Statement (1) textbf I: begin vmatrix 1 & cosα & cosβ \\ cosα & 1 & cosγ \\ cosβ & cosγ & 1 end vmatrix = begin vmatrix 0 & cosα & cosβ \\ cosα & 0 & cosγ \\ cosβ & cosγ & 0 end vmatrix Rightarrow cos 2 α + cos 2 β + cos 2 γ = 3/2 Statement (2) textbf II: 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 + q Rightarrow p 2 = 196\, q 2

Mathematics · Determinants

JEE Main 2026 — 23 January, Morning Shift — Question 11

\begin{vmatrix} 0 & \cos\alpha & \cos\beta \ \cos\alpha & 0 & \cos\gamma \ \cos\beta & \cos\gamma & 0 \end{vmatrix}

\Rightarrow \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = \frac{3}{2}

Statement (2) $ \textbf{II: } $$\begin{vmatrix} x^{2}+x & x+1 & x-2 \\ 2x^{2}+3x-1 & 3x & 3x-3 \\ x^{2}+2x+3 & 2x-1 & 2x-1 \end{vmatrix}$$ = px + q $ $\Rightarrow p^{2} = 196\, q^{2}$

\cos\alpha = x, \quad \cos\beta = y, \quad \cos\gamma = z

\begin{vmatrix} 0 & x & y \\ x & 0 & z \\ y & z & 0 \end{vmatrix} = \begin{vmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1 \end{vmatrix}

\Rightarrow x^2 + y^2 + z^2 = 1

\Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

Statement I is false

\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 + q

Put $x=0$ ,

q= \begin{vmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 3 & -1 & -1 \end{vmatrix} =-12

Put $x=1$ ,

p+q= \begin{vmatrix} 2 & 2 & -1 \\ 4 & 3 & 3 \\ 6 & 1 & 1 \end{vmatrix} =42

\Rightarrow p=54

\frac{p^2}{q^2} = \left(\frac{54}{-12}\right)^2 \neq 196

\Rightarrow p^2 \ne 196\,q^2

Statement II is false

Correct option (1)

Answer key and solution verified before publishing.

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