Question

(a) Consider the linear system $\mathbf{X}\mathbf{\text{'}}\mathbf{=}\mathit{A}\mathit{X}$of three first-order differential equations, where the coefficient matrix is and$\mathit{\lambda }\mathbf{=}\mathbf{2}$ is known to be an eigenvalue of multiplicity two. Find two different solutions of the system corresponding to this eigenvalue without using a special formula (such as (12) of Section 8.2).

$\mathit{A}\mathbf{=}\left(\begin{array}{ccc}5& 3& 3\\ 3& 5& 3\\ 5& -5& -3\end{array}\right)$

(b) Use the procedure of part (a) to solve

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)\mathit{X}$

Short answer

(a)The general solution for the given matrix is $X=c_1\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)e^{2t}+c_2\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)e^{2t}$.

(b)The general solution for the given matrix is $X=c_1\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)+c_2\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)+c_3\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)e^{3t}$.

Step-by-step solution

Definition of Eigenvalue

• Eigenvalues, also known as characteristic roots, characteristic values, appropriate values, or latent roots, are specific sets of scalars associated with a linear system of equations (i.e., a matrix equation).
• The Lanczos algorithm computes the eigenvalues and eigenvectors of large symmetric sparse matrices.

Find the general solution for the given matrix.

(a)

Given that, the matrix is $A=\left(\begin{array}{ccc}5& 3& 3\\ 3& 5& 3\\ 5& -5& -3\end{array}\right)$.

The eigenvalue, which is known to be an eigenvalue of multiplicity two, solving,

localid="1663928348637" role="math" $\begin{array}{rcl}\left(\mathrm{A}-2\mathrm{I}|0\right)& =& \left(\begin{array}{cccc}5-2& 3& 3& 0\\ 3& 5-2& 3& 0\\ -5& -5& -3-2& 0\end{array}\right)\\ & =& \left(\begin{array}{cccc}3& 3& 3& 0\\ 3& 3& 3& 0\\ -5& -5& -5& 0\end{array}\right)\end{array}$

Here it has localid="1663928103007" $3k_1+3k_2+3k_3=0$implies $k_1=-k_2-k_3$.

Choosing $k_2=0$ and $k_3=1$ yields $k_1=-1$. This gives an eigenvector and a corresponding solution vector which is written as:

localid="1663927118500" $K_1=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right),X_1=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)e^{2t}$

Choosing $k_2=-1$and $k_3=0$yields $k_1=1$. This gives an eigenvector and a corresponding solution vector which is written as:

$K_2=\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right),X_2=\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)e^{2t}$

Therefore, the general solution of the system is written as $\mathbf{X}=c_1{\mathbf{X}}_1+c_2{\mathbf{X}}_2$implies $c_1\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)e^{2t}+c_2\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)e^{2t}$.

Use the eigenvalue for the coefficient matrix

(b)

Given that, the system is $X\text{'}=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)X$.

Where the matrix is $A=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)$.

Obtaining the eigenvalues of the coefficient matrix,

role="math" localid="1663927590565" $$\begin{gathered} \det \left(\mathrm{A}-\mathrm{\lambda I}\right)=0 \\ \left|\begin{array}{ccc}1-\mathrm{\lambda }& 1& 1\\ 1& 1-\mathrm{\lambda }& 1\\ 1& 1& 1-\mathrm{\lambda }\end{array}\right|=0 \\ \left(1-\mathrm{\lambda }\right)\left|\begin{array}{cc}1-\mathrm{\lambda }& 1\\ 1& 1-\mathrm{\lambda }\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1-\mathrm{\lambda }& 1\end{array}\right|+1\left|\begin{array}{cc}1& 1-\mathrm{\lambda }\\ 1& 1\end{array}\right|=0 \\ \left(1-\mathrm{\lambda }\right)\left[{\left(1-\mathrm{\lambda }\right)}^2-1\right]-\left(1-\mathrm{\lambda }\right)+1+1-\left(1-\mathrm{\lambda }\right)=0 \end{gathered}$$

Further solve the above expression,

$$\begin{gathered} \left(1-\lambda \right)\left[{\left(1-\lambda \right)}^2-1\right]-1+\lambda +1+1-1+\lambda =0 \\ \left(1-\lambda \right)\left[{\left(1-\lambda \right)}^2-1\right]+2\lambda =0 \\ \left(1-\lambda \right)\left[1-2\lambda +{\lambda }^2-1\right]+2\lambda =0 \\ \left(1-\lambda \right)\left[{\lambda }^2-2\lambda \right]+2\lambda =0 \end{gathered}$$

Further, solve the above expression,

$$\begin{gathered} {\lambda }^2-2\lambda -{\lambda }^3+2{\lambda }^2+2\lambda =0 \\ -{\lambda }^3+3{\lambda }^2=0 \\ -{\lambda }^2\left(\lambda -3\right)=0 \end{gathered}$$

Eigenvalue ${\lambda }_1=0$ of multiplicity two, and ${\lambda }_2=3$.

Find the Eigenvector and solution vector.

For ${\lambda }_1=0$find the eigenvector;

$\begin{array}{rcl}\left(\mathrm{A}+0\mathrm{I}|0\right)& =& \left(\begin{array}{cccc}1-0& 1& 1& 0\\ 1& 1-0& 1& 0\\ 1& 1& 1-0& 0\end{array}\right)\\ & =& \left(\begin{array}{cccc}1& 1& 1& 0\\ 1& 1& 1& 0\\ 1& 1& 1& 0\end{array}\right)\end{array}$

Here it has $k_1+k_2+k_3=0$ implies $k_1=-\left(k_2+k_3\right)$.

Choosing $k_2=0$and $k_3=1$ yields $k_1=-1$. This gives an eigenvector, and a corresponding solution vector is role="math" localid="1663928280498" $K_1=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right),X_1=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)$.

Choosing $k_2=1$ and $k_3=-1$yields $k_1=0$. This gives an eigenvector, and a corresponding solution vector is role="math" localid="1663928501944" $K_2=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right),X_2=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$.

For k=3, the eigenvector is found as;

$\begin{array}{rcl}\left(\mathrm{A}+3\mathrm{I}|0\right)& =& \left(\begin{array}{cccc}1-3& 1& 1& 0\\ 1& 1-3& 1& 0\\ 1& 1& 1-3& 0\end{array}\right)\\ & =& \left(\begin{array}{cccc}-2& 1& 1& 0\\ 1& -2& 1& 0\\ 1& 1& -2& 0\end{array}\right)\end{array}$

Here it has $-2k_1+k_2+k_3=0$.

Choose $k_1=1,k_2=1$ and $k_3=1$. This gives an eigenvector, and a corresponding solution vector is $K_3=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),X_3=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)e^{3t}$.

Therefore, the general solution of the system is $\mathbf{X}=c_1{\mathbf{X}}_1+c_2{\mathbf{X}}_2+c_3{\mathbf{X}}_3$ implies$c_1\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)+c_2\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)+c_3\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)e^{3t}$