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Question

If A=⎡⎢⎣0121233a1⎤⎥⎦ and A−1=⎡⎢⎣1/2−1/21/2−43c5/2−3/21/2⎤⎥⎦, then a+c=

A
0.0
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B
0
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C
0.00
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Solution

We have, I=AA1=120121233a1111862c531

=10c+1012(c+1)4(1a)3(a1)2+ac
a1=0 a=1
c+1=0 c=1
a+c=0

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