Step 1: Compute \( \det(A) \)
\[ \det(A) = 2\begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} - (-3)\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} + 5\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix}. \] Simplify to find \( \det(A) \).Step 2: Compute \( \text{adj}(A) \)
Find the cofactor matrix of \( A \) and transpose it to get \( \text{adj}(A) \).Step 3: Compute \( A^{-1} \)
\[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A). \]The area of a parallelogram whose diagonals are given by $ \vec{u} + \vec{v} $ and $ \vec{v} + \vec{w} $, where:
$ \vec{u} = 2\hat{i} - 3\hat{j} + \hat{k}, \quad \vec{v} = -\hat{i} + \hat{k}, \quad \vec{w} = 2\hat{j} - \hat{k} $ is:
The direction ratios of the normal to the plane passing through the points
$ (1, 2, -3), \quad (1, -2, 1) \quad \text{and parallel to the line} \quad \frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z}{4} \text{ is:} $
(b) Order of the differential equation: $ 5x^3 \frac{d^3y}{dx^3} - 3\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 + y = 0 $