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જો ${A_i} = \left[ {\begin{array}{*{20}{c}}{{a^i}}&{{b^i}}\\{{b^i}}&{{a^i}}\end{array}} \right]$ અને $|a|\, < 1,\,|b|\, < 1$, તો $\sum\limits_{i = 1}^\infty {\det ({A_i})} $= . . .
$\frac{{{a^2}}}{{{{(1 - a)}^2}}} - \frac{{{b^2}}}{{{{(1 - b)}^2}}}$
$\frac{{{a^2} - {b^2}}}{{(1 - {a^2})(1 - {b^2})}}$
$\frac{{{a^2}}}{{{{(1 - a)}^2}}} + \frac{{{b^2}}}{{{{(1 - b)}^2}}}$
$\frac{{{a^2}}}{{{{(1 + a)}^2}}} - \frac{{{b^2}}}{{{{(1 + b)}^2}}}$
Solution
(b) $|{A_i}| = \left| {\,\begin{array}{*{20}{c}}{{a^i}}&{{b^i}}\\{{b^i}}&{{a^i}}\end{array}\,} \right|$ = ${({a^i})^2} – {({b^i})^2}$, $|a|\, < 1,|b|\, < 1$
$\sum\limits_{i = 1}^\infty {|{A_i}|} = ({a^2} – {b^2}) + ({a^4} – {b^4})$$ + ({a^6} – {b^6}) + …….$
$ = ({a^2} + {a^4} + {a^6} + ……)$$ – ({b^2} + {b^4} + {b^6} + …….)$
$ = \frac{{{a^2}}}{{1 – {a^2}}} – \frac{{{b^2}}}{{1 – {b^2}}}$
$ = \frac{{{a^2} – {a^2}{b^2} – {b^2} + {a^2}{b^2}}}{{(1 – {a^2})(1 – {b^2})}}$
$ = \frac{{{a^2} – {b^2}}}{{(1 – {a^2})(1 – {b^2})}}$.
