Task 1

${\displaystyle A=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right],B=\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]}$

• ${\displaystyle A+B=\left[\begin{array}{cc}2+1& 1-1\\ 1+0& 1+2\end{array}\right]=\left[\begin{array}{cc}3& 0\\ 1& 3\end{array}\right]}$

• ${\displaystyle A\ast B=\left[\begin{array}{cc}2\ast 1+1\ast 0& 2\ast (-1)+1\ast 2\\ 1\ast 1+1\ast 0& 1\ast (-1)+1\ast 2\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]}$

• ${\displaystyle (A·B{)}^{-1}}$:

${\displaystyle [(A\ast B{)}^{-1}|I]=\left[\begin{array}{ccccc}2& 0& |& 1& 0\\ 1& 1& |& 0& 1\end{array}\right]=\left[\begin{array}{ccccc}1& 0& |& 0.5& 0\\ 1& 1& |& 0& 1\end{array}\right]=\left[\begin{array}{ccccc}1& 0& |& 0.5& 0\\ 0& 1& |& -0.5& 1\end{array}\right]}$ so, ${\displaystyle (A\ast B{)}^{-1}=\left[\begin{array}{cc}0.5& 0\\ -0.6& 1\end{array}\right]}$

Task 2

${\displaystyle A=\left[\begin{array}{cc}0& 1\\ -6& -5\end{array}\right],B=\left[\begin{array}{c}0\\ 2\end{array}\right],C=\left[\begin{array}{cc}1& 0\end{array}\right]}$

${\displaystyle \{\begin{array}{l}{x}^{\prime }=Ax+Bu\\ y=Cx\end{array}}$, so:

${\displaystyle \{\begin{array}{l}pX(s)=AX(s)+BU(s)\\ Y(s)=CX(s)\end{array}}$

and we have:

${\displaystyle Y(s)=C(pI-A{)}^{-1}BU(s)}$

${\displaystyle W(p)=C(pI-A{)}^{-1}B}$

${\displaystyle pI-A=\left[\begin{array}{cc}p-0& 0-1\\ 0-(-6)& p-(-5)\end{array}\right]}$

${\displaystyle [(pI-A)|I]=\left[\begin{array}{ccccc}p& -1& |& 1& 0\\ 6& p+5& |& 0& 1\end{array}\right]=\left[\begin{array}{ccccc}1& -\frac{1}{p}& |& \frac{1}{p}& 0\\ 6& p+5& |& 0& 1\end{array}\right]=\left[\begin{array}{ccccc}1& 0& |& \frac{p+5}{p^2+5p+6}& \frac{1}{p^2+5p+6}\\ 0& 1& |& -\frac{6}{p^2+5p+6}& \frac{p}{p^2+5p+6}\end{array}\right]}$ so, ${\displaystyle (pI-A{)}^{-1}=\left[\begin{array}{cc}\frac{p+5}{p^2+5p+6}& \frac{1}{p^2+5p+6}\\ -\frac{6}{p^2+5p+6}& \frac{p}{p^2+5p+6}\end{array}\right]}$

and ${\displaystyle W(p)=C(pI-A{)}^{-1}B=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}\frac{p+5}{p^2+5p+6}& \frac{1}{p^2+5p+6}\\ -\frac{6}{p^2+5p+6}& \frac{p}{p^2+5p+6}\end{array}\right]\left[\begin{array}{c}0\\ 2\end{array}\right]=\frac{2}{p^2+5p+6}}$

and the input-output differential equation equals to:

${\displaystyle y^{″}+5y^{\prime }+6y=2u}$

Task 3

${\displaystyle x_1^{\prime }=2x_1-x_1^2-x_1{x}_2}$, ${\displaystyle x_2^{\prime }=3x_2-2x_1{x}_2-x_2^2}$

When it comes to the equilibrium point, we have the cases:

${\displaystyle \{\begin{array}{l}{x}_1^{\prime }=2x_1-x_1^2-x_1{x}_2=0\\ x_2^{\prime }=3x_2-2x_1{x}_2-x_2^2=0\end{array}}$

so,

${\displaystyle \{\begin{array}{l}{x}_1^{\prime }=x_1(2-x_1-x_2)=0\dots \dots {}_1\\ x_2^{\prime }=x_2(3-2x_1-x_2)=0\dots \dots {}_2\end{array}}$

for equation 1, there are two possible solutions:

${\displaystyle \{\begin{array}{l}{x}_1=0\\ x_1=2-x_2\end{array}}$

and for equation 2:

${\displaystyle \{\begin{array}{l}{x}_2=0\\ x_2=3-2x_1\end{array}}$

so the equilibrium points are:

${\displaystyle (0,0),(0,3),(2,0),(1,1)}$

Task 4

${\displaystyle W(p)=\frac{W_1(p)\ast W_2(p)}{1-W_3(p)-W_4(p)}}$

So, we have:

${\displaystyle W(p)=\frac{\frac{p}{2}}{1-\frac{1}{p+2}-\frac{1}{p}}=\frac{2(p+2)}{p^2-2}}$