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Question

For a given matrix P=[4+3iii43i], i=1, the inverse of matrix P is

A
124[43iii4+3i]
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B
125[i43i4+3ii]
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C
124[4+3iii43i]
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D
125[4+3iii43i]
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Solution

The correct option is A 124[43iii4+3i]
P=[4+3iii43i], i = 1

We know that,
[abcd]1=1(adbc)[dbca]
So,

[4+3iii43i]1=1[(4+3i)(43i+i2)]×[43iii4+3i]
=1[169i2+i2][43iii4+3i]
=124[43iii4+3i]

Option (a) is correct.





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