Let A = [[1,−2,α],[α,2,−1]] and B = [[2,α],[−1,2],[4,−5]], α ∈ C. Then the Absolute Value of the Sum of All Values of α for Which det(AB) = 0 Is

JEE Main 2022 — 30 June, Morning Shift. Previous Year Question.

Problem

Let $A=\begin{bmatrix}1&-2&\alpha\\\alpha&2&-1\end{bmatrix}$ and $B=\begin{bmatrix}2&\alpha\\-1&2\\4&-5\end{bmatrix}$, $\alpha\in\mathbb{C}$. Then the absolute value of the sum of all values of $\alpha$ for which $\det(AB)=0$ is:

Key insight. $A$ is $2\times3$ and $B$ is $3\times2$, so their product $AB$ is a plain $2\times2$ matrix — small enough that its determinant can be written out directly as a cubic in $\alpha$. Once one root of that cubic is spotted by inspection, factoring it out and using Vieta’s formula for the sum of roots avoids solving the remaining quadratic factor explicitly.

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Approach

Multiply the $2\times3$ matrix $A$ by the $3\times2$ matrix $B$ directly to get a $2\times2$ matrix with entries in terms of $\alpha$. Compute $\det(AB)$, which will be a cubic polynomial in $\alpha$. Find one root by inspection, factor it out, and use Vieta’s formula (sum of roots $=-\frac{\text{coefficient of }\alpha^2}{\text{coefficient of }\alpha^3}$) on the full cubic to get the sum of all roots without solving the remaining quadratic.

Solution

Step 1 — Compute the entries of AB

$$(AB)_{11} = (1)(2)+(-2)(-1)+(\alpha)(4) = 2+2+4\alpha = 4+4\alpha$$

$$(AB)_{12} = (1)(\alpha)+(-2)(2)+(\alpha)(-5) = \alpha-4-5\alpha = -4\alpha-4$$

$$(AB)_{21} = (\alpha)(2)+(2)(-1)+(-1)(4) = 2\alpha-2-4 = 2\alpha-6$$

$$(AB)_{22} = (\alpha)(\alpha)+(2)(2)+(-1)(-5) = \alpha^2+4+5 = \alpha^2+9$$

Step 2 — Write det(AB)

$$\det(AB) = (4+4\alpha)(\alpha^2+9) – (-4\alpha-4)(2\alpha-6)$$

Step 3 — Expand each product

$$(4+4\alpha)(\alpha^2+9) = 4\alpha^2+36+4\alpha^3+36\alpha = 4\alpha^3+4\alpha^2+36\alpha+36$$

$$(-4\alpha-4)(2\alpha-6) = -8\alpha^2+24\alpha-8\alpha+24 = -8\alpha^2+16\alpha+24$$

Step 4 — Subtract

$$\det(AB) = \big[4\alpha^3+4\alpha^2+36\alpha+36\big] – \big[-8\alpha^2+16\alpha+24\big]$$

$$= 4\alpha^3+12\alpha^2+20\alpha+12$$

Step 5 — Set det(AB) = 0 and simplify

Dividing by 4:

$$\alpha^3+3\alpha^2+5\alpha+3=0$$

Step 6 — Find one root by inspection

Testing $\alpha=-1$: $(-1)^3+3(-1)^2+5(-1)+3 = -1+3-5+3=0$ ✓

Step 7 — Use Vieta’s formula for the sum of all three roots

For a cubic $\alpha^3+3\alpha^2+5\alpha+3=0$, the sum of all roots is:

$$-\frac{\text{coefficient of }\alpha^2}{\text{coefficient of }\alpha^3} = -\frac{3}{1} = -3$$

Step 8 — Compute the absolute value

$$|{-3}| = 3$$

Answer

Common mistakes

• Trying to fully factor and solve the cubic for all three roots. Since the other two roots are complex ($-1\pm i\sqrt2$), finding them individually is unnecessary work — Vieta’s formula gives the sum of all roots directly from the cubic’s coefficients, without solving anything further.
• Making an arithmetic slip when expanding $(4+4\alpha)(\alpha^2+9)$ or the second product. Since both expansions feed into a single subtraction, a small error in either one changes every coefficient of the resulting cubic.

Practise next

• Let $P=\begin{bmatrix}2&-1&\alpha\\\alpha&1&-2\end{bmatrix}$ and $Q=\begin{bmatrix}1&\alpha\\-2&1\\3&-4\end{bmatrix}$, $\alpha\in\mathbb{C}$. Find the absolute value of the sum of all values of $\alpha$ for which $\det(PQ)=0$, using the same direct-multiplication and Vieta’s-formula approach.