Question

For the following question, enter the correct numerical value up to TWO decimal places. (If the numerical value has more than two decimal places, round-off the value to TWO decimal places.) If \[ A=\begin{pmatrix}2&1\\[2pt]1&1\end{pmatrix},\; B=\begin{pmatrix}3&4\\[2pt]1&3\end{pmatrix},\; C=\begin{pmatrix}3&-4\\[2pt]-2&3\end{pmatrix}, \] then the value of \[ \operatorname{tr}(A)+\operatorname{tr}\!\left(\frac{ABC}{2}\right) +\operatorname{tr}\!\left(\frac{A(BC)^2}{4}\right) +\operatorname{tr}\!\left(\frac{A(BC)^3}{8}\right)+\cdots = \_\_\_\_ \]

Hint:

Key ideas used:
Linearity of trace: \(\operatorname{tr}(X+Y)=\operatorname{tr}(X)+\operatorname{tr}(Y)\)
Matrix geometric series: \(\sum (M)^k=(I-M)^{-1}\) when it converges
Convert infinite trace series into a single trace expression

Answer 1

Correct Answer: 5.33

Step 1: Factor the series using linearity of trace: \[ \sum_{k=0}^{\infty}\operatorname{tr}\!\left(A\frac{(BC)^k}{2^k}\right) =\operatorname{tr}\!\left(A\sum_{k=0}^{\infty}\left(\frac{BC}{2}\right)^k\right) \]
Step 2: The series is a matrix geometric series: \[ \sum_{k=0}^{\infty}\left(\frac{BC}{2}\right)^k =\left(I-\frac{BC}{2}\right)^{-1} \quad (\text{since }\rho(BC/2)<1) \]
Step 3: Compute \(BC\): \[ BC= \begin{pmatrix}3&4\\1&3\end{pmatrix} \begin{pmatrix}3&-4\\-2&3\end{pmatrix} = \begin{pmatrix}1&0\\-3&5\end{pmatrix} \] Hence, \[ \frac{BC}{2}= \begin{pmatrix}\tfrac12&0\\-\tfrac32&\tfrac52\end{pmatrix} \]
Step 4: Compute the inverse: \[ I-\frac{BC}{2}= \begin{pmatrix}\tfrac12&0\\ \tfrac32&-\tfrac32\end{pmatrix} \] \[ \left(I-\frac{BC}{2}\right)^{-1} = \begin{pmatrix}2&0\\ 2&-\tfrac23\end{pmatrix} \]
Step 5: Multiply with \(A\): \[ A\left(I-\frac{BC}{2}\right)^{-1} = \begin{pmatrix}2&1\\1&1\end{pmatrix} \begin{pmatrix}2&0\\2&-\tfrac23\end{pmatrix} = \begin{pmatrix}6&-\tfrac23\\4&-\tfrac23\end{pmatrix} \]
Step 6: Take trace: \[ \operatorname{tr} = 6-\frac23=\frac{16}{3}=5.33 \]