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Question

# Solve the given initial-value problem.$\mathbf{X}^{\prime}=\left(\begin{array}{rrr}{1} & {-12} & {-14} \\ {1} & {2} & {-3} \\ {1} & {1} & {-2}\end{array}\right) \mathbf{X}, \quad \mathbf{X}(0)=\left(\begin{array}{r}{4} \\ {6} \\ {-7}\end{array}\right)$

Solution

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Determine the eigenvalues of the matrix $\mathbf{A}$.

\begin{align*} \det\left(\mathbf{A}-\lambda\mathbf{I}\right) &= 0\\ \begin{vmatrix} 1-\lambda & -12 & -14\\ 1 & 2-\lambda & -3\\ 1 & 1 & -2-\lambda \end{vmatrix} &= 0\\ \left(1-\lambda\right)\left[\left(2-\lambda\right)\left(-2-\lambda\right)-\left(-3 \right)\right]-\left(-12\right)\left[\left(-2-\lambda\right)-\left(-3 \right)\right]+\left(-14\right)\left[1-\left(2-\lambda\right)\right]&=0\\ \left(1-\lambda\right)\left(-4-2\lambda+2\lambda+\lambda^2+3\right)+12\left(-2-\lambda+3\right)-14\left(1-2+\lambda\right)&=0\\ \left(1-\lambda\right)\left(-1+\lambda^2\right)+12\left(1-\lambda\right)-14\left(-1+\lambda\right)&=0\\ -1+\lambda^2+\lambda-\lambda^3+12-12\lambda+14-14\lambda&=0\\ -\lambda^3+\lambda^2-25\lambda+25&=0\\ \lambda^3-\lambda^2+25\lambda-25&=0\\ \left(\lambda-1\right)\left(\lambda^2+25\right)&=0\\ \left(\lambda-1\right)\left(\lambda+5i\right)\left(\lambda-5i\right)&=0\\ \lambda_1=1\;\;\text{or}\;\;\lambda_2&=\pm5i \end{align*}

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