If the Determinant |x−2, 2x−3, 3x−4; 2x−3, 3x−4, 4x−5; 3x−5, 5x−8, 10x−17| = Ax³ + Bx² + Cx + D, Then B + C Is Equal To

JEE Main 2018 — Previous Year Question.

Problem

If

$$\Delta = \begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ 2x-3 & 3x-4 & 4x-5 \\ 3x-5 & 5x-8 & 10x-17 \end{vmatrix} = Ax^3+Bx^2+Cx+D$$

then $B+C$ is equal to:

Key insight. Substituting specific numerical values of $x$ into the determinant turns a messy symbolic expansion into simple numerical evaluations — four values of $x$ are enough to pin down all four coefficients $A,B,C,D$ of the resulting cubic.

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Approach

Rather than expanding the determinant symbolically in $x$, evaluate $\Delta$ at four convenient values of $x$ — this gives four numerical equations in the unknown coefficients $A,B,C,D$ of the cubic $\Delta(x)=Ax^3+Bx^2+Cx+D$, which can then be solved as an ordinary system. Choosing $x=1$ and $x=2$ turns out to be especially convenient, since two of the determinant’s rows become identical at those values, making $\Delta$ vanish immediately.

Solution

Step 1 — Evaluate Δ at x = 0

$$\Delta(0) = \begin{vmatrix} -2 & -3 & -4 \\ -3 & -4 & -5 \\ -5 & -8 & -17 \end{vmatrix} = -2(68-40)+3(51-25)-4(24-20) = -56+78-16 = 6$$

So $D=\Delta(0)=6$.

Step 2 — Evaluate Δ at x = 1

$$\Delta(1) = \begin{vmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -2 & -3 & -7 \end{vmatrix}$$

Rows 1 and 2 are identical, so $\Delta(1)=0$.

Step 3 — Evaluate Δ at x = 2

$$\Delta(2) = \begin{vmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{vmatrix}$$

Rows 2 and 3 are identical, so $\Delta(2)=0$.

Step 4 — Evaluate Δ at x = 3

$$\Delta(3) = \begin{vmatrix} 1 & 3 & 5 \\ 3 & 5 & 7 \\ 4 & 7 & 13 \end{vmatrix} = 1(65-49)-3(39-28)+5(21-20) = 16-33+5 = -12$$

Step 5 — Set up the system for A, B, C

With $D=6$, and $\Delta(x)=Ax^3+Bx^2+Cx+6$:

$$\Delta(1)=0: \quad A+B+C=-6 \quad\text{…(i)}$$

$$\Delta(2)=0: \quad 8A+4B+2C=-6 \implies 4A+2B+C=-3 \quad\text{…(ii)}$$

$$\Delta(3)=-12: \quad 27A+9B+3C=-18 \implies 9A+3B+C=-6 \quad\text{…(iii)}$$

Step 6 — Eliminate C and solve

Subtracting (i) from (ii): $3A+B=3$ …(iv)

Subtracting (ii) from (iii): $5A+B=-3$ …(v)

Subtracting (iv) from (v): $2A=-6 \implies A=-3$

From (iv): $3(-3)+B=3 \implies B=12$

From (i): $-3+12+C=-6 \implies C=-15$

Step 7 — Compute B + C

$$B+C = 12+(-15) = -3$$

Answer

Common mistakes

• Trying to expand the determinant symbolically from scratch. The entries are linear in $x$ with awkward coefficients, and a direct cofactor expansion in terms of $x$ is long and error-prone — spotting that specific $x$-values make two rows identical (giving $\Delta=0$ instantly) is what makes this approach fast.
• Choosing x-values that don’t simplify anything. The whole point of picking $x=1$ and $x=2$ is that they create identical rows; picking arbitrary values like $x=4$ or $x=5$ still works in principle but requires computing full $3\times3$ determinants with larger numbers.
• Losing track of which equation came from which x-value when solving the system. With three equations to eliminate two variables in stages (first $C$, then $A$), it helps to label them explicitly, as in Steps 5–6, rather than trying to solve the system all at once.

Practise next

• For the same determinant $\Delta(x)=Ax^3+Bx^2+Cx+D$, find $A+D$.