Detailed Solution

This problem combines concepts from matrices and determinants, specifically focusing on the relationship between a matrix and its inverse. We'll use the fundamental property of determinants of inverse matrices and then apply determinant calculation and quadratic equation solving techniques.

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Key Concept: Determinant of an Inverse Matrix

The most crucial concept for this problem is the relationship between the determinant of an invertible matrix $A$ and the determinant of its inverse, $A^{-1}$. This relationship is given by: $\det(A^{-1}) = \frac{1}{\det(A)}$ This formula is valid provided that $\det(A) \neq 0$, which must be true for $A$ to be invertible and thus for its inverse $B$ to exist.

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Problem Breakdown

We are given a $3 \times 3$ matrix $B$ and told that it is the inverse of another $3 \times 3$ matrix $A$. This means $B = A^{-1}$. The matrix $B$ contains an unknown parameter $\alpha$: $B = \left[ {\begin{matrix} 5 & {2\alpha} & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & { - 1} \\ \end{matrix} } \right]$ We are also given a condition involving the determinant of $A$: $\det(A) + 1 = 0$. Our goal is to find the sum of all possible values of $\alpha$ that satisfy these conditions.

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Step-by-Step Solution

Step 1: Determine the value of $\det(A)$

• We are given the condition $\det(A) + 1 = 0$.
• To find $\det(A)$, we simply rearrange this equation: $\det(A) = -1$
• Why this step? This is our starting point. Knowing the numerical value of $\det(A)$ is essential because it allows us to use the key concept to find the numerical value of $\det(B)$.

Step 2: Relate $\det(B)$ to $\det(A)$ and find $\det(B)$

• We are given that $B = A^{-1}$.
• Using the key concept mentioned above, we can write the relationship between their determinants: $\det(B) = \det(A^{-1}) = \frac{1}{\det(A)}$
• Now, substitute the value of $\det(A) = -1$ (found in Step 1) into this equation: $\det(B) = \frac{1}{-1} = -1$
• Why this step? This step establishes the specific numerical value that the determinant of matrix $B$ must equal. Our next task will be to calculate $\det(B)$ in terms of $\alpha$ and then equate it to this numerical value to form an equation for $\alpha$.

Step 3: Calculate $\det(B)$ in terms of $\alpha$

• The matrix $B$ is: $B = \left[ {\begin{matrix} 5 & {2\alpha} & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & { - 1} \\ \end{matrix} } \right]$
• We will calculate the determinant of $B$ by expanding along the first row. The general formula for expanding a $3 \times 3$ determinant $\left| {\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} } \right|$ along the first row is $a(ei - fh) - b(di - fg) + c(dh - eg)$.
• Applying this formula to matrix $B$: $\det(B) = 5 \left| {\begin{matrix} 2 & 1 \\ 3 & { - 1} \end{matrix} } \right| - 2\alpha \left| {\begin{matrix} 0 & 1 \\ \alpha & { - 1} \end{matrix} } \right| + 1 \left| {\begin{matrix} 0 & 2 \\ \alpha & 3 \end{matrix} } \right|$
• Now, we calculate each $2 \times 2$ determinant:
  • First term: $5 \times ((2)(-1) - (1)(3)) = 5 \times (-2 - 3) = 5 \times (-5) = -25$
  • Second term: $-2\alpha \times ((0)(-1) - (1)(\alpha)) = -2\alpha \times (0 - \alpha) = -2\alpha \times (-\alpha) = 2\alpha^2$
  • Third term: $1 \times ((0)(3) - (2)(\alpha)) = 1 \times (0 - 2\alpha) = -2\alpha$
• Summing these results, we get the expression for $\det(B)$: $\det(B) = -25 + 2\alpha^2 - 2\alpha$ $\det(B) = 2\alpha^2 - 2\alpha - 25$
• Why this step? This step provides an algebraic expression for $\det(B)$ in terms of $\alpha$. By equating this expression to the numerical value of $\det(B)$ found in Step 2, we can form an equation to solve for $\alpha$.

Step 4: Form and Solve the Quadratic Equation for $\alpha$}

• From Step 2, we know that $\det(B)$ must be equal to $-1$.
• From Step 3, we found that $\det(B) = 2\alpha^2 - 2\alpha - 25$.
• Equating these two expressions for $\det(B)$: $2\alpha^2 - 2\alpha - 25 = -1$
• To solve this quadratic equation, we first rearrange it into the standard form $a\alpha^2 + b\alpha + c = 0$: $2\alpha^2 - 2\alpha - 25 + 1 = 0$ $2\alpha^2 - 2\alpha - 24 = 0$
• We can simplify this quadratic equation by dividing all terms by 2: $\alpha^2 - \alpha - 12 = 0$
• Why this step? This step translates the determinant equality into a standard algebraic equation. Solving this equation will give us the possible values of $\alpha$ that satisfy all the given conditions.

Step 5: Find the Sum of all Values of $\alpha$

• The quadratic equation we need to solve is $\alpha^2 - \alpha - 12 = 0$.
• This equation is in the standard form $a\alpha^2 + b\alpha + c = 0$, where $a=1$, $b=-1$, and $c=-12$.
• According to Vieta's formulas, for a quadratic equation $ax^2 + bx + c = 0$, the sum of its roots (values of $x$) is given by $-\frac{b}{a}$.
• In our case, the sum of all values of $\alpha$ is: $\text{Sum of } \alpha = - \frac{b}{a} = - \frac{(-1)}{1} = 1$
• Why this step? The question specifically asks for the sum of the values of $\alpha$, not the individual values. Vieta's formulas provide an efficient way to find this sum directly from the coefficients of the quadratic equation without needing to factorize it or use the quadratic formula to find the roots individually.

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Important Tips and Common Mistakes

1. Determinant Calculation Accuracy: Carefully expand the determinant, paying close attention to signs and arithmetic. A common mistake is an error in multiplying or subtracting terms, especially with negative numbers or variables.
2. Correct Determinant Property: Ensure you use $\det(A^{-1}) = \frac{1}{\det(A)}$ correctly. Do not confuse it with $\det(A^{-1}) = -\det(A)$ or other incorrect relationships.
3. Algebraic Manipulation: Be meticulous when rearranging the equation into standard quadratic form. Sign errors are frequent when moving terms across the equality.
4. Vieta's Formulas: Remember Vieta's formulas for quadratic equations. For $ax^2 + bx + c = 0$, the sum of roots is $-b/a$ and the product of roots is $c/a$. Using these can save significant time compared to finding individual roots and then summing them.

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Summary and Key Takeaway

This problem demonstrates a typical application of determinant properties in JEE Mathematics. By leveraging the relationship $\det(A^{-1}) = \frac{1}{\det(A)}$, we could translate the condition on matrix $A$ into a condition on matrix $B$. Calculating $\det(B)$ in terms of $\alpha$ led to a quadratic equation, whose roots represented the possible values of $\alpha$. Finally, Vieta's formulas provided an elegant way to find the sum of these values directly.

The final answer is $\boxed{1}$.