Questions/Maths/Matrices & DeterminantsMathsJEE Main 23 Jan 2026, Shift 1·Q11·MCQ·ModerateMatrices & Determinants · Evaluation of determinantsAmong the statements:I: If ∣1cosαcosβcosα1cosγcosβcosγ1∣=∣0cosαcosβcosα0cosγcosβcosγ0∣\begin{vmatrix} 1 & \cos\alpha & \cos\beta\\ \cos\alpha & 1 & \cos\gamma\\ \cos\beta & \cos\gamma & 1\end{vmatrix} = \begin{vmatrix} 0 & \cos\alpha & \cos\beta\\ \cos\alpha & 0 & \cos\gamma\\ \cos\beta & \cos\gamma & 0\end{vmatrix}1cosαcosβcosα1cosγcosβcosγ1=0cosαcosβcosα0cosγcosβcosγ0, then cos2α+cos2β+cos2γ=32\cos^2\alpha + \cos^2\beta + \cos^2\gamma = \dfrac32cos2α+cos2β+cos2γ=23, andII: If ∣x2+xx+1x−22x2+3x−13x3x−3x2+2x+32x−12x−1∣=px+q\begin{vmatrix} x^2 + x & x + 1 & x - 2\\ 2x^2 + 3x - 1 & 3x & 3x - 3\\ x^2 + 2x + 3 & 2x - 1 & 2x - 1\end{vmatrix} = px + qx2+x2x2+3x−1x2+2x+3x+13x2x−1x−23x−32x−1=px+q, then p2=196q2p^2 = 196q^2p2=196q2,(1)both are false(2)only II is true(3)both are true(4)only I is trueShow answerVideo solution coming soonWe're animating every PYQ step by step. Subscribe to get them first. Subscribe Previous in paperMaths Q10 · Definite IntegrationNext in paper Maths Q12 · Binomial TheoremSee the full 23 Jan 2026, Shift 1 paper