If A_i = begin{bmatrix} a^i & b^i b^i & a^i | vedclass

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(B) Given $A_i = \begin{bmatrix} a^i & b^i \ b^i & a^i \end{bmatrix}$.The determinant of $A_i$ is $\det(A_i) = (a^i)(a^i) - (b^i)(b^i) = a^{2i} - b^{

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$\frac{a^2 - b^2}{(1-a^2)(1-b^2)}$

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(B) Given $A_i = \begin{bmatrix} a^i & b^i \ b^i & a^i \end{bmatrix}$.The determinant of $A_i$ is $\det(A_i) = (a^i)(a^i) - (b^i)(b^i) = a^{2i} - b^{2i}$.We need to calculate the sum $S = \sum_{i=1}^{\infty} \det(A_i) = \sum_{i=1}^{\infty} (a^{2i} - b^{2i})$.Since $|a| $S = \sum_{i=1}^{\infty} a^{2i} - \sum_{i=1}^{\infty} b^{2i}$.The sum of an infinite geometric series $\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}$ for $|r| So,$\sum_{i=1}^{\infty} a^{2i} = \frac{a^2}{1-a^2}$ and $\sum_{i=1}^{\infty} b^{2i} = \frac{b^2}{1-b^2}$.Therefore,$S = \frac{a^2}{1-a^2} - \frac{b^2}{1-b^2}$.Taking the common denominator: $S = \frac{a^2(1-b^2) - b^2(1-a^2)}{(1-a^2)(1-b^2)} = \frac{a^2 - a^2b^2 - b^2 + a^2b^2}{(1-a^2)(1-b^2)} = \frac{a^2 - b^2}{(1-a^2)(1-b^2)}$.

If $A_i = \begin{bmatrix} a^i & b^i \\ b^i & a^i \end{bmatrix}$ and if $|a| < 1, |b| < 1$,then $\sum_{i=1}^{\infty} \det(A_i)$ is equal to

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