The matrix \[\left[ {\begin{array}{*{20}{c}} 1&2&4 \\ 3&0&6 \\ 1&1&{\text{p}} \end{array}} \right]\] has one eigen value equal to 3. The sum of the other two eigen values is A. p B. p – 1 C. p – 2 D. p – 3

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

To find the eigenvalues of a matrix, we can use the characteristic polynomial. The characteristic polynomial of a matrix $A$ is given by

$$p(x) = \det(xI – A)$$

where $I$ is the identity matrix.

In this case, the characteristic polynomial is

$$p(x) = \det \begin{bmatrix} x – 1 & 2 & 4 \\ 3 & x – 0 & 6 \\ 1 & 1 & x – p \end{bmatrix} = (x – 1)(x – p)(x – 3)$$

Therefore, the eigenvalues of the matrix are $1$, $p$, and $3$.

Since one of the eigenvalues is $3$, the sum of the other two eigenvalues is $p$.