If
\[ A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \]
then verify that \( A(\text{adj} A) = |A| I \) and find \( A^{-1} \).
Step 1: To verify that \( A(\text{adj} A) = |A| I \), we first calculate the determinant \( |A| \) of matrix \( A \).
The determinant is given by:
\[ |A| = \begin{vmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{vmatrix} = 1 \cdot \begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} - 3 \cdot \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} \]
After performing the calculation, \( |A| = 2 \).
Step 2: The adjugate of \( A \), denoted \( \text{adj} A \), is the transpose of the cofactor matrix of \( A \). Calculate the cofactor matrix and then its transpose to find \( \text{adj} A \).
Step 3: Verify the identity \( A(\text{adj} A) = |A| I \), where \( I \) is the identity matrix.
Step 4: To find \( A^{-1} \), use the formula:
\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj} A \]
Substitute \( |A| = 2 \) and the calculated adjugate to find \( A^{-1} \).
Thus, the inverse of \( A \) is:
\[ A^{-1} = \frac{1}{2} \begin{bmatrix} -1 & 0 & 1 \\ -1 & 1 & 1 \\ 1 & -1 & -1 \end{bmatrix} \]
(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 $