A) \[\frac{4}{9}\]
B) \[\frac{1}{2}\]
C) 1
D) 2
Correct Answer: D
Solution :
\[{{(ad-bc)}^{2}}+{{(ac+bd)}^{2}}\] \[={{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}-2abcd+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}-2abcd\] \[={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}\] \[={{a}^{2}}{{d}^{2}}+{{b}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}\] \[={{d}^{2}}({{a}^{2}}+{{b}^{2}})+{{c}^{2}}({{b}^{2}}+{{a}^{2}})\] \[=({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})\] \[=2\times 1=2\]You need to login to perform this action.
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