Q.There are two values of $a$ which make the determinant $\Delta = \begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = 86$, then the sum of these numbers is
(A) $4$
(B) $5$
(C) $-4$
(D) $9$

Expand the determinant, set it equal to $86$, and read off the sum of the roots.

Expanding along the first column (its bottom entry is $0$):

Set $\Delta = 86$: $\;2a^2+8a+44 = 86 \Rightarrow 2a^2+8a-42 = 0 \Rightarrow a^2+4a-21 = 0.$ …

The idea

A determinant with an unknown inside is just a polynomial in that unknown. Evaluate it, set it equal to the given number, and solve the resulting equation. Here that equation is a quadratic, so there are two values of $a$ and we only need their sum.

Expand the determinant

Because the $(3,1)$ entry is $0$, expanding along the first column is quickest:

The two $2\times2$ minors are

So …

Method: Solving a "Determinant Equals a Number" Equation for a Sum of Unknowns

When a determinant containing one unknown is set equal to a given number and the question only asks for the sum of the solutions (not each individual value), expand the determinant into a polynomial equation and read the sum off its coefficients — don't solve for each root separately if you don't have to.

Steps

Step 1: Expand the determinant along the row or column with the most zeros

Pick whichever row or column has a $0$ entry (or create one with a row/column operation) to shorten the cofactor expansion. Keep careful track of the alternating sign pattern for a column expansion — the cofactor of the $i$-th entry down a column carries sign $(-1)^{i+1}$, so the middle term is subtracted, not added.

Step 2: Set the resulting polynomial equal to the given value and simplify to standard form

Move everything to one side to get a polynomial equation in the unknown, typically a quadratic $pa^2+qa+r=0$ once you subtract the given determinant value from both sides.

Step 3: Use Vieta's formula instead of finding each root, when only the sum is needed …

Common Mistakes

Mistake 1: Using the wrong sign for the middle cofactor in a column/row expansion

Why it's wrong: expanding along a column, the cofactor signs alternate $+,-,+,\dots$ down the column, not all $+$. Treating the middle term's cofactor as $+$ instead of $-$ (or vice versa) flips the sign of one term in the polynomial and produces a wrong quadratic — and hence a wrong sum of roots. Correct approach: write out the $(-1)^{i+j}$ sign for each term explicitly before substituting the minors.

Mistake 2: Solving the full quadratic for individual roots and then mis-adding them …