A) \[|A{{|}^{n-1}}A\]
B) \[|A{{|}^{n}}A\]
C) \[|A{{|}^{n-2}}A\]
D) None of these
Correct Answer: C
Solution :
[c] For any square matrix X, we have \[X(adj\,\,X)=\left| X \right|{{I}_{n}}\] Taking X = adj A, we get \[(adj\,\,A)[adj(adj\,\,A)]=\left| adj\,\,A \right|\,{{I}_{n}}\] \[\Rightarrow \,\,\,(adj\,\,A)\,[adj\,(adj\,\,A)]={{\left| A \right|}^{n-1}}{{I}_{n}}\] \[[\because \left| adj\,A \right|={{\left| A \right|}^{n-1}}]\] \[\Rightarrow (A\,\,adj\,\,A)[adj(adj\,\,A)]={{\left| A \right|}^{n-1}}A\] \[[\because \,A\,\,{{I}_{n}}=A]\] \[(\left| A \right|{{I}_{n}})(adj(adj\,A))={{\left| A \right|}^{n-1}}A\] \[\Rightarrow adj(adj\,\,A)={{\left| A \right|}^{n-2}}A\]
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