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. 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 \( \text{det}(A) + 1 = 0 \), given that \( B \) is the inverse of matrix \( A \).

### Step 1: Understand the relationship between the determinants
Since \( B \) is the inverse of \( A \), we have:
\[ \text{det}(A) \cdot \text{det}(B) = 1 \]
Thus, we can express \( \text{det}(A) \) as:
\[ \text{det}(A) = \frac{1}{\text{det}(B)} \]

### Step 2: Calculate the determinant of matrix \( B \)
Given:
\[ B = \begin{pmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{pmatrix} \]
We will calculate \( \text{det}(B) \) 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 \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows.

Calculating:
\[ \text{det}(B) = 5 \left( 2 \cdot (-1) - 1 \cdot 3 \right) - 2\alpha \left( 0 \cdot (-1) - 1 \cdot \alpha \right) + 1 \left( 0 \cdot 3 - 2 \cdot \alpha \right) \]
\[ = 5(-2 - 3) - 2\alpha(0 - \alpha) + 1(0 - 2\alpha) \]
\[ = 5(-5) + 2\alpha^2 - 2\alpha \]
\[ = -25 + 2\alpha^2 - 2\alpha \]

### Step 3: Set up the equation based on the condition \( \text{det}(A) + 1 = 0 \)
From the relationship established earlier:
\[ \text{det}(A) + 1 = 0 \implies \frac{1}{\text{det}(B)} + 1 = 0 \]
This implies:
\[ \text{det}(B) = -1 \]

### Step 4: Set the determinant of \( B \) equal to -1
Now we set the determinant we calculated equal to -1:
\[ -25 + 2\alpha^2 - 2\alpha = -1 \]
Rearranging gives:
\[ 2\alpha^2 - 2\alpha - 24 = 0 \]
Dividing through by 2:
\[ \alpha^2 - \alpha - 12 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:
\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -12 \):
\[ \alpha = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \]
\[ = \frac{1 \pm \sqrt{1 + 48}}{2} \]
\[ = \frac{1 \pm \sqrt{49}}{2} \]
\[ = \frac{1 \pm 7}{2} \]
Thus, the solutions are:
\[ \alpha = \frac{8}{2} = 4 \quad \text{and} \quad \alpha = \frac{-6}{2} = -3 \]

### Step 6: Find the sum of all 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 \( \text{det}(A) + 1 = 0 \) is:
\[ \boxed{1} \]