\( 8 \)
The matrix is given as: \[ A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \] The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is calculated as \( ad - bc \). For matrix \( A \): \[ a = 2, \ b = 3, \ c = 1, \ d = 4 \] \[ \det(A) = (2 \cdot 4) - (3 \cdot 1) = 8 - 3 = 5 \] Thus, the determinant of \( A \) is: \[ {5} \]
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to:
If $ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\27 & 28 & 27 \\1 & 1 & 1 \end{vmatrix} $, $ x \in \mathbb{R} $, then $ \frac{d^2y}{dx^2} + y $ is equal to