By simplifying the determinant using row operations:
\( R_2 \rightarrow R_2 - R_1 \), \( R_3 \rightarrow R_3 - R_1 \)
we find that \( f(x) \) is constant. Therefore, \( f'(x) = 0 \).
Thus,
\[ \frac{1}{5} f'(0) = 0 \]
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
Let I be the identity matrix of order 3 × 3 and for the matrix $ A = \begin{pmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} $, $ |A| = -1 $. Let B be the inverse of the matrix $ \text{adj}(A \cdot \text{adj}(A^2)) $. Then $ |(\lambda B + I)| $ 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
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is: