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Question
Given that ⎡⎢⎣413⎤⎥⎦A=⎡⎢⎣−484−121−363⎤⎥⎦, if A=[abc]. Find a+b+c.
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Solution
Given, ⎡⎢⎣413⎤⎥⎦A=⎡⎢⎣−484−121−363⎤⎥⎦
⇒⎡⎢⎣413⎤⎥⎦[abc]=⎡⎢⎣−484−121−363⎤⎥⎦
⇒⎡⎢⎣4a4b4cabc3a3b3c⎤⎥⎦=⎡⎢⎣−484−121−363⎤⎥⎦
Comparing elements, we get a=−1,b=2,c=1
Hence, a+b+c=2.
⇒⎡⎢⎣413⎤⎥⎦[abc]=⎡⎢⎣−484−121−363⎤⎥⎦
⇒⎡⎢⎣4a4b4cabc3a3b3c⎤⎥⎦=⎡⎢⎣−484−121−363⎤⎥⎦
Comparing elements, we get a=−1,b=2,c=1
Hence, a+b+c=2.
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