यदि {A_i} = left[ {begin{array}{*{20}{c}}{{a^ | vedclass

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(b) $|{A_i}| = \left| {\,\begin{array}{*{20}{c}}{{a^i}}&{{b^i}}\{{b^i}}&{{a^i}}\end{array}\,} ight|$ = ${({a^i})^2} - {({b^i})^2}$, $|a|\,

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $|{A_i}| = \left| {\,\begin{array}{*{20}{c}}{{a^i}}&{{b^i}}\{{b^i}}&{{a^i}}\end{array}\,} ight|$ = ${({a^i})^2} - {({b^i})^2}$, $|a|\, < 1,|b|\, < 1$

$\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})}}$.

यदि ${A_i} = \left[ {\begin{array}{*{20}{c}}{{a^i}}&{{b^i}}\\{{b^i}}&{{a^i}}\end{array}} \right]$ तथा $|a|\, < 1,\,|b|\, < 1$, तो $\sum\limits_{i = 1}^\infty {\det ({A_i})} $ का मान है

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