\[ \begin{pmatrix} 3 & i & 0 \\ -i & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]
\[ MN = \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} = 2 \]
\[ M^2 = M + 2I, \quad \text{where} \quad I \text{ denotes the } 3 \times 3 \text{ identity matrix.} \]
Let \( y(x) = x v(x) \) be a solution of the differential equation \[ x^2 \frac{d^2y}{dx^2} - 3x \frac{dy}{dx} + 3y = 0. \] If \( v(0) = 0 \) and \( v(1) = 1, \) then \( v(-2) \) is equal to .................
If \( y(x) \) is the solution of the initial value problem \[ \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} + 4y = 0, \quad y(0) = 2, \quad \frac{dy}{dx}(0) = 0, \] then \( y(\ln 2) \) is (round off to 2 decimal places) equal to ...............
\[ \begin{pmatrix} z \\ y \end{pmatrix} \]
\[ \begin{pmatrix} 0 & 0 & 2 \\ 1 & 0 & -4 \\ 0 & 1 & 3 \end{pmatrix} \]
\[ f(x, y) = \begin{cases} \frac{x^3 + y^3}{x^2 - y^2}, & x^2 - y^2 \neq 0 \\ 0, & x^2 - y^2 = 0 \end{cases} \]