CUET UG 2025 — Mathematics Algebra

The determinant equation is:

$$\begin{vmatrix} y & 2 & 3 \\ 1 & y & 1 \\ 3 & 2 & y \end{vmatrix} = 0$$​y13​2y2​31y​​=0

Expanding along the first row:

$$= y \begin{vmatrix} y & 1 \\ 2 & y \end{vmatrix} - 2 \begin{vmatrix} 1 & 1 \\ 3 & y \end{vmatrix} + 3 \begin{vmatrix} 1 & y \\ 3 & 2 \end{vmatrix}$$=y​y2​1y​​−2​13​1y​​+3​13​y2​​

$$= y(y^2 - 2) - 2(y - 3) + 3(2 - 3y)$$=y(y2−2)−2(y−3)+3(2−3y)

$$= y^3 - 2y - 2y + 6 + 6 - 9y$$=y3−2y−2y+6+6−9y

$$= y^3 - 13y + 12$$=y3−13y+12

$$= 0$$=0

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For a cubic equation $y^3 + ay^2 + by + c = 0$y3+ay2+by+c=0 with roots $\alpha, \beta, \gamma$α,β,γ, Vieta's formula states:

$$\alpha \times \beta \times \gamma = -c$$α×β×γ=−c

The equation is $y^3 + 0y^2 - 13y + 12 = 0$y3+0y2−13y+12=0, so $c = 12$c=12.

Product of all three roots $= -12$=−12

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If the three roots are $-4, \alpha, \beta$−4,α,β, then:

$$(-4) \times \alpha \times \beta = -12$$(−4)×α×β=−12

$$\alpha \times \beta = \dfrac{-12}{-4}$$α×β=−4−12​

$$\alpha \times \beta = 3$$α×β=3

Therefore, the product of the other two roots is $3$3.