Step 1: Find \( A^2 \)
To find \( A^2 \), we compute the matrix product of \( A \) with itself: \[ A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ 1 & 3 \end{bmatrix} \]
Step 2: Substitute \( A^2 \) into the equation \( A^2 + 7I = kA \)
The equation becomes: \[ \begin{bmatrix} 8 & 5 \\ 1 & 3 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = k \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \] Simplifying the left side: \[ \begin{bmatrix} 8 & 5 \\ 1 & 3 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ 1 & 10 \end{bmatrix} \] Now equating to \( kA \), we have: \[ \begin{bmatrix} 15 & 5 \\ 1 & 10 \end{bmatrix} = k \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \] Equating corresponding elements gives \( k = 5 \).
Step 3: Verify the options
The correct value of \( k \) is \( 5 \), matching option (C).
An instructor at the astronomical centre shows three among the brightest stars in a particular constellation. Assume that the telescope is located at \( O(0,0,0) \) and the three stars have their locations at points \( D, A, \) and \( V \), having position vectors: \[ 2\hat{i} + 3\hat{j} + 4\hat{k}, \quad 7\hat{i} + 5\hat{j} + 8\hat{k}, \quad -3\hat{i} + 7\hat{j} + 11\hat{k} \] respectively. Based on the above information, answer the following questions: