If `|(x,x^(2),1+x^(2)),(y,y^(2),1+y^(2)),(z,z^(2),1+z^(2))|` where x, y, z are distinct what is |A| ?

Options

A. A. 0
B. B. `x^(2) y-y^(2) x + xyz`
C. C. `(x - y) (y - z) (z - x)`
D. D. xyz

Answer

To find the determinant of the matrix

\[ A = \begin{pmatrix} x & x^2 & 1 + x^2 \\ y & y^2 & 1 + y^2 \\ z & z^2 & 1 + z^2 \end{pmatrix} \]

where \(x\), \(y\), and \(z\) are distinct, we will perform a series of row operations and simplifications.

### Step 1: Write the matrix
We start with the matrix:

\[ A = \begin{pmatrix} x & x^2 & 1 + x^2 \\ y & y^2 & 1 + y^2 \\ z & z^2 & 1 + z^2 \end{pmatrix} \]

### Step 2: Perform row operations
We will perform the following row operations:
- \(R_1 \leftarrow R_1 - R_2\)
- \(R_3 \leftarrow R_3 - R_2\)

This gives us:

\[ \begin{pmatrix} x - y & x^2 - y^2 & (1 + x^2) - (1 + y^2) \\ y & y^2 & 1 + y^2 \\ z - y & z^2 - y^2 & (1 + z^2) - (1 + y^2) \end{pmatrix} \]

Simplifying the entries, we have:

\[ \begin{pmatrix} x - y & x^2 - y^2 & x^2 - y^2 \\ y & y^2 & 1 + y^2 \\ z - y & z^2 - y^2 & z^2 - y^2 \end{pmatrix} \]

### Step 3: Factor out common terms
Notice that \(x^2 - y^2 = (x - y)(x + y)\) and \(z^2 - y^2 = (z - y)(z + y)\). Thus, we can factor out:

\[ \begin{pmatrix} x - y & (x - y)(x + y) & (x - y)(x + y) \\ y & y^2 & 1 + y^2 \\ z - y & (z - y)(z + y) & (z - y)(z + y) \end{pmatrix} \]

Taking \(x - y\) from the first row and \(z - y\) from the third row, we have:

\[ (x - y)(z - y) \begin{pmatrix} 1 & x + y & x + y \\ y & y^2 & 1 + y^2 \\ 1 & z + y & z + y \end{pmatrix} \]

### Step 4: Make the third row zero
Next, we perform the operation \(R_3 \leftarrow R_3 - R_1\):

\[ \begin{pmatrix} 1 & x + y & x + y \\ y & y^2 & 1 + y^2 \\ 0 & (z + y) - (x + y) & (z + y) - (x + y) \end{pmatrix} \]

This simplifies to:

\[ \begin{pmatrix} 1 & x + y & x + y \\ y & y^2 & 1 + y^2 \\ 0 & z - x & z - x \end{pmatrix} \]

### Step 5: Calculate the determinant
Now, we can compute the determinant of the resulting matrix. The determinant can be calculated along the third row:

\[ \text{Det}(A) = (x - y)(z - y)(z - x) \cdot \begin{vmatrix} 1 & x + y & x + y \\ y & y^2 & 1 + y^2 \\ 0 & z - x & z - x \end{vmatrix} \]

Calculating the determinant yields:

\[ = (x - y)(z - y)(z - x) \cdot 0 = 0 \]

### Conclusion
Thus, the determinant of the matrix \(A\) is:

\[ |A| = 0 \]