First, calculate \( 6A \): \[ 6A = 6 \begin{bmatrix} 3 & 6 \\ -5 & 4 \end{bmatrix} = \begin{bmatrix} 18 & 36 \\ -30 & 24 \end{bmatrix}. \]
Next, calculate \( 5B \): \[ 5B = 5 \begin{bmatrix} 7 & 8 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 35 & 40 \\ 25 & 30 \end{bmatrix}. \]
Now subtract \( 5B \) from \( 6A \): \[ 6A - 5B = \begin{bmatrix} 18 & 36 \\ -30 & 24 \end{bmatrix} - \begin{bmatrix} 35 & 40 \\ 25 & 30 \end{bmatrix}. \]
Simplify by subtracting corresponding elements: \[ 6A - 5B = \begin{bmatrix} 18 - 35 & 36 - 40 \\ -30 - 25 & 24 - 30 \end{bmatrix} = \begin{bmatrix} -17 & -4 \\ -55 & -6 \end{bmatrix}. \]
Therefore, the result is: \[ \boxed{ 6A - 5B = \begin{bmatrix} -17 & -4 \\ -55 & -6 \end{bmatrix}. } \]
Correct Answer:
(C) \( \begin{bmatrix} -17 & -4 \\ -55 & -6 \end{bmatrix} \)
If \( A \), \( B \), and \( \left( \text{adj}(A^{-1}) + \text{adj}(B^{-1}) \right) \) are non-singular matrices of the same order, then the inverse of \[ A \left( \text{adj}(A^{-1}) + \text{adj}(B^{-1}) \right) B \] is equal to:
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to: