Question
If $$A$$ is a square matrix of order $$n,$$ then $$adj\left( {adj\,A} \right)$$ is equal to
A.
$${\left| A \right|^{n - 1}}A$$
B.
$${\left| A \right|^{n}}A$$
C.
$${\left| A \right|^{n - 2}}A$$
D.
None of these
Answer :
$${\left| A \right|^{n - 2}}A$$
Solution :
For any square matrix $$X,$$ we have $$X\left( {adj\,X} \right) = \left| X \right|{I_n}$$
Taking $$X = adj\,A,$$ we get
$$\eqalign{
& \left( {adj\,A} \right)\left[ {adj\left( {adj\,A} \right)} \right] = \left| {adj\,A} \right|{I_n} \cr
& \Rightarrow \left( {adj\,A} \right)\left[ {adj\left( {adj\,A} \right)} \right] = {\left| A \right|^{n - 1}}{I_n} \cr
& \left[ {\because \left| {adj\,A} \right| = {{\left| A \right|}^{n - 1}}} \right] \cr
& \Rightarrow \left( {A\,adj\,A} \right)\left[ {adj\left( {adj\,A} \right)} \right] = {\left| A \right|^{n - 1}}A \cr
& \left[ {\because A\,{I_n} = A} \right] \cr
& \left( {\left| A \right|{I_n}} \right)\left( {adj\left( {adj\,A} \right)} \right) = {\left| A \right|^{n - 1}}A \cr
& \Rightarrow adj\left( {adj\,A} \right) = {\left| A \right|^{n - 2}}A \cr} $$