If `B=[{:(5,2alpha,1),(0,2,1),(alpha,3,-1):}]` is the inverse of a `3xx3` matrix A, then the sum of all values of `alpha` for which det `(A)+1=0` is:

Options

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

Explanation

To solve the problem, we need to find the sum of all values of \( \alpha \) for which the determinant of matrix \( A \) plus 1 equals zero, given that \( B \) is the inverse of \( A \).

### Step-by-Step Solution:

1. **Identify the Matrix \( B \)**:
\[ B = \begin{pmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{pmatrix} \]

2. **Find the Determinant of Matrix \( B \)**:
We will calculate the determinant of \( B \) using the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ \text{det}(B) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is represented as:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix \( B \):
- \( a = 5, b = 2\alpha, c = 1 \)
- \( d = 0, e = 2, f = 1 \)
- \( g = \alpha, h = 3, i = -1 \)

Plugging in these values:
\[ \text{det}(B) = 5(2 \cdot (-1) - 1 \cdot 3) - 2\alpha(0 \cdot (-1) - 1 \cdot \alpha) + 1(0 \cdot 3 - 2 \cdot \alpha) \]
Simplifying this:
\[ = 5(-2 - 3) - 2\alpha(0 - \alpha) + 1(0 - 2\alpha) \]
\[ = 5(-5) + 2\alpha^2 - 2\alpha \]
\[ = -25 + 2\alpha^2 - 2\alpha \]

3. **Relate Determinant of \( A \) and \( B \)**:
Since \( B \) is the inverse of \( A \), we have:
\[ \text{det}(A) = \text{det}(B) \]
Therefore:
\[ \text{det}(A) = -25 + 2\alpha^2 - 2\alpha \]

4. **Set Up the Equation**:
We need to find \( \alpha \) such that:
\[ \text{det}(A) + 1 = 0 \]
Substituting for \( \text{det}(A) \):
\[ -25 + 2\alpha^2 - 2\alpha + 1 = 0 \]
Simplifying this gives:
\[ 2\alpha^2 - 2\alpha - 24 = 0 \]

5. **Divide by 2**:
\[ \alpha^2 - \alpha - 12 = 0 \]

6. **Factor the Quadratic**:
We can factor this quadratic equation:
\[ (\alpha - 4)(\alpha + 3) = 0 \]

7. **Find the Roots**:
Setting each factor to zero gives:
\[ \alpha - 4 = 0 \quad \Rightarrow \quad \alpha = 4 \]
\[ \alpha + 3 = 0 \quad \Rightarrow \quad \alpha = -3 \]

8. **Sum of the Roots**:
The sum of the values of \( \alpha \) is:
\[ 4 + (-3) = 1 \]

### Final Answer:
The sum of all values of \( \alpha \) for which \( \text{det}(A) + 1 = 0 \) is \( \boxed{1} \).