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).
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.