A system is represented in state-space form as follows:  $\stackrel{.}{x}=\left[\begin{array}{cc}1& 2\\ -3& 0\end{array}\right]x+\left[\begin{array}{c}1\\ 2\end{array}\right]u,y=\left[\begin{array}{cc}1& 2\end{array}\right]x.$  Consider the new state vector  $z=\left[\begin{array}{cc}2& 1\\ -1& 0\end{array}\right]x$ . What is the state-space representation in terms of the new vector  $z$ ?

The correct answer is the option shown in the third image (image_2.webp):
$$\stackrel{.}{z}=\left[\begin{array}{cc}4& 9\\ -2& -3\end{array}\right]z+\left[\begin{array}{c}4\\ -1\end{array}\right]u,y=\left[\begin{array}{cc}2& 3\end{array}\right]z$$

Step-by-Step Explanation:

1. Identify the system matrices:
From the given state-space representation:
$$\stackrel{.}{x}=Ax+Bu$$
$$y=Cx$$
We extract the matrices:
$$A=\left[\begin{array}{cc}1& 2\\ -3& 0\end{array}\right],B=\left[\begin{array}{c}1\\ 2\end{array}\right],C=\left[\begin{array}{cc}1& 2\end{array}\right]$$

2. Understand the state transformation:
We are given a new state vector:
$$z=Tx$$
where the transformation matrix $T$ is:
$$T=\left[\begin{array}{cc}2& 1\\ -1& 0\end{array}\right]$$
Therefore, the relationship between the old state $x$ and the new state $z$ is given by:
$$x=T^{-1}z$$

3. Compute the inverse of the transformation matrix, $T^{-1}$:
First, find the determinant of $T$:
$$det\left(T\right)=(2\times 0)-(1\times (-1\left)\right)=0+1=1$$
Using the formula for the inverse of a 2 × 2 matrix:
$$T^{-1}=\frac{1}{det\left(T\right)}\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right]$$

4. Find the new state-space system matrices:
Substituting $x=T^{-1}z$ into the state equations yields:
$$\stackrel{.}{z}=\stackrel{.}{Tx}=T\stackrel{.}{x}=T(Ax+Bu)=TA{T}^{-1}z+TBu$$
$$y=C{T}^{-1}z$$
Thus, the new system matrices are:
$$A_{\text{new}}=TA{T}^{-1}$$
$$B_{\text{new}}=TB$$
$$C_{\text{new}}=C{T}^{-1}$$

5. Perform the matrix multiplications:
• For $A_{\text{new}}$:
First compute $TA$:
$$TA=\left[\begin{array}{cc}2& 1\\ -1& 0\end{array}\right]\left[\begin{array}{cc}1& 2\\ -3& 0\end{array}\right]=\left[\begin{array}{cc}(2\times 1)+(1\times (-3\left)\right)& (2\times 2)+(1\times 0)\\ (-1\times 1)+(0\times (-3\left)\right)& (-1\times 2)+(0\times 0)\end{array}\right]=\left[\begin{array}{cc}-1& 4\\ -1& -2\end{array}\right]$$
Now compute $\left(TA\right)T^{-1}$:
$$A_{\text{new}}=\left[\begin{array}{cc}-1& 4\\ -1& -2\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}(-1\times 0)+(4\times 1)& (-1\times (-1\left)\right)+(4\times 2)\\ (-1\times 0)+(-2\times 1)& (-1\times (-1\left)\right)+(-2\times 2)\end{array}\right]=\left[\begin{array}{cc}4& 9\\ -2& -3\end{array}\right]$$

• For $B_{\text{new}}$:
$$B_{\text{new}}=\left[\begin{array}{cc}2& 1\\ -1& 0\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}(2\times 1)+(1\times 2)\\ (-1\times 1)+(0\times 2)\end{array}\right]=\left[\begin{array}{c}4\\ -1\end{array}\right]$$

• For $C_{\text{new}}$:
$$C_{\text{new}}=\left[\begin{array}{cc}1& 2\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}(1\times 0)+(2\times 1)& (1\times (-1\left)\right)+(2\times 2)\end{array}\right]=\left[\begin{array}{cc}2& 3\end{array}\right]$$

Thus, the transformed state-space equations match the result in image_2.webp.