Let x and y be two vectors in a 3 dimensional space and denote their dot product. Then the determinant det \[\left[ {\begin{array}{*{20}{c}} { < {\text{x}},{\text{x}} > }&{ < {\text{x}},{\text{y}} > } \\ { < {\text{y}},{\text{x}} > }&{ < {\text{y}},{\text{y}} > } \end{array}} \right].\] A. is zero when x and y are linearly independent B. is positive when x and yare linearly independent C. is non-zero for all non-zero x and y D. is zero only when either x or y is zero

The correct answer is $\boxed{\text{C}}$.

The determinant of a 2×2 matrix can be computed using the formula:

$$\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = (a \times d) – (b \times c)$$

In this case, we have:

$$\det \begin{bmatrix} <{\text{x}},{\text{x}} > & <{\text{x}},{\text{y}} > \\ <{\text{y}},{\text{x}} > & <{\text{y}},{\text{y}} > \end{bmatrix} = (<{\text{x}},{\text{x}} > \times <{\text{y}},{\text{y}} >) – (<{\text{x}},{\text{y}} > \times <{\text{y}},{\text{x}} >)$$

We can use the definition of the dot product to rewrite this as:

$$\det \begin{bmatrix} <{\text{x}},{\text{x}} > & <{\text{x}},{\text{y}} > \\ <{\text{y}},{\text{x}} > & <{\text{y}},{\text{y}} > \end{bmatrix} = \|{\text{x}}\|^2 \|{\text{y}}\|^2 – <{\text{x}},{\text{y}} >^2$$

The norm of a vector is its length, so $\|{\text{x}}\|^2$ and $\|{\text{y}}\|^2$ are always positive. The dot product of two vectors is zero if and only if the vectors are orthogonal (i.e., they are perpendicular). Therefore, the determinant is zero if and only if either $x$ or $y$ is zero, or if $x$ and $y$ are orthogonal.

When $x$ and $y$ are linearly independent, they cannot be orthogonal, so the determinant cannot be zero. Therefore, the determinant is non-zero for all non-zero $x$ and $y$.