question_answer

Assertion: \[adj\,\,\left( adj\,\,A \right)={{\left( \det \,\,A \right)}^{n-2}}\,A\].

Reason: \[\left| adj\,\,A \right|={{\left| A \right|}^{n-1}}\].

A) Both A and R are individually true and R is the correct explanation of A.

B) Both A and R are individually true and R is not the correct explanation of A.

C) 'A' is true but 'R' is false

D) 'A' is false but 'R' is true

E) Both A and R are false.

Correct Answer: A

Solution :

Given Assertion is \[adj\,\left( adj\,\,A \right)={{\left| A \right|}^{n-2}}\,A\]

We know that if A is an invertible matrix then

\[{{A}^{-1}}=\frac{adj\,A}{\left| A \right|}\,\Rightarrow adj\,\,A={{A}^{-1}}\,\left| A \right|\]

\[\therefore \,\,A\left( adj\,\,A \right)=A.{{A}^{-1}}\,\left| A \right|=I\left| A \right|=\left| A \right|\,I\]

\[\Rightarrow \,\,\left| A \right|\,\,\left| adj\,\,A \right|={{\left| A \right|}^{n}}\]\[\Rightarrow \,\,\,\left| adj\,\,A \right|={{\left| A \right|}^{n-1}}\]  (1)

Given Reason (R) is true

Now \[{{A}^{-1}}=\frac{adj\,\,A}{\left| A \right|}\,\,\Rightarrow adj\,\,A={{A}^{-1}}\,\left| A \right|\]

Multiply both sides with \[{{\left( adj\,\,A \right)}^{-1}}\]

\[\therefore \,\,\,\,\left( adj\,\,\,A \right)\,\,{{\left( adj\,\,A \right)}^{-1}}={{A}^{-1}}\,\left| A \right|\,\,{{\left( adj\,\,A \right)}^{-1}}\]

\[\Rightarrow \,\,I={{A}^{-1}}\,\left| A \right|\,\,{{\left( adj\,\,A \right)}^{-1}}\]

Multiply both sides with A

\[\therefore \,\,\,A=\left| A \right|\,\,{{\left( adj\,\,\,A \right)}^{-1}}\]

\[\Rightarrow \,\,\,{{\left( adj\,\,\,A \right)}^{-1}}=\frac{A}{\left| A \right|}\]                      (2)

Also \[{{\left( adj\,\,\,A \right)}^{-1}}=\frac{adj\,\left( adj\,\,A \right)}{\left| adj\,\,A \right|}\]

\[\Rightarrow \,\,\,{{\left( adj\,\,A \right)}^{-1}}.\,\left| adj\,\,A \right|=adj\,\,\left( adj\,\,A \right)\]

\[\Rightarrow \,\,\frac{A}{\left| A \right|}.{{\left| A \right|}^{n-1}}=adj\,\,\left( adj\,\,A \right)\]

{using (1) and (2)}

\[\therefore \,\,\,adj\,\,\left( adj\,\,A \right)={{\left| A \right|}^{n-2}}.\,A\]

\[\Rightarrow \]Assertion [A] is true

\[\therefore \]Both A and R are true and R is the correct explanation of A

Hence option [A] is the correct answer.