Step 1: Characteristic Equation. We substitute \( A \) into the given equation: \[ A^2 - kA - 5I = 0 \] Calculating \( A^2 \):
\[A^2 = \begin{bmatrix} 1 & 3\\ 3 & 4 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix}\]Using matrix identity \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), substituting in the equation and solving for \( k \), we obtain: \[ k = 5 \]
Conclusion: Thus, the required value is \( 5 \), which corresponds to option \( \mathbf{(B)} \).
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 $ (1 + A)^5 $ 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
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: