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Question: Let A be a non-singular matrix. Then |adjA| is equal to A.\({\text{|A}}{{\text{|}}^{\text{n}}}\) ...

Let A be a non-singular matrix. Then |adjA| is equal to
A.|A|n{\text{|A}}{{\text{|}}^{\text{n}}}
B.|A|n - 1{\text{|A}}{{\text{|}}^{{\text{n - 1}}}}
C.|A|n - 1{\text{|A}}{{\text{|}}^{{\text{n - 1}}}}
D. None of these

Explanation

Solution

Hint – In order to solve this problem we need to know the formula A(adjA)=|A|I then solve further to get the asked term.

Complete step-by-step answer:

As we know that A(adjA)=|A|I
Where I is the identity matrix.
A(adjA)=|A|I

On taking determinant both sides we get,
⇒\Rightarrow|A(adjA)|=||A|I|
⇒|A||adjA| = |A|n\Rightarrow {\text{|A||adjA| = |A}}{{\text{|}}^{\text{n}}}

Where n is the order of matrix A.

⇒|A||adjA| = |A|n On cancelling ∣A∣ from both sides we can say ⇒|adjA| = |A|n - 1  \Rightarrow {\text{|A||adjA| = |A}}{{\text{|}}^{\text{n}}} \\\ {\text{On cancelling }}\left| {\text{A}} \right|{\text{ from both sides we can say}} \\\ \Rightarrow {\text{|adjA| = |A}}{{\text{|}}^{{\text{n - 1}}}} \\\

Therefore the correct option is B.

Note – In this problem we have used the formula A(adjA)=|A|I and then solved it to get what was asked. Here we have raised determinant of A to the power n because when you have multiplied the determinant with identity matrix then it is multiplied with all of its diagonal element and when we take determinant of that matrix then the determinant of A gets multiplied the number of times equal to that of order, that’s why we have raised it to the power n. Proceeding like this carefully will solve your problem.