The correct option is(D): skew-symmetric.
Given, \(A=\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]\)
Then, \(A'=\left[ \begin{matrix} 3 & 1 \\ -4 & -1 \\ \end{matrix} \right]\)
Now, \(A-A'=\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 1 \\ -4 & -1 \\ \end{matrix} \right]\)
\(\Rightarrow\) \(A-A'=\left[ \begin{matrix} 0 & -5 \\ 5 & 0 \\ \end{matrix} \right]\) .. (i)
Now, we have \(A'-A=\left[ \begin{matrix} 3 & 1 \\ -4 & -1 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 5 \\ -5 & 0 \\ \end{matrix} \right]\)
\(\Rightarrow\) \((A'-A)'=\left[ \begin{matrix} 0 & -5 \\ 5 & 0 \\ \end{matrix} \right]=(A-A')\)
[From E (i)] which represent that \((A-A')\) is skew-symmetric matrix.
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:
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:
A matrix is a rectangular array of numbers, variables, symbols, or expressions that are defined for the operations like subtraction, addition, and multiplications. The size of a matrix is determined by the number of rows and columns in the matrix.
