Question

Solve the IVP.

Answer 1

I'm having difficulty computing the eigenvalues

Answer 2

since x(t) is a function of t
let x' = A x
for the characteristic equation use : A - k I = 0 here I is the 3 x 3 identity matrix

Answer 3

using det(A-kI) = 0
you'll get the equation: (1-k)(2-k)(3-k) = 0 so k = 1, 2, 3

Answer 4

now form a matrix equation : (A- k*I ) B = 0 here B is the column matrix which shall be used in the final solution in the form X = BC
\[\left[\begin{matrix}1-k & 1&2 \\ 0 &2-k&2\\-1 &1&3-k\end{matrix}\right]\left[\begin{matrix}b_1 \\ b_2\\b_3\end{matrix}\right]= 0\]
plug in the values of k we got one at a time and you'll get three sets of equations in b1,b2 and b3 from each of these sets
for example: when you plug in k = 1 \[\left[\begin{matrix}0 & 1&2 \\ 0 &0&2\\-1 &1&2\end{matrix}\right]\left[\begin{matrix}b_1 \\ b_2\\b_3\end{matrix}\right]= 0\]lets write these values of b1, b2, b3 obtained for k = 1 as \[b^1_1 , b^1_2, b^1_3\]similarly when you plug k = 2 you'll get \[b^2 _1,b^2 _2,b^2 _3\] and for k = 3 you'll get \[b^3 _1,b^3 _2,b^3 _3\]

Answer 5

\[\left[\begin{matrix}x_1\\x_2 \\x_3\end{matrix}\right]=\left[\begin{matrix}b^1_1&b^2_1&b^3_1\\b^1_2&b_2^2&b_2^3 \\b^1_3&b^2_3&b^3_3\end{matrix}\right]\left[\begin{matrix}C_1e^{k_1 t}\\C_2e^{k_2 t} \\C_3e^{k_3 t}\end{matrix}\right]\]

Answer 6

once you have the solution in this form you can use the initial values to find the coefficients C1, C2, C3

Answer 7

did you understand the method?