A scalar matrix is a matrix in which all the diagonal elements are equal, and all the off-diagonal elements are zero.
The given matrix $A$ is: \[ A = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{pmatrix} \] This matrix is not a scalar matrix because its diagonal elements are not all equal. Thus, we rule out the scalar matrix option.
Next, we observe that a symmetric matrix is one where $A = A^T$, and a skew-symmetric matrix is one where $A = -A^T$.
The given matrix $A$ is neither symmetric nor skew-symmetric, as it is not equal to its transpose and also not equal to the negative of its transpose. Therefore, the correct answer is that $A$ is a scalar matrix.
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 $ 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 \( A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix} , \ \alpha > 0 \), such that \( \det(A) = 0 \) and \( \alpha + \beta = 1. \) If \( I \) denotes the \( 2 \times 2 \) identity matrix, then the matrix \( (I + A)^8 \) is:
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
For the curve \( \sqrt{x} + \sqrt{y} = 1 \), find the value of \( \frac{dy}{dx} \) at the point \( \left(\frac{1}{9}, \frac{1}{9}\right) \).