Question

If \[ \begin{vmatrix} -1 & -2 & 5 \\ -2 & a & 4 \\ 0 & 4 & 2a \end{vmatrix} = -86 \] then the sum of all possible values of \( a \) is

• A. 4
• B. 5
• C. $-4$
• D. 9

Hint:

For quadratic equations \(ax^2+bx+c=0\), sum of roots = \(-\frac{b}{a}\).

Answer 1

The Correct Option is A

To find the sum of all possible values of \( a \) for the determinant equation, we need to solve the following equation:

\[ \begin{vmatrix} -1 & -2 & 5 \\ -2 & a & 4 \\ 0 & 4 & 2a \end{vmatrix} = -86 \]

We will use the cofactor expansion along the first row to calculate the determinant:

1. Expand across the first row: \[ \det = (-1)((a)(2a) - (4)(4)) - (-2)((-2)(2a) - (4)(0)) + (5)((-2)(4) - (0)(a)) \]
2. Simplify the terms:
   • \((-1)(2a^2 - 16)\)
   • \(+ (-2)(-4a)\)
   • \(+ (5)(-8)\)
3. Calculate the terms:
   • \((-1)(2a^2 - 16) = -2a^2 + 16\)
   • \(8a\)
   • \(-40\)

Sum the terms to form the equation:

\[ -2a^2 + 16 + 8a - 40 = -86 \]

Simplifying the equation, we get:

\[-2a^2 + 8a - 24 = -86\]

Bring all terms to one side:

\[-2a^2 + 8a + 62 = 0\]

Divide the entire equation by \(-2\) to simplify:

\[a^2 - 4a - 31 = 0\]

Using the quadratic formula to solve for \( a \):

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substitute \( a = 1 \), \( b = -4 \), \( c = -31 \):

\[ a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-31)}}{2 \cdot 1} \]

Simplifying further gives:

\[ a = \frac{4 \pm \sqrt{16 + 124}}{2} \]

Calculate the discriminant:

\[ a = \frac{4 \pm \sqrt{140}}{2} \]

Therefore:

\[ a = \frac{4 \pm \sqrt{4 \cdot 35}}{2} = \frac{4 \pm 2\sqrt{35}}{2} \]

Hence the values of \( a \) are:

\[ a = 2 \pm \sqrt{35} \]

The sum of these values is simply:

\[ a = 2 + \sqrt{35} + 2 - \sqrt{35} = 4 \]

The correct answer is therefore that the sum of all possible values of \( a \) is 4.