Let `A=[(cos^(- 1)x,cos^(- 1)y,cos^(- 1)z),(cos^(- 1)y,cos^(- 1)z,cos^(- 1)x),(cos^(- 1)z,cos^(- 1)x,cos^(- 1)y)]` such that `|A| = 0`, then maximum value of `x + y + z` is

Options

A. 3
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to find the maximum value of \( x + y + z \) given that the determinant of the matrix \( A \) is zero. The matrix \( A \) is defined as follows:

\[ A = \begin{pmatrix} \cos^{-1} x & \cos^{-1} y & \cos^{-1} z \\ \cos^{-1} y & \cos^{-1} z & \cos^{-1} x \\ \cos^{-1} z & \cos^{-1} x & \cos^{-1} y \end{pmatrix} \]

### Step 1: Write the determinant condition
We know that \( |A| = 0 \). This means that the rows (or columns) of the matrix \( A \) are linearly dependent.

### Step 2: Perform a column operation
We can perform a column operation to simplify the determinant. Let's change the first column by adding the second and third columns to it:

\[ \text{New Column 1} = \cos^{-1} x + \cos^{-1} y + \cos^{-1} z \]

This gives us:

\[ A = \begin{pmatrix} \cos^{-1} x + \cos^{-1} y + \cos^{-1} z & \cos^{-1} y & \cos^{-1} z \\ \cos^{-1} y + \cos^{-1} z + \cos^{-1} x & \cos^{-1} z & \cos^{-1} x \\ \cos^{-1} z + \cos^{-1} x + \cos^{-1} y & \cos^{-1} x & \cos^{-1} y \end{pmatrix} \]

### Step 3: Factor out the common term
We can factor out \( \cos^{-1} x + \cos^{-1} y + \cos^{-1} z \) from the first column:

\[ |A| = (\cos^{-1} x + \cos^{-1} y + \cos^{-1} z) \cdot |B| \]

Where \( B \) is the remaining \( 2 \times 2 \) matrix formed by the other columns. Since \( |A| = 0 \), we have:

\[ \cos^{-1} x + \cos^{-1} y + \cos^{-1} z = 0 \quad \text{or} \quad |B| = 0 \]

### Step 4: Analyze the case \( \cos^{-1} x + \cos^{-1} y + \cos^{-1} z = 0 \)
The only way for \( \cos^{-1} x + \cos^{-1} y + \cos^{-1} z \) to be zero is if:

\[ \cos^{-1} x = 0, \quad \cos^{-1} y = 0, \quad \cos^{-1} z = 0 \]

This implies:

\[ x = 1, \quad y = 1, \quad z = 1 \]

### Step 5: Calculate the maximum value of \( x + y + z \)
Now, we can find the maximum value of \( x + y + z \):

\[ x + y + z = 1 + 1 + 1 = 3 \]

### Conclusion
Thus, the maximum value of \( x + y + z \) is:

\[ \boxed{3} \]