(z,x,2y)]` and `A A^(T)=9I.` If `Tr(A) gt0` and `xyz=(1)/(6)`, then the vlaue of `x^(3)+8y^(3)+z^(3)` is equal to (where, `Tr(A), I and A^(T)` denote the trace of matrix A i.e. the sum of all the principal diagonal elements, the identity matrix of the same order of matrix A and the transpose of matrix A respectively)

Options

A. 20
B. 22
C. 26
D. 28

Answer

To solve the problem, we need to analyze the given matrix \( A \) and the conditions provided.

The matrix \( A \) is given as:

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

We know that \( AA^T = 9I \), where \( I \) is the identity matrix. The identity matrix \( I \) for a 3x3 matrix is:

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

### Step 1: Calculate \( A^T \)

The transpose of matrix \( A \) is:

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

### Step 2: Calculate \( AA^T \)

Now we compute \( AA^T \):

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

Calculating the elements of \( AA^T \):

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

2. First row, second column:
\[ 2yx + 2yz + xz \]

3. First row, third column:
\[ zx + 2xy + 2y^2 \]

4. Second row, first column (similar to first row, second column):
\[ 2yx + 2yz + xz \]

5. Second row, second column:
\[ 4y^2 + z^2 + x^2 \]

6. Second row, third column:
\[ 2yz + zx + 2xy \]

7. Third row, first column (similar to first row, third column):
\[ zx + 2xy + 2y^2 \]

8. Third row, second column (similar to second row, third column):
\[ 2yz + zx + 2xy \]

9. Third row, third column:
\[ z^2 + 4y^2 + x^2 \]

Thus, we can write:

\[ AA^T = \begin{pmatrix} x^2 + 4y^2 + z^2 & 2yx + 2yz + xz & zx + 2xy + 2y^2 \\ 2yx + 2yz + xz & 4y^2 + z^2 + x^2 & 2yz + zx + 2xy \\ zx + 2xy + 2y^2 & 2yz + zx + 2xy & z^2 + 4y^2 + x^2 \end{pmatrix} \]

### Step 3: Set \( AA^T = 9I \)

From the equation \( AA^T = 9I \), we have:

\[ AA^T = \begin{pmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{pmatrix} \]

This gives us the following equations:

1. \( x^2 + 4y^2 + z^2 = 9 \)
2. \( 2xy + 2yz + xz = 0 \)
3. \( z^2 + 4y^2 + x^2 = 9 \)

### Step 4: Use the Trace Condition

The trace of \( A \) is given by:

\[ \text{Tr}(A) = x + z + 2y > 0 \]

### Step 5: Use the Product Condition

We are also given that \( xyz = \frac{1}{6} \).

### Step 6: Find \( x^3 + 8y^3 + z^3 \)

Using the identity:

\[ x^3 + 8y^3 + z^3 - 3xyz = (x + 2y + z)(x^2 + 4y^2 + z^2 - xz - 2xy - 2yz) \]

From our previous calculations, we know:

- \( x + 2y + z = 3 \)
- \( x^2 + 4y^2 + z^2 = 9 \)
- \( xz + 2xy + 2yz = 0 \)

Substituting these into the identity gives:

\[ x^3 + 8y^3 + z^3 - 3xyz = 3(9 - 0) = 27 \]

Thus, we have:

\[ x^3 + 8y^3 + z^3 = 27 + 3xyz \]

Substituting \( xyz = \frac{1}{6} \):

\[ x^3 + 8y^3 + z^3 = 27 + 3 \cdot \frac{1}{6} = 27 + \frac{1}{2} = 27.5 \]

### Final Answer

Thus, the value of \( x^3 + 8y^3 + z^3 \) is:

\[ \boxed{28} \]