Let's carefully solve this problem step-by-step. --- ### **Step 1: Understand the problem** - The **system** is: \[ x' = A x \] where \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \\ -8 & -5 & 1 \\ 5 & 0 & 0 \end{bmatrix} \] (Note: The matrix appears to be 5x3, which suggests some confusion, but based on the typical structure, it's likely that the matrix is 3x3. Let's assume the correct matrix is a 3x3 matrix given the context and the initial condition vector size.) - The **general solution** is given as: \[ x(t) = c_1 e^{\alpha_0 t} \begin{bmatrix} \alpha \\ 0 \\ 2 \end{bmatrix} + c_2 e^{\lambda_1 t} \begin{bmatrix} 0 \\ \beta \\ 1 \end{bmatrix} + c_3 e^{\lambda_2 t} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \] - The **initial condition** is: \[ x(0) = \begin{bmatrix} \lambda_1 + \lambda_2 \\ \alpha \\ \beta \end{bmatrix} \] - The goal: **Find \( c_1 + c_2 + c_3 \), \( \lambda_1 \), \( \lambda_2 \), \( \alpha \), and \( \beta \).** --- ### **Step 2: Use initial condition to find relations** At \( t=0 \), the solution becomes: \[ x(0) = c_1 \begin{bmatrix} \alpha \\ 0 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ \beta \\ 1 \end{bmatrix} + c_3 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \] which equals: \[ \begin{bmatrix} \lambda_1 + \lambda_2 \\ \alpha \\ \beta \end{bmatrix} \] Matching components: 1. **First component:** \[ c_1 \alpha + c_2 \cdot 0 + c_3 \cdot 0 = \lambda_1 + \lambda_2 \] \[ c_1 \alpha = \lambda_1 + \lambda_2 \] **(Equation 1)** 2. **Second component:** \[ c_1 \cdot 0 + c_2 \beta + c_3 \cdot 1 = \alpha \] \[ c_2 \beta + c_3 = \alpha \] **(Equation 2)** 3. **Third component:** \[ c_1 \cdot 2 + c_2 \cdot 1 + c_3 \cdot 1 = \beta \] \[ 2 c_1 + c_2 + c_3 = \beta \] **(Equation 3)** --- ### **Step 3: Express \( c_1 \), \( c_2 \), \( c_3 \)** From Equation 1: \[ c_1 = \frac{\lambda_1 + \lambda_2}{\alpha} \] --- ### **Step 4: Express \( c_2 \) and \( c_3 \)** From Equation 2: \[ c_2 \beta + c_3 = \alpha \Rightarrow c_3 = \alpha - c_2 \beta \] From Equation 3: \[ 2 c_1 + c_2 + c_3 = \beta \] Substitute \( c_1 \) and \( c_3 \): \[ 2 \cdot \frac{\lambda_1 + \lambda_2}{\alpha} + c_2 + (\alpha - c_2 \beta) = \beta \] Simplify: \[ \frac{2 (\lambda_1 + \lambda_2)}{\alpha} + c_2 + \alpha - c_2 \beta = \beta \] Group \( c_2 \) terms: \[ \left( c_2 - c_2 \beta \right) + \left( \frac{2 (\lambda_1 + \lambda_2)}{\alpha} + \alpha \right) = \beta \] \[ c_2 (1 - \beta) + \left( \frac{2 (\lambda_1 + \lambda_2)}{\alpha} + \alpha \right) = \beta \] Solve for \( c_2 \): \[ c_2 (1 - \beta) = \beta - \frac{2 (\lambda_1 + \lambda_2)}{\alpha} - \alpha \] \[ c_2 = \frac{\beta - \frac{2 (\lambda_1 + \lambda_2)}{\alpha} - \alpha}{1 - \beta} \] Similarly, \( c_3 \): \[ c_3 = \alpha - c_2 \beta \] --- ### **Step 5: Find the sum \( c_1 + c_2 + c_3 \)** \[ c_1 + c_2 + c_3 = \frac{\lambda_1 + \lambda_2}{\alpha} + c_2 + (\alpha - c_2 \beta) \] Combine: \[ = \frac{\lambda_1 + \lambda_2}{\alpha} + c_2 (1 - \beta) + \alpha \] Recall from earlier: \[ c_2 (1 - \beta) = \beta - \frac{2 (\lambda_1 + \lambda_2)}{\alpha} - \alpha \] So: \[ c_1 + c_2 + c_3 = \frac{\lambda_1 + \lambda_2}{\alpha} + \left( \beta - \frac{2 (\lambda_1 + \lambda_2)}{\alpha} - \alpha \right) + \alpha \] Simplify: \[ = \frac{\lambda_1 + \lambda_2}{\alpha} + \beta - \frac{2 (\lambda_1 + \lambda_2)}{\alpha} \] \[ = \left( \frac{\lambda_1 + \lambda_2}{\alpha} - \frac{2 (\lambda_1 + \lambda_2)}{\alpha} \right) + \beta \] \[ = - \frac{\lambda_1 + \lambda_2}{\alpha} + \beta \] --- ### **### **Final expressions:** \[ \boxed{ \begin{aligned} c_1 + c_2 + c_3 &= - \frac{\lambda_1 + \lambda_2}{\alpha} + \beta \\ c_1 &= \frac{\lambda_1 + \lambda_2}{\alpha} \end{aligned} } \] --- ### **Step 6: Additional unknowns** From the problem statement, the eigenvalues are: - \( \lambda_1 \), - \( \lambda_2 \), - and the parameters \( \alpha \), \( \beta \). **Note:** To fully determine these parameters, typically, the eigenvalues \( \lambda_1,\lambda_2 \) and the eigenvector components relate to the matrix \( A \). --- ### **### **Conclusion:** - The sum of constants: \[ \boxed{ c_1 + c_2 + c_3 = - \frac{\lambda_1 + \lambda_2}{\alpha} + \beta } \] - The individual constants \( c_1, c_2, c_3 \) depend on the eigenvalues \( \lambda_1, \lambda_2 \) and parameters \( \alpha, \beta \). - Further steps require explicit eigenvalues of \( A \) to compute numerical values. --- ### **Summary:** - The problem involves matching initial conditions to the general solution to find relationships among constants. - Key relationships involve expressing constants in terms of eigenvalues and parameters. - The final expression links these constants via the initial condition. --- **If you need further clarification, specific eigenvalues, or numerical solutions, please provide the matrix \( A \) explicitly or additional context!**

Let's carefully solve the revised problem step-by-step. --- ## **Restated Problem** Given the system: \[ x' = A x \] where \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \\ -8 & -5 & 1 \\ 5 & 0 & 0 \end{bmatrix} \] But this matrix is **5x3**, which is not standard for a system of first-order ODEs for a 3-vector \( x \). Most likely, this is a typo or formatting issue, and the intended matrix is: \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \end{bmatrix} \] or \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ -2 & 8 & -5 \end{bmatrix} \] But let's proceed with the **standard 3x3 case**. The general solution is given as: \[ x(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2 + c_3 e^{\lambda_3 t} v_3 \] where \( v_i \) are eigenvectors, \( \lambda_i \) the eigenvalues. Given initial condition: \[ x(0) = \alpha \] (Assume this means a vector with entries involving \( \alpha \), \( \beta \), \( \lambda_1 \), \( \lambda_2 \), etc.) --- ## **Step 1: Find Eigenvalues (\( \lambda_i \))** Let’s use the matrix: \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \end{bmatrix} \] The characteristic equation is: \[ \left| A - \lambda I \right| = 0 \] \[ \left| \begin{array}{ccc} 1-\lambda & 0 & 0 \\ -2 & -\lambda & 0 \\ 1 & 0 & -2-\lambda \end{array} \right| = 0 \] This determinant can be computed as follows (expand along row 1): \[ (1-\lambda)\left| \begin{array}{cc} -\lambda & 0 \\ 0 & -2-\lambda \end{array} \right| - 0 + 0 \] \[ = (1-\lambda)[(-\lambda)(-2-\lambda) - 0] = (1-\lambda)(\lambda)(2+\lambda) \] So the characteristic equation is: \[ (1-\lambda)(\lambda)(2+\lambda) = 0 \] Thus the eigenvalues are: \[ \lambda_1 = 1,\quad \lambda_2 = 0,\quad \lambda_3 = -2 \] --- ## **Step 2: Find Eigenvectors (\( v_i \))** #### **For \( \lambda_1 = 1 \):** \[ A - I = \begin{bmatrix} 0 & 0 & 0 \\ -2 & -1 & 0 \\ 1 & 0 & -3 \end{bmatrix} \] Let \( v_1 = [a, b, c]^T \): - First row: \( 0\cdot a + 0\cdot b + 0\cdot c = 0 \) ⇒ free variable (let's set \( a = k \)) - Second row: \( -2a - b = 0 \) ⇒ \( b = -2a \) - Third row: \( a - 3c = 0 \) ⇒ \( a = 3c \) ⇒ \( c = a/3 \) Let \( a = 3 \), then \( c = 1 \), \( b = -6 \). So eigenvector: \[ v_1 = \begin{bmatrix} 3 \\ -6 \\ 1 \end{bmatrix} \] #### **For \( \lambda_2 = 0 \):** \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \end{bmatrix} \] Let \( v_2 = [a, b, c]^T \): - First row: \( a = 0 \) ⇒ \( a = 0 \) - Second row: \( -2a = 0 \) ⇒ (already satisfied) - Third row: \( 0 + 0 -2c = 0 \) ⇒ \( c = 0 \) So only \( b \) is free, let's set \( b = 1 \): \[ v_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \] #### **For \( \lambda_3 = -2 \):** \[ A + 2I = \begin{bmatrix} 3 & 0 & 0 \\ -2 & 2 & 0 \\ 1 & 0 & 0 \end{bmatrix} \] Let \( v_3 = [a, b, c]^T \): - First row: \( 3a = 0 \) ⇒ \( a = 0 \) - Second row: \( -2a + 2b = 0 \) ⇒ \( b = 0 \) - Third row: \( 0 + 0 = 0 \) (always satisfied) - \( c \) free, set \( c = 1 \): \[ v_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \] --- ## **Step 3: Write General Solution** \[ x(t) = c_1 e^{t} \begin{bmatrix} 3 \\ -6 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + c_3 e^{-2t} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \] --- ## **Step 4: Apply Initial Condition** Let \( x(0) = \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} \) (since it was just "α" in the question, but solution is a 3-vector). At \( t = 0 \): \[ x(0) = c_1 \begin{bmatrix} 3 \\ -6 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + c_3 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} \] So, - First entry: \( 3c_1 = \alpha \) ⇒ \( c_1 = \frac{\alpha}{3} \) - Second entry: \( -6c_1 + c_2 = \beta \) - Third entry: \( c_1 + c_3 = \gamma \) Plug in \( c_1 \): Second entry: \[ -6c_1 + c_2 = \beta \implies c_2 = \beta + 6c_1 = \beta + 6(\frac{\alpha}{3}) = \beta + 2\alpha \] Third entry: \[ c_1 + c_3 = \gamma \implies c_3 = \gamma - c_1 = \gamma - \frac{\alpha}{3} \] --- ## **Step 5: Compute \( c_1 + c_2 + c_3 \)** \[ c_1 + c_2 + c_3 = \frac{\alpha}{3} + (\beta + 2\alpha) + (\gamma - \frac{\alpha}{3}) \] \[ = \frac{\alpha}{3} - \frac{\alpha}{3} + 2\alpha + \beta + \gamma \] \[ = 2\alpha + \beta + \gamma \] --- ## **Step 6: Conclusion** ### **Summary Table** | Parameter | Value | |-------------------|----------------------------------------| | λ₁ | 1 | | λ₂ | 0 | | λ₃ | -2 | | c₁ | α/3 | | c₂ | β + 2α | | c₃ | γ - α/3 | | c₁ + c₂ + c₃ | **2α + β + γ** | --- ## **Final Answer** \[ \boxed{ c_1 + c_2 + c_3 = 2 \alpha + \beta + \gamma } \] - The eigenvalues are \( \lambda_1 = 1, \lambda_2 = 0, \lambda_3 = -2 \). - The constants \( c_1, c_2, c_3 \) are as above, found by matching the initial condition. - The sum \( c_1 + c_2 + c_3 = 2\alpha + \beta + \gamma \). **If you have a specific initial condition vector, plug in the values for \( \alpha, \beta, \gamma \) to get your final answer!**

Let's solve the problem step by step. --- ## **Step 1: Clarify the System** You wrote: > Let X = ce + Credit + Ce be the general solution of the system > 1 0 X-200 10 X -28-5 15 [ []] > Given initial condition X(0) = α > Solve to find c₁ + c₂ + c₃ > (Hint: First find λ₁, λ₂, α, and β) It appears the system is a 3×3 system of the form \[ \mathbf{X}' = A \mathbf{X} \] where \( A \) is a 3×3 matrix. From your numbers, the matrix is likely: \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \\ \end{bmatrix} \] (If you meant a different matrix, the method is the same.) --- ## **Step 2: Find Eigenvalues (\( \lambda \))** Find the characteristic polynomial: \[ \det(A - \lambda I) = 0 \] \[ = \begin{vmatrix} 1-\lambda & 0 & 0 \\ -2 & -\lambda & 0 \\ 1 & 0 & -2-\lambda \end{vmatrix} \] Expand along the first row: \[ (1-\lambda)\begin{vmatrix} -\lambda & 0 \\ 0 & -2-\lambda \end{vmatrix} = (1-\lambda)[(-\lambda)(-2-\lambda) - 0] = (1-\lambda)\lambda(2+\lambda) \] So the eigenvalues are: \[ \boxed{ \lambda_1 = 1,\quad \lambda_2 = 0,\quad \lambda_3 = -2 } \] --- ## **Step 3: Find Eigenvectors** ### For \( \lambda_1 = 1 \): \[ (A - I) \mathbf{v} = 0 \] \[ \begin{bmatrix} 0 & 0 & 0 \\ -2 & -1 & 0 \\ 1 & 0 & -3 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = 0 \] - First row: always 0 - Second: \(-2a - b = 0 \implies b = -2a\) - Third: \(a - 3c = 0 \implies a = 3c\) Let \( c = 1 \): \( a = 3, b = -6 \) > So, eigenvector: \( \mathbf{v}_1 = \begin{bmatrix}3\\-6\\1\end{bmatrix} \) --- ### For \( \lambda_2 = 0 \): \[ A \mathbf{v} = 0 \] \[ \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = 0 \] - First: \( a=0 \) - Second: \(-2a = 0\) (OK) - Third: \(0 - 2c = 0 \implies c=0\) Let \( b = 1 \): \( \mathbf{v}_2 = \begin{bmatrix}0\\1\\0\end{bmatrix} \) --- ### For \( \lambda_3 = -2 \): \[ (A + 2I)\mathbf{v} = 0 \] \[ \begin{bmatrix} 3 & 0 & 0 \\ -2 & 2 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = 0 \] - First: \(3a=0 \implies a=0\) - Second: \(2b=0 \implies b=0\) - Third: \(0=0\), so \( c \) is free. Let \( c=1 \): \( \mathbf{v}_3 = \begin{bmatrix}0\\0\\1\end{bmatrix} \) --- ## **Step 4: General Solution** \[ \mathbf{X}(t) = c_1 e^{t} \begin{bmatrix}3\\-6\\1\end{bmatrix} + c_2 \begin{bmatrix}0\\1\\0\end{bmatrix} + c_3 e^{-2t} \begin{bmatrix}0\\0\\1\end{bmatrix} \] --- ## **Step 5: Apply Initial Condition** Suppose \( \mathbf{X}(0) = \begin{bmatrix}\alpha \\ \beta \\ \gamma \end{bmatrix} \) (since it says "α", but it must be a 3-vector). At \( t=0 \): \[ \mathbf{X}(0) = c_1 \begin{bmatrix}3\\-6\\1\end{bmatrix} + c_2 \begin{bmatrix}0\\1\\0\end{bmatrix} + c_3 \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3c_1\\-6c_1 + c_2\\c_1 + c_3\end{bmatrix} = \begin{bmatrix}\alpha \\ \beta \\ \gamma \end{bmatrix} \] So, - \( 3c_1 = \alpha \implies c_1 = \frac{\alpha}{3} \) - \( -6c_1 + c_2 = \beta \implies c_2 = \beta + 6c_1 = \beta + 2\alpha \) - \( c_1 + c_3 = \gamma \implies c_3 = \gamma - c_1 = \gamma - \frac{\alpha}{3} \) --- ## **Step 6: Compute \( c_1 + c_2 + c_3 \)** \[ c_1 + c_2 + c_3 = \frac{\alpha}{3} + (\beta + 2\alpha) + (\gamma - \frac{\alpha}{3}) = \beta + 2\alpha + \gamma \] --- ## **Conclusion** **Step-by-step summary:** 1. Write the system and the general solution using eigenvalues/eigenvectors. 2. Find eigenvalues: \(1, 0, -2\). 3. Find eigenvectors: \([3, -6, 1]^T\), \([0, 1, 0]^T\), \([0, 0, 1]^T\). 4. Write the general solution as a linear combination. 5. Apply the initial condition to solve for \(c_1, c_2, c_3\). 6. The sum \(c_1 + c_2 + c_3 = 2\alpha + \beta + \gamma\). \[ \boxed{c_1 + c_2 + c_3 = 2\alpha + \beta + \gamma} \] If your initial condition is \(X(0) = \alpha\), then the answer is just \(2\alpha\). If your initial condition is \(X(0) = \begin{bmatrix}\alpha \\ \beta \\ \gamma \end{bmatrix}\), use the formula above. **This completes the step-by-step solution.**

Let's solve your problem **step by step**. --- ## **Step 1: Clarify the System** From your message, the system matrix appears to be: \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \end{bmatrix} \] and the general solution is a linear combination of exponentials times eigenvectors. --- ## **Step 2: Find the Eigenvalues (\( \lambda \))** The characteristic polynomial is: \[ \det(A - \lambda I) = \begin{vmatrix} 1-\lambda & 0 & 0 \\ -2 & -\lambda & 0 \\ 1 & 0 & -2-\lambda \end{vmatrix} \] Expand along the first row: \[ (1-\lambda) \cdot \begin{vmatrix} -\lambda & 0 \\ 0 & -2-\lambda \end{vmatrix} = (1-\lambda)\big[(-\lambda)(-2-\lambda)\big] = (1-\lambda)\lambda(2+\lambda) \] Set to zero: \[ (1-\lambda)\lambda(2+\lambda) = 0 \] So the **eigenvalues are:** \[ \boxed{ \lambda_1 = 1,\quad \lambda_2 = 0,\quad \lambda_3 = -2 } \] --- ## **Step 3: Find Eigenvectors** ### 1. For \( \lambda_1 = 1 \): \[ A - I = \begin{bmatrix} 0 & 0 & 0 \\ -2 & -1 & 0 \\ 1 & 0 & -3 \end{bmatrix} \] Let \( v_1 = [a, b, c]^T \): - Row 1: always 0. - Row 2: \(-2a - b = 0 \implies b = -2a\) - Row 3: \(a - 3c = 0 \implies a = 3c\) Let \( c = 1 \Rightarrow a = 3, b = -6 \). So, \[ v_1 = \begin{bmatrix}3\\-6\\1\end{bmatrix} \] --- ### 2. For \( \lambda_2 = 0 \): \[ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 1 & 0 & -2 \end{bmatrix} \] Let \( v_2 = [a, b, c]^T \): - Row 1: \(a = 0\) - Row 2: \(-2a = 0\) (OK) - Row 3: \(-2c = 0 \implies c = 0\) Let \( b = 1 \Rightarrow v_2 = [0, 1, 0]^T \) --- ### 3. For \( \lambda_3 = -2 \): \[ A + 2I = \begin{bmatrix} 3 & 0 & 0 \\ -2 & 2 & 0 \\ 1 & 0 & 0 \end{bmatrix} \] Let \( v_3 = [a, b, c]^T \): - Row 1: \(3a = 0 \implies a = 0\) - Row 2: \(2b = 0 \implies b = 0\) - Row 3: always 0. Let \( c = 1 \Rightarrow v_3 = [0, 0, 1]^T \) --- ## **Step 4: Write the General Solution** \[ X(t) = c_1 e^{t} \begin{bmatrix}3\\-6\\1\end{bmatrix} + c_2 \begin{bmatrix}0\\1\\0\end{bmatrix} + c_3 e^{-2t} \begin{bmatrix}0\\0\\1\end{bmatrix} \] --- ## **Step 5: Apply the Initial Condition** Suppose \( X(0) = \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} \) (if you are given only \( X(0) = \alpha \), you need all three components for a 3D system; let's proceed with variables). At \( t = 0 \): \[ X(0) = c_1 \begin{bmatrix}3\\-6\\1\end{bmatrix} + c_2 \begin{bmatrix}0\\1\\0\end{bmatrix} + c_3 \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix} 3c_1 \\ -6c_1 + c_2 \\ c_1 + c_3 \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} \] So, - \( 3c_1 = \alpha \implies c_1 = \frac{\alpha}{3} \) - \( -6c_1 + c_2 = \beta \implies c_2 = \beta + 6c_1 = \beta + 2\alpha \) - \( c_1 + c_3 = \gamma \implies c_3 = \gamma - c_1 = \gamma - \frac{\alpha}{3} \) --- ## **Step 6: Find \( c_1 + c_2 + c_3 \)** \[ c_1 + c_2 + c_3 = \frac{\alpha}{3} + (\beta + 2\alpha) + \left(\gamma - \frac{\alpha}{3}\right) = (\frac{\alpha}{3} - \frac{\alpha}{3} + 2\alpha) + \beta + \gamma = 2\alpha + \beta + \gamma \] --- ## **Conclusion** \[ \boxed{c_1 + c_2 + c_3 = 2\alpha + \beta + \gamma} \] --- **Summary of steps:** 1. Found the eigenvalues and eigenvectors of the matrix. 2. Wrote the general solution. 3. Matched the initial condition to get \( c_1, c_2, c_3 \). 4. Summed them for the answer. **If you have a specific initial condition vector (for example, \( X(0) = [5, 1, 4]^T \)), substitute those values into the formula.** If your initial condition is only \( X(0) = \alpha \) (a scalar), then the answer is \( 2\alpha \) (since it fills all three entries). --- **Final Answer:** \[ \boxed{c_1 + c_2 + c_3 = 2\alpha + \beta + \gamma} \] where \( X(0) = [\alpha, \beta, \gamma]^T \). --- If you clarify your initial condition, I can give the explicit value!