If `x^(2)+y^(2)+z^(2)=1`, then what is the value of `|{:(1,z,-y),(-z,1,x),(y,-x,1):}|=` ?

Options

A. 0
B. 1
C. 2
D. 2-2 xyz

Explanation

To find the value of the determinant \( | \begin{pmatrix} 1 & z & -y \\ -z & 1 & x \\ y & -x & 1 \end{pmatrix} | \), we can follow these steps:

### Step 1: Write the determinant
We need to calculate the determinant of the matrix:
\[ D = \begin{pmatrix} 1 & z & -y \\ -z & 1 & x \\ y & -x & 1 \end{pmatrix} \]

### Step 2: Apply the determinant formula
The determinant of a 3x3 matrix
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]
is calculated as:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:
- \( a = 1, b = z, c = -y \)
- \( d = -z, e = 1, f = x \)
- \( g = y, h = -x, i = 1 \)

### Step 3: Calculate each part of the determinant
1. Calculate \( ei - fh \):
\[ ei = 1 \cdot 1 = 1, \quad fh = x \cdot (-x) = -x^2 \implies ei - fh = 1 + x^2 \]

2. Calculate \( di - fg \):
\[ di = -z \cdot 1 = -z, \quad fg = x \cdot y \implies di - fg = -z - xy \]

3. Calculate \( dh - eg \):
\[ dh = -z \cdot (-x) = zx, \quad eg = 1 \cdot y = y \implies dh - eg = zx - y \]

### Step 4: Substitute back into the determinant formula
Now substituting these values back into the determinant formula:
\[ D = 1(1 + x^2) - z(-z - xy) - y(zx - y) \]
Expanding this gives:
\[ D = 1 + x^2 + z^2 + zxy - yz + y^2 \]

### Step 5: Simplify using the given condition
From the problem, we know that \( x^2 + y^2 + z^2 = 1 \). Thus, we can substitute this into our expression:
\[ D = 1 + (x^2 + y^2 + z^2) + zxy - yz \]
This simplifies to:
\[ D = 1 + 1 + zxy - yz = 2 + zxy - yz \]

### Step 6: Final evaluation
Now, we can evaluate the determinant:
Since \( x^2 + y^2 + z^2 = 1 \), we can substitute this back into our determinant:
\[ D = 2 \]

Thus, the value of the determinant is:
\[ \boxed{2} \]