Let $A$ be a $2 \times 2$ matrix with non-zero entries and let $A^2 = I$,where $I$ is the $2 \times 2$ identity matrix. Define $tr(A) = \text{sum of diagonal elements of } A$ and $|A| = \text{determinant of matrix } A$.
Statement $-1: tr(A) = 0$
Statement $-2: \det(A) = 1$
Statement $-1: tr(A) = 0$
Statement $-2: \det(A) = 1$
- AStatement $-1$ is false,Statement $-2$ is true.
- BStatement $-1$ is true,Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$.
- CStatement $-1$ is true,Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$.
- DStatement $-1$ is true,Statement $-2$ is false.
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View SolutionAmong the statements:
$I$: If $\begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} = \begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{vmatrix}$, then $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=\frac{3}{2}$
$II$: If $\begin{vmatrix} x^{2}+x & x+1 & x-2 \\ 2x^{2}+3x-1 & 3x & 3x-3 \\ x^{2}+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = px+q$, then $p^{2}=196q^{2}$
$I$: If $\begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} = \begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{vmatrix}$, then $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=\frac{3}{2}$
$II$: If $\begin{vmatrix} x^{2}+x & x+1 & x-2 \\ 2x^{2}+3x-1 & 3x & 3x-3 \\ x^{2}+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = px+q$, then $p^{2}=196q^{2}$
DifficultJEE MAIN 2026
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