(Created page with "AC-2 P1.")
 
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AC-2
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AC-2 P1.
P1.
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<math>\mathbf{a)} \quad C=\begin{bmatrix}
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B & AB & A^2B
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\end{bmatrix}=\begin{bmatrix}
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1 & 2 & 2 \\
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1 & 0 & 0 \\
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0 & 0 & 0
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\end{bmatrix}</math>
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<math>\quad rank=2\ne \mbox 3</math>
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<math>\quad Not\quad controllable</math>
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<math>\mathbf{b)} \quad Subspace\quad is \begin{Bmatrix}
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\begin{bmatrix}
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1 \\
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1 \\
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0
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\end{bmatrix} ,& \begin{bmatrix}
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2 \\
 +
0  \\
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0
 +
\end{bmatrix}
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\end{Bmatrix}.</math>
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<math>\mathbf{c)} \quad 0=\begin{bmatrix}
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0 & 1 & 1 \\
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0 & 0 & 0 \\
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0 & 0 & 0 \\
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0 & 0 & 0
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\end{bmatrix}
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\quad Not \quad observable</math>
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<math>\mathbf{d)} \quad X_1=r,X_2=-s,X_3=s  \quad
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X=r\begin{bmatrix}
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1 \\
 +
0 \\
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0
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\end{bmatrix}+s\begin{bmatrix}
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0 \\
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-1 \\
 +
1
 +
\end{bmatrix}\\</math>
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<math>Subspace\quad is \begin{Bmatrix}
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\begin{bmatrix}
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1 \\
 +
0 \\
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0
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\end{bmatrix} ,& \begin{bmatrix}
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0 \\
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-1  \\
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1
 +
\end{bmatrix}
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\end{Bmatrix}.\\</math>
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<math>\mathbf{e)} \quad \lambda I-A=\begin{bmatrix}
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\lambda-1 & -1 & -1 \\
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0 & \lambda & -1 \\
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0 & 0 & \lambda+1
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\end{bmatrix}\quad \lambda_1=1,\lambda_2=0,\lambda_3=-1\\</math>
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<math>\qquad for \;\lambda_1=1 \qquad\begin{bmatrix}
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\lambda I-A & B \end{bmatrix}=\begin{bmatrix}
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0 & -1 & -1 & 1 \\
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0 & 1 & -1 & 1 \\
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0 & 0 & 2 & 0
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\end{bmatrix}\\</math>
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<math>\qquad for \;\lambda_2=0 \qquad\begin{bmatrix}
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\lambda I-A & B \end{bmatrix}=\begin{bmatrix}
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-1 & -1 & -1 & 1 \\
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0 & 0 & -1 & 1 \\
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0 & 0 & 1 & 0
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\end{bmatrix}\\</math>
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<math>\qquad for \;\lambda_3=-1 \qquad\begin{bmatrix}
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\lambda I-A & B \end{bmatrix}=\begin{bmatrix}
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-2 & -1 & -1 & 1 \\
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0 & -1 & -1 & 1 \\
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0 & 0 & 0 & 0
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\end{bmatrix}
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\qquad rank<3 \qquad must\;contain\;\lambda=-1
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\qquad so\qquad No.\\</math>
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<math>\mathbf{f)} \quad \begin{bmatrix}
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\lambda I-A \\
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C \end{bmatrix}=\begin{bmatrix}
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\lambda-1 & -1 & -1 \\
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0 & \lambda & -1 \\
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0 & 0 & \lambda+1 \\
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0 & 1 & 1
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\end{bmatrix}</math>
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<math>for\; \lambda_1=1 \quad \begin{bmatrix}
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\lambda I-A \\
 +
C \end{bmatrix}=\begin{bmatrix}
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0 & -1 & -1 \\
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0 & 1 & -1 \\
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0 & 0 & 2 \\
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0 & 1 & 1
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\end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_1=1</math>
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 +
<math>for\; \lambda_2=0 \quad \begin{bmatrix}
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\lambda I-A \\
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C \end{bmatrix}=\begin{bmatrix}
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-1 & -1 & -1 \\
 +
0 & 0 & -1 \\
 +
0 & 0 & 1 \\
 +
0 & 1 & 1
 +
\end{bmatrix}\\</math>
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 +
<math>for \lambda-3=-1\quad \begin{bmatrix}
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\lambda I-A \\
 +
C \end{bmatrix}=\begin{bmatrix}
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-2 & -1 & -1 \\
 +
0 & -1 & -1 \\
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0 & 0 & 0 \\
 +
0 & 1 & 1
 +
\end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_3=-1\\</math>

Revision as of 08:57, 16 May 2017

AC-2 P1.

$ \mathbf{a)} \quad C=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 1 & 2 & 2 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $ $ \quad rank=2\ne \mbox 3 $

$ \quad Not\quad controllable $

$ \mathbf{b)} \quad Subspace\quad is \begin{Bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} ,& \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} \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\\ $

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