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

Options

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

Answer

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 matrix \( B \) is the inverse of matrix \( A \).

### Step-by-Step Solution:

1. **Understanding the relationship between \( A \) and \( B \)**:
Since \( B \) is the inverse of \( A \), we have:
\[ \det(B) = \frac{1}{\det(A)} \]
Therefore, we can express the determinant of \( A \) in terms of the determinant of \( B \):
\[ \det(A) = \frac{1}{\det(B)} \]

2. **Setting up the equation**:
We are given that:
\[ \det(A) + 1 = 0 \]
This implies:
\[ \det(A) = -1 \]
Substituting the expression for \( \det(A) \):
\[ \frac{1}{\det(B)} = -1 \implies \det(B) = -1 \]

3. **Finding the determinant of matrix \( B \)**:
The matrix \( B \) is given as:
\[ B = \begin{pmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{pmatrix} \]
We need to calculate \( \det(B) \).

4. **Calculating \( \det(B) \)**:
Using the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ \det(B) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]
For our matrix \( B \):
\[ a = 5, b = 2\alpha, c = 1, d = 0, e = 2, f = 1, g = \alpha, h = 3, i = -1 \]
Therefore:
\[ \det(B) = 5(2 \cdot (-1) - 1 \cdot 3) - (2\alpha)(0 \cdot (-1) - 1 \cdot \alpha) + 1(0 \cdot 3 - 2 \cdot \alpha) \]
Simplifying each term:
- First term: \( 5(-2 - 3) = 5(-5) = -25 \)
- Second term: \( -2\alpha(-\alpha) = 2\alpha^2 \)
- Third term: \( -2\alpha \)

Combining these:
\[ \det(B) = -25 + 2\alpha^2 - 2\alpha \]

5. **Setting the determinant equal to -1**:
Now, we set the determinant equal to -1:
\[ -25 + 2\alpha^2 - 2\alpha = -1 \]
Rearranging gives:
\[ 2\alpha^2 - 2\alpha - 24 = 0 \]
Dividing the entire equation by 2:
\[ \alpha^2 - \alpha - 12 = 0 \]

6. **Factoring the quadratic equation**:
The quadratic can be factored as:
\[ (\alpha - 4)(\alpha + 3) = 0 \]
Thus, the solutions are:
\[ \alpha = 4 \quad \text{and} \quad \alpha = -3 \]

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

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