Line 6: | Line 6: | ||
-1& -1 & -1 | -1& -1 & -1 | ||
\end{bmatrix}</math> | \end{bmatrix}</math> | ||
+ | <math>\quad rank=1\ne \mbox 3</math><math>\quad ,Not\quad controllable</math> | ||
+ | |||
+ | <math>\mathbf{b)} \quad The reachable Subspace\quad is \begin{Bmatrix} | ||
+ | \begin{bmatrix} | ||
+ | 1 \\ | ||
+ | 1 \\ | ||
+ | -1 | ||
+ | \end{Bmatrix}.</math> | ||
+ | |||
+ | <math>\mathbf{c)} \quad 0=\begin{bmatrix} | ||
+ | 0 & 1 & 1 \\ | ||
+ | 0 & 0 & 0 \\ | ||
+ | 0 & 0 & 0 \\ | ||
+ | 0 & 0 & 0 | ||
+ | \end{bmatrix} | ||
+ | \quad Not \quad observable</math> | ||
+ | |||
+ | |||
+ | <math>\mathbf{d)} \quad X_1=r,X_2=-s,X_3=s \quad | ||
+ | |||
+ | X=r\begin{bmatrix} | ||
+ | 1 \\ | ||
+ | 0 \\ | ||
+ | 0 | ||
+ | \end{bmatrix}+s\begin{bmatrix} | ||
+ | 0 \\ | ||
+ | -1 \\ | ||
+ | 1 | ||
+ | \end{bmatrix}\\</math> | ||
+ | <math>Subspace\quad is \begin{Bmatrix} | ||
+ | \begin{bmatrix} | ||
+ | 1 \\ | ||
+ | 0 \\ | ||
+ | 0 | ||
+ | \end{bmatrix} ,& \begin{bmatrix} | ||
+ | 0 \\ | ||
+ | -1 \\ | ||
+ | 1 | ||
+ | \end{bmatrix} | ||
+ | \end{Bmatrix}.\\</math> | ||
+ | <math>\mathbf{e)} \quad \lambda I-A=\begin{bmatrix} | ||
+ | \lambda-1 & -1 & -1 \\ | ||
+ | 0 & \lambda & -1 \\ | ||
+ | 0 & 0 & \lambda+1 | ||
+ | \end{bmatrix}\quad \lambda_1=1,\lambda_2=0,\lambda_3=-1\\</math> | ||
+ | <math>\qquad for \;\lambda_1=1 \qquad\begin{bmatrix} | ||
+ | \lambda I-A & B \end{bmatrix}=\begin{bmatrix} | ||
+ | 0 & -1 & -1 & 1 \\ | ||
+ | 0 & 1 & -1 & 1 \\ | ||
+ | 0 & 0 & 2 & 0 | ||
+ | \end{bmatrix}\\</math> | ||
+ | <math>\qquad for \;\lambda_2=0 \qquad\begin{bmatrix} | ||
+ | \lambda I-A & B \end{bmatrix}=\begin{bmatrix} | ||
+ | -1 & -1 & -1 & 1 \\ | ||
+ | 0 & 0 & -1 & 1 \\ | ||
+ | 0 & 0 & 1 & 0 | ||
+ | \end{bmatrix}\\</math> | ||
+ | <math>\qquad for \;\lambda_3=-1 \qquad\begin{bmatrix} | ||
+ | \lambda I-A & B \end{bmatrix}=\begin{bmatrix} | ||
+ | -2 & -1 & -1 & 1 \\ | ||
+ | 0 & -1 & -1 & 1 \\ | ||
+ | 0 & 0 & 0 & 0 | ||
+ | \end{bmatrix} | ||
+ | \qquad rank<3 \qquad must\;contain\;\lambda=-1 | ||
+ | \qquad so\qquad No.\\</math> | ||
+ | <math>\mathbf{f)} \quad \begin{bmatrix} | ||
+ | \lambda I-A \\ | ||
+ | C \end{bmatrix}=\begin{bmatrix} | ||
+ | \lambda-1 & -1 & -1 \\ | ||
+ | 0 & \lambda & -1 \\ | ||
+ | 0 & 0 & \lambda+1 \\ | ||
+ | 0 & 1 & 1 | ||
+ | \end{bmatrix}</math> | ||
+ | |||
+ | <math>for\; \lambda_1=1 \quad \begin{bmatrix} | ||
+ | \lambda I-A \\ | ||
+ | C \end{bmatrix}=\begin{bmatrix} | ||
+ | 0 & -1 & -1 \\ | ||
+ | 0 & 1 & -1 \\ | ||
+ | 0 & 0 & 2 \\ | ||
+ | 0 & 1 & 1 | ||
+ | \end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_1=1</math> | ||
+ | |||
+ | <math>for\; \lambda_2=0 \quad \begin{bmatrix} | ||
+ | \lambda I-A \\ | ||
+ | C \end{bmatrix}=\begin{bmatrix} | ||
+ | -1 & -1 & -1 \\ | ||
+ | 0 & 0 & -1 \\ | ||
+ | 0 & 0 & 1 \\ | ||
+ | 0 & 1 & 1 | ||
+ | \end{bmatrix}\\</math> | ||
+ | |||
+ | <math>for \lambda-3=-1\quad \begin{bmatrix} | ||
+ | \lambda I-A \\ | ||
+ | C \end{bmatrix}=\begin{bmatrix} | ||
+ | -2 & -1 & -1 \\ | ||
+ | 0 & -1 & -1 \\ | ||
+ | 0 & 0 & 0 \\ | ||
+ | 0 & 1 & 1 | ||
+ | \end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_3=-1\\</math> |
Revision as of 23:29, 16 May 2017
$ \mathbf{a)} \quad C=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ -1& -1 & -1 \end{bmatrix} $ $ \quad rank=1\ne \mbox 3 $$ \quad ,Not\quad controllable $
$ \mathbf{b)} \quad The reachable Subspace\quad is \begin{Bmatrix} \begin{bmatrix} 1 \\ 1 \\ -1 \end{Bmatrix}. $
$ \mathbf{c)} \quad 0=\begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \quad Not \quad observable $
$ \mathbf{d)} \quad X_1=r,X_2=-s,X_3=s \quad X=r\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}+s\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}\\ $
$ Subspace\quad is \begin{Bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} ,& \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} \end{Bmatrix}.\\ $
$ \mathbf{e)} \quad \lambda I-A=\begin{bmatrix} \lambda-1 & -1 & -1 \\ 0 & \lambda & -1 \\ 0 & 0 & \lambda+1 \end{bmatrix}\quad \lambda_1=1,\lambda_2=0,\lambda_3=-1\\ $
$ \qquad for \;\lambda_1=1 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} 0 & -1 & -1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 2 & 0 \end{bmatrix}\\ $
$ \qquad for \;\lambda_2=0 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} -1 & -1 & -1 & 1 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}\\ $
$ \qquad for \;\lambda_3=-1 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} -2 & -1 & -1 & 1 \\ 0 & -1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \qquad rank<3 \qquad must\;contain\;\lambda=-1 \qquad so\qquad No.\\ $
$ \mathbf{f)} \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} \lambda-1 & -1 & -1 \\ 0 & \lambda & -1 \\ 0 & 0 & \lambda+1 \\ 0 & 1 & 1 \end{bmatrix} $
$ for\; \lambda_1=1 \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} 0 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \\ 0 & 1 & 1 \end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_1=1 $
$ for\; \lambda_2=0 \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} -1 & -1 & -1 \\ 0 & 0 & -1 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}\\ $
$ for \lambda-3=-1\quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} -2 & -1 & -1 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_3=-1\\ $