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If A is a square matrix of order 3 such that `abs(A)=2,` then `abs((adjA^(-1))^(-1))` isA. 1B. 2C. 4D. 8 |
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Answer» Correct Answer - C `because abs(adj A^(-1)) = abs(A^(-1))^(2) = 1/(abs(A))^(2)` `therefore abs((adj A^(-1) )^(-1) ) = 1/(abs(adjA^(-1)))=abs(A)^(2) = 2^(2) = 4` |
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