A) If det\[\left( \text{A} \right)=\pm \text{ 1},\], then\[{{A}^{-1}}\] need not exist
B) lf det, \[\left( \text{A} \right)=\pm \text{ 1},\] then \[{{A}^{-1}}\] exists but all its entrie not necessarily integers
C) lf det \[\text{(A)}\ne \pm 1,\] then \[{{A}^{-1}}\]exists and all its entrie non-integers
D) If det\[\left( \text{A} \right)=\pm \text{ 1,}\], then\[{{A}^{-1}}\] exists and all its entries ?integers
Correct Answer: D
Solution :
As det \[(A)=\pm 1,\] then \[{{A}^{-1}}\]exists and \[{{A}^{-1}}=\frac{1}{\det (A)}(adjA)=\pm adj(A)\] Since, all entries in adj (A) are integers. \[\therefore {{A}^{-1}}\]has integers entries.
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