Consider three matrices $A = \begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix}$ and $C = \begin{bmatrix} 3 & - 4 \ - 2 & 3 \end{bmatrix}$. Then the value of the sum $tr(A) + tr\left( \frac{ABC}{2} ight) + tr\left( \frac{A(BC)^{2}}{4} ight) + tr\left( \frac{A(BC)^{3}}{8} ight) + \ \cdots + \ \infty$ is

Question

Consider three matrices $A = \begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix}$ and $C = \begin{bmatrix} 3 & - 4 \ - 2 & 3 \end{bmatrix}$. Then the value of the sum $tr(A) + tr\left( \frac{ABC}{2} ight) + tr\left( \frac{A(BC)^{2}}{4} ight) + tr\left( \frac{A(BC)^{3}}{8} ight) + \ \cdots + \ \infty$ is

Dr. Abubacker · Accepted Answer

(a)
$BC = \begin{bmatrix}
3 & 4 \
2 & 3
\end{bmatrix}\begin{bmatrix}
3 & - 4 \
 - 2 & 3
\end{bmatrix} \Rightarrow B = \begin{bmatrix}
1 & 0 \
0 & 1
\end{bmatrix}$
$\Rightarrow tr(A) + tr\left( \frac{ABC}{2} ight) + tr\left( \frac{A(BC)^{2}}{4} ight) + tr\left( \frac{A(BC)^{3}}{8} ight) + \cdots + \infty$
$= tr(A) + \ tr\left( \frac{A}{2} ight)$+$\ tr\left( \frac{A}{2^{2}} ight)$+$\cdots$
$= tr(A) + \frac{1}{2}tr(A) + \frac{1}{2^{2}}tr(A)\ldots$
$= \frac{tr(A)}{1 - \left( \frac{1}{2} ight)}$
$= 2tr(A) = 2(2 + 1) = 6$

Explanation

Reviewed by Dr. Abubacker

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