A) If det. \[(A)=\pm 1\], then \[{{A}^{-1}}\] need not exist.
B) If det. \[(A)=\pm 1\], then \[{{A}^{-1}}\] exist but all its entries are not necessarily integers.
C) If det. \[(A)\ne \pm 1\], then \[{{A}^{-1}}\] exists and all its entries are non-integers.
D) If det. \[(A)=\pm 1\], then \[{{A}^{-1}}\] exists and all its entries are integers.
Correct Answer: D
Solution :
As, \[\det .(A)=\pm 1,\] so\[{{A}^{-1}}\]will exists \[{{A}^{-1}}=\frac{adj.A}{|A|}=\pm (adj.A)\] \[\therefore \,\,\frac{{{x}^{2}}}{{{a}^{2}}}\,+\frac{{{y}^{2}}}{{{b}^{2}}}=\frac{1}{9}\] All entries in adj. A are integers. So, \[{{A}^{-1}}\] has integer entries.
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