\( 8 \)
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: \[ {5} \]
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |