Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Show Hint
Correct Answer: 6
Solution and Explanation
1. Given that $A$ is an orthogonal matrix: \[ A^T = A^{-1} \] \[ A^2 = A^{-1} \]
2. Given $A^2 = A^T$: \[ A^3 = I \]
3. Calculate $(A + I)^3 + (A - I)^3 - 6A$: \[ (A + I)^3 + (A - I)^3 - 6A = 2(A^3 + 3A) - 6A = 2A^3 = 2I \]
4. Sum of the diagonal elements of $2I$: \[ 2I = \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \] \[ \text{Sum of diagonal elements} = 2 + 2 + 2 = 6 \] Therefore, the correct answer is (1) 6.
Top Questions on Matrices
- If \(A\) is a square matrix such that \(A^2 = A\), then \((I - A)^3\) is:
Let $ A $ be a $ 3 \times 3 $ matrix such that $ | \text{adj} (\text{adj} A) | = 81.
$ If $ S = \left\{ n \in \mathbb{Z}: \left| \text{adj} (\text{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)} \right\}, $ then the value of $ \sum_{n \in S} |A| (n^2 + n) $ is:- Let $ A $ be a matrix of order $ 3 \times 3 $ and $ |A| = 5 $. If $ |2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}, \quad \alpha, \beta, \gamma \in \mathbb{N} $ then $ \alpha + \beta + \gamma $ is equal to
- If \( A \) is a square matrix of order \( 3 \times 3 \), \( \det A = 3 \), then the value of \( \det(3A^{-1}) \) is:
- Which of the following statements is not correct?
Questions Asked in JEE Main exam
If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:
- JEE Main - 2025
- Trigonometric Identities
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
- JEE Main - 2025
- Area between Two Curves
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
- JEE Main - 2025
- Determinants
Let $C$ be the circle $x^2 + (y - 1)^2 = 2$, $E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on the $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$, and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$ is equal to
- JEE Main - 2025
- Coordinate Geometry
Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:
- JEE Main - 2025
- 3D Geometry