Let's solve **2(i)** (system of equations for \( e^{it} \) and \( e^{-it} \) by row-reduction) in **2 steps**, with explanations after each step. --- ### **Step 1: Write the system as a matrix equation** Given: \[ \begin{cases} e^{it} + e^{-it} = 2\cosh(it) \\ e^{it} - e^{-it} = 2\sinh(it) \end{cases} \] Write this as a matrix equation \( Ax = b \): Let \( x = \begin{bmatrix} e^{it} \\ e^{-it} \end{bmatrix} \), \( b = \begin{bmatrix} 2\cosh(it) \\ 2\sinh(it) \end{bmatrix} \), and \( A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \): \[ \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^{it} \\ e^{-it} \end{bmatrix} = \begin{bmatrix} 2\cosh(it) \\ 2\sinh(it) \end{bmatrix} \] **Explanation:** We rewrote the given system of equations in matrix form \( Ax = b \), where \( x \) is the vector of unknowns and \( b \) is the vector of constants. --- ### **Step 2: Solve for \( e^{it} \) and \( e^{-it} \) by row-reduction** Let's add and subtract the two equations: - **Add:** \( (e^{it} + e^{-it}) + (e^{it} - e^{-it}) = 2\cosh(it) + 2\sinh(it) \) \[ 2e^{it} = 2\cosh(it) + 2\sinh(it) \implies e^{it} = \cosh(it) + \sinh(it) \] - **Subtract:** \( (e^{it} + e^{-it}) - (e^{it} - e^{-it}) = 2\cosh(it) - 2\sinh(it) \) \[ 2e^{-it} = 2\cosh(it) - 2\sinh(it) \implies e^{-it} = \cosh(it) - \sinh(it) \] **Explanation:** By adding and subtracting the equations, we isolate each variable. This method is equivalent to row-reduction in this 2x2 case. --- ### **Final Answer:** \[ \boxed{ e^{it} = \cosh(it) + \sinh(it), \qquad e^{-it} = \cosh(it) - \sinh(it) } \] Let me know if you want the next parts (ii)–(v) solved as well!

Let's proceed step-by-step with the additional parts, following your instruction to do it in 2 steps with explanations after each, and the final answer at the end. --- ### Part 6(2b) — Express the set of equations in matrix form \(A x = b\) **Step 1: Write the equations in matrix form** Given the equations: \[ \begin{cases} e^{it} + e^{-it} = 2 \cosh(it) \\ e^{it} - e^{-it} = 2 \sinh(it) \end{cases} \] Define the unknown vector: \[ x = \begin{bmatrix} e^{it} \\ e^{-it} \end{bmatrix} \] And the constants: \[ b = \begin{bmatrix} 2 \cosh(it) \\ 2 \sinh(it) \end{bmatrix} \] The coefficient matrix \(A\) is: \[ A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \] Therefore, the matrix equation is: \[ A x = b \] **Explanation:** This form consolidates the system into a compact matrix equation, suitable for solving via row-reduction or other matrix methods. --- ### Part 6(2b) — **Compute \(A^{-1}\) and solve \(A x = b\)** **Step 2: Find the inverse \(A^{-1}\) and solve for \(x\).** First, compute the determinant of \(A\): \[ \det(A) = (1)(-1) - (1)(1) = -1 - 1 = -2 \] The inverse of \(A\) is: \[ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} \] Simplify: \[ A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \] Now, multiply \(A^{-1}\) by \(b\): \[ x = A^{-1} b = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 2 \cosh(it) \\ 2 \sinh(it) \end{bmatrix} \] Compute each component: - \( e^{it} = \frac{1}{2} [1 \cdot 2 \cosh(it) + 1 \cdot 2 \sinh(it)] = \frac{1}{2} [2 \cosh(it) + 2 \sinh(it)] = \cosh(it) + \sinh(it) \) - \( e^{-it} = \frac{1}{2} [1 \cdot 2 \cosh(it) - 1 \cdot 2 \sinh(it)] = \frac{1}{2} [2 \cosh(it) - 2 \sinh(it)] = \cosh(it) - \sinh(it) \) --- ### **Final answers:** \[ \boxed{ \begin{aligned} e^{it} &= \cosh(it) + \sinh(it) \\ e^{-it} &= \cosh(it) - \sinh(it) \end{aligned} } \] **Note:** These are consistent with the identities for exponential and hyperbolic functions. --- ### **Summary:** - **Part 6(2b):** The system can be written as \(A x = b\) with the matrix \(A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\). - **Part 6(2b):** The inverse \(A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\). - **Part 6(2b):** The solutions for the unknowns are \( e^{it} = \cosh(it) + \sinh(it) \) and \( e^{-it} = \cosh(it) - \sinh(it) \). Let me know if you'd like to proceed with the remaining parts (iii), (iv), and (v).

Let's proceed systematically as per your instructions, doing each part in 2 steps with explanations after each, and providing the final answer at the end. --- ### Part 6(2b): Express the set of equations in matrix form \(A x = b\) **Step 1: Write the equations in matrix form** Given the system: \[ \begin{cases} e^{it} + e^{-it} = 2 \cosh(it) \\ e^{it} - e^{-it} = 2 \sinh(it) \end{cases} \] Define the unknown vector: \[ x = \begin{bmatrix} e^{it} \\ e^{-it} \end{bmatrix} \] Constants: \[ b = \begin{bmatrix} 2 \cosh(it) \\ 2 \sinh(it) \end{bmatrix} \] Coefficient matrix: \[ A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \] Thus, the system in matrix form is: \[ A x = b \] **Explanation:** This consolidates the two equations into a compact matrix notation \(A x = b\), suitable for solving via matrix methods. --- **Step 2: Compute \(A^{-1}\) and solve for \(x\)** Calculate the determinant of \(A\): \[ \det(A) = (1)(-1) - (1)(1) = -1 - 1 = -2 \] Find \(A^{-1}\): \[ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} \] Simplify: \[ A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \] Now, solve for \(x\): \[ x = A^{-1} b = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 2 \cosh(it) \\ 2 \sinh(it) \end{bmatrix} \] Calculate each component: - \( e^{it} = \frac{1}{2} [1 \cdot 2 \cosh(it) + 1 \cdot 2 \sinh(it)] = \cosh(it) + \sinh(it) \) - \( e^{-it} = \frac{1}{2} [1 \cdot 2 \cosh(it) - 1 \cdot 2 \sinh(it)] = \cosh(it) - \sinh(it) \) --- ### **Final answers:** \[ \boxed{ \begin{aligned} e^{it} &= \cosh(it) + \sinh(it) \\ e^{-it} &= \cosh(it) - \sinh(it) \end{aligned} } \] --- ### **Summary:** - The equations are put into matrix form \(A x = b\) with \(A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\). - The inverse \(A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\). - The solutions: \[ e^{it} = \cosh(it) + \sinh(it), \quad e^{-it} = \cosh(it) - \sinh(it) \] --- If you'd like, I can proceed similarly to cover parts (iii), (iv), and (v).