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यदि $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$ और $B = \left[ {\begin{array}{*{20}{c}}{\cos \beta }&{ - \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$, तो कौन सा सम्बन्ध सत्य है
${A^2} = {B^2}$
$A + B = B - A$
$AB = BA$
इनमें से कोई नहीं
Solution
(c) स्पष्टत:, $AB = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ – \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \beta }&{ – \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}{\cos (\alpha + \beta )}&{ – \sin (\alpha + \beta )}\\{\sin (\alpha + \beta )}&{\cos (\alpha + \beta )}\end{array}} \right] = BA$ (सत्यापित).
