Question

16 A and B are matrices of
order 3x3 and IA-B1 =0, then​

Answer 1

Answer:

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.