The value of the determinant \[ \begin{vmatrix} 2 & 3 & 5 \\ 1 & 0 & 4 \\ 7 & 2 & 1 \end{vmatrix} \] is:
Let the determinant be \(\Delta\): \[ \Delta = \begin{vmatrix} 2 & 3 & 5 \\ 1 & 0 & 4 \\ 7 & 2 & 1 \end{vmatrix} \] We expand along the first row (Row 1): \[ \Delta = 2 \cdot \begin{vmatrix} 0 & 4 \\ 2 & 1 \end{vmatrix} - 3 \cdot \begin{vmatrix} 1 & 4 \\ 7 & 1 \end{vmatrix} + 5 \cdot \begin{vmatrix} 1 & 0 \\ 7 & 2 \end{vmatrix} \] Calculate each minor: \[ \begin{vmatrix} 0 & 4 \\ 2 & 1 \end{vmatrix} = (0)(1) - (4)(2) = 0 - 8 = -8 \] \[ \begin{vmatrix} 1 & 4 \\ 7 & 1 \end{vmatrix} = (1)(1) - (4)(7) = 1 - 28 = -27 \] \[ \begin{vmatrix} 1 & 0 \\ 7 & 2 \end{vmatrix} = (1)(2) - (0)(7) = 2 - 0 = 2 \] Substitute back: \[ \Delta = 2 \times (-8) - 3 \times (-27) + 5 \times 2 = -16 + 81 + 10 = 75 \] Note: The calculated determinant is \(\boxed{75}\), which does not match any of the given options (69, -69, 87, -87). Rechecking with expansion along the second row confirms the same result: \[ \Delta = -1 \cdot \begin{vmatrix} 3 & 5 \\ 2 & 1 \end{vmatrix} + 0 - 4 \cdot \begin{vmatrix} 2 & 3 \\ 7 & 2 \end{vmatrix} \] Calculate minors: \[ \begin{vmatrix} 3 & 5 \\ 2 & 1 \end{vmatrix} = 3 \times 1 - 5 \times 2 = 3 - 10 = -7 \] \[ \begin{vmatrix} 2 & 3 \\ 7 & 2 \end{vmatrix} = 2 \times 2 - 3 \times 7 = 4 - 21 = -17 \] Then, \[ \Delta = -1 \times (-7) - 4 \times (-17) = 7 + 68 = 75 \] Therefore, the determinant is \(\boxed{75}\). If the answer key states (c) 87, there may be a typographical error in the matrix or options. \[ \boxed{75 \text{ (calculated determinant)}} \]
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: