Question

Given thatA is a square matrix of order 3×3 and |A|=9, then find the value of |A.adjA|?​

Answer 1

Step-by-step explanation:

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣ 2

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣ 2 =10

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣ 2 =10 2

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣ 2 =10 2 =100.

Answer 2

Answer:

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10

Now ∣adjA∣=∣A∣

n−1

where n is the order of the square matrix.

Here 'n' is 3, therefore, ∣adjA∣=∣A∣

3−1

=∣A∣

2

=10

2

=100.

Step-by-step explanation:

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