Product of roots of equation `|(1+2x,1,1-x),(2-x,2+x,3+x),(x,1+x,1-x^2)|=0` is

Options

A. `1//2`
B. `3//4`
C. `4//3`
D. `1//4`

Explanation

To solve the problem, we need to find the product of the roots of the equation given by the determinant:

\[ \left| \begin{array}{ccc} 1 + 2x & 1 & 1 - x \\ 2 - x & 2 + x & 3 + x \\ x & 1 + x & 1 - x^2 \end{array} \right| = 0 \]

### Step 1: Simplify the Determinant
We can simplify the determinant by performing column operations. First, we can change column 1 to column 1 minus column 2:

\[ C_1 \rightarrow C_1 - C_2 \]

This gives us:

\[ \left| \begin{array}{ccc} (1 + 2x - 1) & 1 & 1 - x \\ (2 - x - (2 + x)) & (2 + x) & (3 + x) \\ (x - (1 + x)) & (1 + x) & (1 - x^2) \end{array} \right| = \left| \begin{array}{ccc} 2x & 1 & 1 - x \\ -x & 2 + x & 3 + x \\ 0 & 1 + x & 1 - x^2 \end{array} \right| \]

### Step 2: Further Simplify the Determinant
Next, we can change column 3 to column 3 minus column 2:

\[ C_3 \rightarrow C_3 - C_2 \]

This results in:

\[ \left| \begin{array}{ccc} 2x & 1 & (1 - x) - (2 + x) \\ -x & (2 + x) & (3 + x) - (2 + x) \\ 0 & (1 + x) & (1 - x^2) - (1 + x) \end{array} \right| = \left| \begin{array}{ccc} 2x & 1 & -1 - 2x \\ -x & 2 + x & 1 \\ 0 & 1 + x & -x^2 - x \end{array} \right| \]

### Step 3: Expand the Determinant
Now we can expand the determinant along the first row:

\[ = 2x \left| \begin{array}{cc} 2 + x & 1 \\ 1 + x & -x^2 - x \end{array} \right| - 1 \left| \begin{array}{cc} -x & 1 \\ 0 & -x^2 - x \end{array} \right| + (-1 - 2x) \left| \begin{array}{cc} -x & 2 + x \\ 0 & 1 + x \end{array} \right| \]

Calculating the 2x determinant:

\[ = 2x \left( (2 + x)(-x^2 - x) - 1(1 + x) \right) \]

Calculating the second determinant:

\[ = -(-x)(-x^2 - x) = x^3 + x^2 \]

Calculating the third determinant:

\[ = (-1 - 2x)(-x(1 + x)) = (1 + 2x)x(1 + x) \]

### Step 4: Combine and Set to Zero
Now we combine all these results and set the determinant equal to zero:

\[ 2x \left( -2x^2 - 3x - 1 \right) + x^3 + x^2 + (1 + 2x)x(1 + x) = 0 \]

This simplifies to a polynomial in \(x\):

\[ -2x^4 - 6x^3 - 3x^2 - 1 = 0 \]

### Step 5: Find the Product of Roots
The product of the roots of a polynomial \(ax^n + bx^{n-1} + ... + k = 0\) is given by \(-\frac{k}{a}\). Here, \(a = -2\) and \(k = -1\):

\[ \text{Product of roots} = -\frac{-1}{-2} = \frac{1}{2} \]

### Final Answer
The product of the roots of the equation is:

\[ \frac{1}{2} \]