matrix - 5

logistic_guy

logistic_guy

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\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \bold{X}' = \begin{bmatrix}2 & 1 & 6 \\0 & 2 & 5 \\0 & 0 & 2\end{bmatrix}\bold{X}\)
 
logistic_guy said:
\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \bold{X}' = \begin{bmatrix}2 & 1 & 6 \\0 & 2 & 5 \\0 & 0 & 2\end{bmatrix}\bold{X}\)
Click to expand...

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
We start by finding the eigenvalues.

\(\displaystyle \left| \begin{array}{ccc}2 - \lambda & 1 & 6 \\0 & 2 - \lambda & 5 \\0 & 0 & 2 - \lambda\end{array}\right| = 0\)

Or

\(\displaystyle (2 - \lambda)\bigg[(2 - \lambda)(2 - \lambda) - 5(0)\bigg] - 1\bigg[0(2 - \lambda) - 5(0)\bigg] + 6\bigg[0(0) - (2 - \lambda)(0)\bigg] = 0\)

Or

\(\displaystyle = (2 - \lambda)(2 - \lambda)(2 - \lambda) = 0\)

Or

\(\displaystyle = (2 - \lambda)^3 = 0\)

This gives:

\(\displaystyle \lambda_1 = \lambda_2 = \lambda_3 = 2\)
 
At this point, the idea is to find the matrices \(\displaystyle \bold{K}\), \(\displaystyle \bold{P}\), and \(\displaystyle \bold{Q}\).

Since \(\displaystyle \lambda\) has multiplicity three, some of the matrices above will depend on the one before them.

We start with this system of equations:

\(\displaystyle k_2 + 6k_3 = 0\)
\(\displaystyle 5k_3 = 0\)

This gives:

\(\displaystyle k_2 = 0\)
\(\displaystyle k_3 = 0\)

Then

\(\displaystyle k_1\) can be anything. We will choose \(\displaystyle k_1 = 1\).

Then, we got the first matrix.

\(\displaystyle \bold{K} = \begin{bmatrix}1\\0\\0\end{bmatrix}\)
 
Now the matrix \(\displaystyle \bold{P}\) depends on the previous matrix \(\displaystyle \bold{K}\).

We have this system to solve.

\(\displaystyle p_2 + 6p_3 = k_1\)
\(\displaystyle 5p_3 = k_2\)

Or

\(\displaystyle p_2 + 6p_3 = 1\)
\(\displaystyle 5p_3 = 0\)

This gives:

\(\displaystyle p_3 = 0\)

then,

\(\displaystyle p_2 = 1\)

We can choose anything for \(\displaystyle p_1\). The simplest choice is \(\displaystyle p_1 = 0 \longrightarrow\) we are okay with this choice as long as our \(\displaystyle \bold{P}\) matrix is not trivial, ie \(\displaystyle \bold{P} \neq (0,0,0)\).

Then, we have:

\(\displaystyle \bold{P} = \begin{bmatrix}0\\1\\0\end{bmatrix}\)
 
Now the matrix \(\displaystyle \bold{Q}\) depends on the previous matrix \(\displaystyle \bold{P}\).

We have this system to solve.

\(\displaystyle q_2 + 6q_3 = p_1\)
\(\displaystyle 5q_3 = p_2\)

Or

\(\displaystyle q_2 + 6q_3 = 0\)
\(\displaystyle 5q_3 = 1\)

This gives:

\(\displaystyle q_3 = \frac{1}{5}\)

\(\displaystyle q_2 = -\frac{6}{5}\)

\(\displaystyle q_1\) can be anything. We will choose \(\displaystyle q_1 = 0\).

Then, we have:

[imath]\mathbf{Q} = \left[ \begin{array}{c} \phantom{-}0 \\[5pt] -\frac{6}{5} \\[5pt] \phantom{-}\frac{1}{5} \end{array} \right][/imath]
 
Now is the crucial part. We have a beautiful \(\displaystyle \textcolor{black}{\bold{Theorem}}\) which says:

If \(\displaystyle \lambda\) has multiplicity \(\displaystyle 2 \longrightarrow \lambda_1 = \lambda_2\)

Then,

The first solution is:
\(\displaystyle \bold{X}_1 = c_1\bold{K}e^{\lambda_1t}\)

The second solution is:
\(\displaystyle \bold{X}_2 = c_2\left(\bold{K}te^{\lambda_1t} + \bold{P}e^{\lambda_1t}\right)\)

If \(\displaystyle \lambda\) has multiplicity \(\displaystyle 3 \longrightarrow \lambda_1 = \lambda_2 = \lambda_3\)

Then,

The third solution is:
\(\displaystyle \bold{X}_3 = c_3\left(\bold{K}\frac{t^2}{2}e^{\lambda_1t} + \bold{P}te^{\lambda_1t} + \bold{Q}e^{\lambda_1t}\right)\)

If you look carefully at the pattern, you will be a able to solve a system even if \(\displaystyle \lambda\) has multiplicity \(\displaystyle \textcolor{red}{\bold{1}} \ \bold{million}\).

logistic_guy said:
\(\displaystyle \bold{X}' = \begin{bmatrix}2 & 1 & 6 \\0 & 2 & 5 \\0 & 0 & 2\end{bmatrix}\bold{X}\)
Click to expand...
Then, the general solution to this system is:

\(\displaystyle \bold{X} = \bold{X}_1 + \bold{X}_2 + \bold{X}_3\)

Or

\(\displaystyle \bold{X} = c_1\bold{K}e^{\lambda_1t} + c_2\left(\bold{K}te^{\lambda_1t} + \bold{P}e^{\lambda_1t}\right) + c_3\left(\bold{K}\frac{t^2}{2}e^{\lambda_1t} + \bold{P}te^{\lambda_1t} + \bold{Q}e^{\lambda_1t}\right)\)

Or

[imath]\mathbf{X} = \textcolor{red}{c_1}\begin{bmatrix}1\\[5pt]0\\[5pt]0\end{bmatrix}e^{2t} + \textcolor{blue}{c_2} \left( \begin{bmatrix}1\\[5pt]0\\[5pt]0\end{bmatrix} t e^{2t} + \begin{bmatrix}0\\[5pt]1\\[5pt]0\end{bmatrix} e^{2t} \right) + \textcolor{green}{c_3} \left( \begin{bmatrix}1\\[5pt]0\\[5pt]0\end{bmatrix} \frac{t^2}{2} e^{2t} + \begin{bmatrix}0\\[5pt]1\\[5pt]0\end{bmatrix} t e^{2t} + \begin{bmatrix} \phantom{-}0 \\[5pt] -\frac{6}{5} \\[5pt] \phantom{-}\frac{1}{5} \end{bmatrix} e^{2t} \right)[/imath]
 
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