Question

Find \( A^{-1} \), if \[ A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 3 & -1 \\ 1 & 0 & 1 \end{pmatrix} \] Hence, solve the following system of equations: \[ x + 2y + z = 5 \\ 2x + 3y = 1 \\ x - y + z = 8 \]

Hint:

When solving systems of linear equations, finding the inverse of the coefficient matrix allows you to solve for the variable matrix. The inverse is calculated using the adjugate and determinant.

Answer 1

Step 1: The given integral can be written as: \[ I = \int e^x \left( \frac{x}{\sqrt{1+x^2}} + \frac{1}{(1+x^2)^{\frac{3}{2}}} \right) dx \] Let: \[ f(x) = \frac{x}{\sqrt{1+x^2}} \] Step 2: Now, calculate the derivative of \( f(x) \): \[ f'(x) = \frac{\sqrt{1+x^2} - \frac{x \cdot x}{\sqrt{1+x^2}}}{1+x^2} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} \] Simplify the numerator: \[ f'(x) = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} = \frac{1}{(1+x^2)^{\frac{3}{2}}} \] Thus, the integral becomes: \[ I = \int e^x \left( f(x) + f'(x) \right) dx \] Step 3: Using the standard result: \[ \int e^x \left( f(x) + f'(x) \right) dx = e^x f(x) + C \] Substitute \( f(x) = \frac{x}{\sqrt{1+x^2}} \): \[ I = e^x \frac{x}{\sqrt{1+x^2}} + C \] Final Answer: \[ \boxed{I = e^x \frac{x}{\sqrt{1+x^2}} + C} \] Explanation: 1. Splitting the Integral: The given integral is split into terms containing \( \frac{x}{\sqrt{1+x^2}} \) and \( \frac{1}{(1+x^2)^{\frac{3}{2}}} \). 2. Defining \( f(x) \): The function \( f(x) \) is chosen as \( \frac{x}{\sqrt{1+x^2}} \) because its derivative results in the second term, \( \frac{1}{(1+x^2)^{\frac{3}{2}}} \). 3. Applying the Formula: The integral formula for \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \) is directly applied. 4. Substitution: Finally, substituting \( f(x) \) into the formula gives the result.