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 sumlimits_{i = 1}^infty {det ({A

Q: 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 $\sum\limits_{i = 1}^\infty {\det ({A_i})} =\ . . .$

$|{A_i}| = \left| {\,\begin{array}{*{20}{c}}{{a^i}}&{{b^i}}\\{{b^i}}&{{a^i}}\end{array}\,} \right| = {({a^i})^2} - {({b^i})^2}$, $|a|\, < 1,|b|\, < 1$ $\sum\limits_{i = 1}^\infty {|{A_i}|} = ({a^2} - {b^2}) + ({a^4} - {b^4}) + ({a^6} - {b^6}) + .......$ $ = ({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})}}$.

MCQ

ગુજરાતી માધ્યમ

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 $\sum\limits_{i = 1}^\infty {\det ({A_i})} =\ . . .$

A$\frac{{{a^2}}}{{{{(1 - a)}^2}}} - \frac{{{b^2}}}{{{{(1 - b)}^2}}}$
B$\frac{{{a^2} - {b^2}}}{{(1 - {a^2})(1 - {b^2})}}$
C$\frac{{{a^2}}}{{{{(1 - a)}^2}}} + \frac{{{b^2}}}{{{{(1 - b)}^2}}}$
D$\frac{{{a^2}}}{{{{(1 + a)}^2}}} - \frac{{{b^2}}}{{{{(1 + b)}^2}}}$

Difficult

ધોરણ 12 સાયન્સ
ગણિત
3 and 4 . determinant and metrices
JEE

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