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Question

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

A
a=2,c=1/2
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B
a=1,c=1
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C
a=1,c=1
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D
a=1/2,c=1/2
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Solution

The correct option is B a=1,c=1
A=0123123a1A1⎢ ⎢ ⎢ ⎢ ⎢12121243c523212⎥ ⎥ ⎥ ⎥ ⎥
AA1=I
0123123a1⎢ ⎢ ⎢ ⎢ ⎢12121243c523212⎥ ⎥ ⎥ ⎥ ⎥=I
12+2C+32=0
2C=2C=1
Also32+3a32=0
3a=3
a=1

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