If $A_{i} = \left\lbrack \begin{array}{r} a^{i}\ \ b^{i} \ b^{i}\ \ a^{i}\ \end{array} ight brack$ and if $|a|

Question

If $A_{i} = \left\lbrack \begin{array}{r} a^{i}\ \ b^{i} \ b^{i}\ \ a^{i}\ \end{array} ightbrack$ and if $|a| < 1,|b| < 1,$then $\sum_{i = 1}^{\infty}{det(A_{i})}\$is equal to

Dr. Abubacker · Accepted Answer

(b)
$\because\det{\left( A_{1} ight) = \left| \begin{array}{r}
a\ \ b \
b\ \ a
\end{array} ight| = a^{2} - b^{2}}$
$\det{\left( A_{2} ight) = \left| \begin{matrix}
a^{2} & b^{2} \
b^{2} & a^{2}
\end{matrix} ight| = a^{4} - b^{4}}$
$	herefore\sum_{i = 1}^{\infty}\det\ \left( A_{i} ight) = \det{\left( A_{1} ight) + \det\left( A_{2} ight) + ...}$
$= a^{2} - b^{2} + a^{4} - b^{4} + \$…
$= \frac{a^{2}}{1 - a^{2}} - \frac{b^{2}}{1 - b^{2}} = \frac{a^{2} - b^{2}}{\left( 1 - a^{2} ight)(1 - b^{2})}$

Q. If $A_{i} = \left\lbrack \begin{array}{r} a^{i}\ \ b^{i} \\ b^{i}\ \ a^{i}\ \end{array} \right\rbrack$ and if $|a| < 1,|b| < 1,$then $\sum_{i = 1}^{\infty}{det(A_{i})}\$is equal to

Explanation

Reviewed by Dr. Abubacker

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