Question

A system is represented in state-space form as follows (\(u\): input, \(\mathbf{x}\): state vector, \(y\): output)
\[ \dot{\mathbf{x}}=\begin{bmatrix}1&2\\-3&0\end{bmatrix}\mathbf{x}+\begin{bmatrix}1\\2\end{bmatrix}u \]
\[ y=\begin{bmatrix}1&2\end{bmatrix}\mathbf{x} \]
Consider the new state vector \(\mathbf{z}=\begin{bmatrix}2&1\\-1&0\end{bmatrix}\mathbf{x}\)
What is the state-space representation of the system in terms of the new state vector \(\mathbf{z}\)?

• A. \(\dot{\mathbf{z}}=\begin{bmatrix}-1&4\\-1&-2\end{bmatrix}\mathbf{z}+\begin{bmatrix}4\\-1\end{bmatrix}u\)
  \(y=\begin{bmatrix}2&3\end{bmatrix}\mathbf{z}\)
• B. \(\dot{\mathbf{z}}=\begin{bmatrix}2&3\\0&3\end{bmatrix}\mathbf{z}+\begin{bmatrix}3\\5\end{bmatrix}u\)
  \(y=\begin{bmatrix}2&3\end{bmatrix}\mathbf{z}\)
• C. \(\dot{\mathbf{z}}=\begin{bmatrix}4&9\\-2&-3\end{bmatrix}\mathbf{z}+\begin{bmatrix}4\\-1\end{bmatrix}u\)
  \(y=\begin{bmatrix}2&3\end{bmatrix}\mathbf{z}\)
• D. \(\dot{\mathbf{z}}=\begin{bmatrix}2&1\\-4&1\end{bmatrix}\mathbf{z}+\begin{bmatrix}4\\-1\end{bmatrix}u\)
  \(y=\begin{bmatrix}4&-1\end{bmatrix}\mathbf{z}\)

Hint:

Use the similarity transform rules A' = TAT^-1, B' = TB, C' = CT^-1 with T being the given transformation matrix.

Answer 1

The Correct Option is C

Step 1: Use an invariant that survives any similarity transform.
A similarity transform never changes the trace or determinant of the state matrix, since $A'=TAT^{-1}$ has the same eigenvalues as $A$. So first find the trace and determinant of the original $A$.

Step 2: Compute trace and determinant of $A$.
\[ A=\begin{bmatrix}1&2\\-3&0\end{bmatrix} \]
\[ \text{trace}(A)=1+0=1 \]
\[ \det(A)=1(0)-2(-3)=6 \]
So any correct $A'$ must also have trace $1$ and determinant $6$.

Step 3: Test each option's matrix against this invariant.
Option (A): $\begin{bmatrix}-1&4\\-1&-2\end{bmatrix}$ has trace $-3$. Fails.
Option (B): $\begin{bmatrix}2&3\\0&3\end{bmatrix}$ has trace $5$. Fails.
Option (C): $\begin{bmatrix}4&9\\-2&-3\end{bmatrix}$ has trace $4+(-3)=1$ and determinant $4(-3)-9(-2)=-12+18=6$. Passes both checks.
Option (D): $\begin{bmatrix}2&1\\-4&1\end{bmatrix}$ has trace $3$. Fails.
Only option (C) survives.

Step 4: Confirm with the transformation matrix directly.
With $T=\begin{bmatrix}2&1\\-1&0\end{bmatrix}$, $\det(T)=1$, so $T^{-1}=\begin{bmatrix}0&-1\\1&2\end{bmatrix}$. Then $B'=TB=\begin{bmatrix}2&1\\-1&0\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}4\\-1\end{bmatrix}$, which matches option (C)'s input vector exactly.

Step 5: Confirm the output row.
$C'=CT^{-1}=\begin{bmatrix}1&2\end{bmatrix}\begin{bmatrix}0&-1\\1&2\end{bmatrix}=\begin{bmatrix}2&3\end{bmatrix}$, which also matches option (C).

Step 6: Conclude.
\[ \boxed{\dot{\mathbf{z}}=\begin{bmatrix}4&9\\-2&-3\end{bmatrix}\mathbf{z}+\begin{bmatrix}4\\-1\end{bmatrix}u,\qquad y=\begin{bmatrix}2&3\end{bmatrix}\mathbf{z}} \]