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 integral is represented 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: The derivative of \( f(x) \) is calculated: \[ f'(x) = \frac{\sqrt{1+x^2} - \frac{x \cdot x}{\sqrt{1+x^2}}}{1+x^2} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} \] Simplifying the numerator yields: \[ f'(x) = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} = \frac{1}{(1+x^2)^{\frac{3}{2}}} \] Consequently, the integral transforms to: \[ I = \int e^x \left( f(x) + f'(x) \right) dx \] Step 3: Applying the standard integral formula: \[ \int e^x \left( f(x) + f'(x) \right) dx = e^x f(x) + C \] Substituting \( f(x) = \frac{x}{\sqrt{1+x^2}} \) gives: \[ I = e^x \frac{x}{\sqrt{1+x^2}} + C \] Final Answer: \[ \boxed{I = e^x \frac{x}{\sqrt{1+x^2}} + C} \] Explanation: 1. Integral Decomposition: The initial integral is separated into terms involving \( \frac{x}{\sqrt{1+x^2}} \) and \( \frac{1}{(1+x^2)^{\frac{3}{2}}} \). 2. Function Definition: \( f(x) \) is defined as \( \frac{x}{\sqrt{1+x^2}} \) because its derivative is the second term of the integral, \( \frac{1}{(1+x^2)^{\frac{3}{2}}} \). 3. Formula Application: The integral formula \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \) is directly utilized. 4. Result Derivation: The final result is obtained by substituting the defined \( f(x) \) into the applied formula.