If the value of the determinant `|(x,1,1),(1,y,1),(1,1,z)|` is positive, then

Options

A. `xyz gt 1`
B. `xyz gt - 8`
C. `xyz lt - 8`
D. `xyz gt - 2`

Answer

To solve the determinant \( D = \begin{vmatrix} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{vmatrix} \) and find the conditions under which it is positive, we will follow these steps:

### Step 1: Calculate the Determinant
We can calculate the determinant using the formula for a 3x3 matrix:

\[ D = x \begin{vmatrix} y & 1 \\ 1 & z \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & z \end{vmatrix} + 1 \begin{vmatrix} 1 & y \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

\[ \begin{vmatrix} y & 1 \\ 1 & z \end{vmatrix} = yz - 1 \]

\[ \begin{vmatrix} 1 & 1 \\ 1 & z \end{vmatrix} = z - 1 \]

\[ \begin{vmatrix} 1 & y \\ 1 & 1 \end{vmatrix} = 1 - y \]

Substituting these back into the determinant formula:

\[ D = x(yz - 1) - (z - 1) + (1 - y) \]

### Step 2: Simplify the Expression
Now, we simplify \( D \):

\[ D = xyz - x - z + 1 + 1 - y \]

Combining like terms:

\[ D = xyz - x - y - z + 2 \]

### Step 3: Set the Condition for Positivity
We want \( D > 0 \):

\[ xyz - x - y - z + 2 > 0 \]

Rearranging gives:

\[ xyz > x + y + z - 2 \]

### Step 4: Analyze the Condition
To analyze this inequality, we can use the AM-GM inequality, which states that for non-negative numbers \( a, b, c \):

\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]

Setting \( a = x \), \( b = y \), and \( c = z \):

\[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \]

Multiplying both sides by 3 gives:

\[ x + y + z \geq 3\sqrt[3]{xyz} \]

### Step 5: Combine the Results
From our earlier inequality \( xyz > x + y + z - 2 \), we can substitute \( x + y + z \) from the AM-GM result:

\[ xyz > 3\sqrt[3]{xyz} - 2 \]

Let \( k = \sqrt[3]{xyz} \), then \( xyz = k^3 \):

\[ k^3 > 3k - 2 \]

### Step 6: Solve the Cubic Inequality
Rearranging gives:

\[ k^3 - 3k + 2 > 0 \]

Factoring the cubic:

\[ (k - 1)^2(k + 2) > 0 \]

### Step 7: Analyze the Roots
The roots of the equation are \( k = 1 \) and \( k = -2 \). The intervals to test are:

1. \( k < -2 \)
2. \( -2 < k < 1 \)
3. \( k > 1 \)

Testing these intervals, we find that the inequality holds for:

\[ k > 1 \quad \text{or} \quad xyz > 1 \]

### Final Conclusion
Thus, the condition for the determinant to be positive is:

\[ xyz > 1 \]